Why Arithmetic Mean Greater Than Geometric? (Correct answer)

The arithmetic mean is always higher than the geometric mean as it is calculated as a simple average. It is applicable only to only a positive set of numbers. It can be calculated with both positive and negative sets of numbers. Geometric mean can be more useful when the dataset is logarithmic.

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Why is arithmetic mean more than geometric mean?

The geometric mean differs from the arithmetic average, or arithmetic mean, in how it is calculated because it takes into account the compounding that occurs from period to period. Because of this, investors usually consider the geometric mean a more accurate measure of returns than the arithmetic mean.

Is arithmetic average always greater than geometric average?

The case where not all the terms are equal It remains to show that if not all the terms are equal, then the arithmetic mean is greater than the geometric mean.

Which is better arithmetic or geometric mean?

The arithmetic mean is more useful and accurate when it is used to calculate the average of a data set where numbers are not skewed and not dependent on each other. However, in the scenario where there is a lot of volatility in a data set, a geometric mean is more effective and more accurate.

Is am greater than GM?

Arithmetic mean (A.M.) is greater than geometric mean (G.M.) for three valuables is as follows: Equality sign holds if and only if x = y = z.

What is the relation between arithmetic mean and geometric mean?

Let A and G be the Arithmetic Means and Geometric Means respectively of two positive numbers m and n. Then, we have A = m + n/2 and G = ±√mn. Since, m and n are positive numbers, hence it is evident that A > G when G = -√mn.

What is the difference between arithmetic and geometric returns?

Arithmetic returns are the everyday calculation of the average. The geometric mean is calculated by multiplying all the (1+ returns), taking the n-th root and subtracting the initial capital (1). The result is the same as compounding the returns across the years.

What is difference between geometric and arithmetic?

An arithmetic sequence has a constant difference between each consecutive pair of terms. A geometric sequence has a constant ratio between each pair of consecutive terms. This would create the effect of a constant multiplier.

Which statements are true when comparing arithmetic mean return to geometric mean return?

Which statements are TRUE when comparing arithmetic mean return to geometric mean return? The best answer is C. Arithmetic mean return is simply the average annual return of an investment over its time horizon.

Why is geometric mean always less than arithmetic mean?

The geometric mean is always lower than the arithmetic means due to the compounding effect. The arithmetic mean is always higher than the geometric mean as it is calculated as a simple average. It is applicable only to only a positive set of numbers. It can be calculated with both positive and negative sets of numbers.

What is the difference between arithmetic mean and mean?

Average, also called the arithmetic mean, is the sum of all the values divided by the number of values. Whereas, mean is the average in the given data.

What is the purpose of geometric mean?

The geometric mean is used in finance to calculate average growth rates and is referred to as the compounded annual growth rate. Consider a stock that grows by 10% in year one, declines by 20% in year two, and then grows by 30% in year three.

What is the difference between arithmetic mean geometric mean and harmonic mean?

The arithmetic mean is appropriate if the values have the same units, whereas the geometric mean is appropriate if the values have differing units. The harmonic mean is appropriate if the data values are ratios of two variables with different measures, called rates.

Is harmonic mean greater than arithmetic mean?

& (2) Harmonic mean is always lower than arithmetic mean and geometric mean. only if the values (or the numbers or the observations) whose means are to calculated are real and strictly positive.

Can a geometric mean be negative?

Like zero, it is impossible to calculate Geometric Mean with negative numbers.

Arithmetic Mean – Geometric Mean

The Arithmetic Mean – Geometric Mean inequality, sometimes known as the AM-GM inequality, is defined as follows: The geometric mean cannot be greater than the arithmetic mean, and they will be equal if and only if all of the numbers picked are equal in the first instance. That is, a1+a2+anna1a2.annfracgesqrtwith equality if and only ifa1=a2=ana 1=a 2=cdots =a n, and a1=a2+anna1a2.annfracgesqrtwith equality if and only ifa1=a2=ana 1=a 2=cdots =a n To put it another way, i=1nain i=1nain. ge _sqrt Take, for example, the scenario where n=2,n=2, i.e.

The AM-GM inequality therefore saysx+y2xy.dfracge xy.dfracge xy.dfracge sqrt.

It is also possible to have equality whenx=ysqrt x=sqrt yor if the variables are equal to each other.

Inducting variables based on the number of variables is a frequent strategy.

Given the AM-GM disparity, here is a straightforward illustration.

Allow the two numbers to beaa and bb to be used.

The AM-GM inequality asserts thata+b2ab,dfracgeq sqrt,implying for this issue thata+b2100=20.a + bgeq 2 sqrt= 20.a + bgeq 2 sqrt= 20.a + bgeq 2 sqrt= 20.a + bgeq 2 sqrt= 20.a + bgeq 2 sqrt= 20.a + bgeq 2 sqrt Note:When a=b=10a = b=10a = b=10a = b=10a = b=10a = b=10a = b=10a = b=10a = b=10a = b=10a = b=10a = b=10a = b=10a = b=10a = b=10 You can experiment with an issue similar to the one shown above: The sum of two positive real numbers equals one hundred.

  1. Find the most profitable product for them.
  2. The three geen jellies are a trio of geen jellies.
  3. There is no way to establish a definitive relationship.
  4. The three red cubes have side lengths of abc, a b, and c, while the three green cuboids have dimensions of abc, a times b times c, as illustrated in the diagram above.
  5. AM-GM inequality may be demonstrated using a variety of approaches.
  6. The first item on the list is to demonstrate using some type of inductive reasoning.

Conventional inductions examine a base case and then demonstrate thatP(n)P(n+1)P(n)P(n)P(n)P(n+1)P(n)P(n)P(n)P(n)P(n)P(n)P(n)P(n)P(n)P(n)P(n)P(n)P(n)P(n)P(n) However, we will demonstrate the following:

  • P(2)P(2)holds
  • sP(n)⟹P(2n). P(n) implies P(2n)
  • P(n) implies P(n1). P(n) implies P(n1). P(n) is equivalent to P(n-1)

What makes you think this will work? It is important to note that we leap fromnnto2n2nby demonstrating thatP(n)P(2n)P(n)implies P(n) (2n). In this case, we may induct backwards from2n2nton+1,n+1, usingP(n)P(n1)P(n)implies P(n-1), in order to check that all integers in the range of nand 2n2n(inclusive) fulfill the claim. Forward-backward induction is the term used to describe this. We’ll now proceed to demonstrate our points. It has previously been demonstrated that P(2)P(2) holds, and now we will demonstrate that P(n)P(2n)P(n)implies P.

  1. Positive realsa1, A2,.,a2na 1, A 2,.,a_ are taken into consideration.
  2. For anynnpositive reals, we can assume thatP(n)P(n)is true.
  3. In addition, we’ve utilized the base example, which has 22 variables.
  4. It is still necessary to demonstrate that P(n)P(n1)P(n)implies P(n) (n-1).
  5. Then you’ll observe that 1ni=1nai=1n(i=1n1ai+1n1i=1n1ai)=1n1i=1n1ai=1n(i=1n1ai+1n1i=1n1ai)=1n1i=1n1ai.
  6. dfracsum_ n a i =dfracleft(sum_ a i + dfracsum_ a iright)=dfracsum_ a i.
  7. As a result, the third element of the evidence has been completed, as has the induction.
  8. So let’s use that method to solving the problem.
  9. f(x) = log x for every x0 is defined as f(x).
  10. Thus, f(x)f(x) turns out to be a concave function when x = 1.

As a result, according to Jensen’s inequality, we havef(i=1nain)i=1nf(ai)nlog(i=1nain)i=1nlog(ai)n.beginfleft(dfrac n a i)right)f(i=1nain)f(i=1nain)f(i=1nain)f(i=1nain)f(i= eq dfrac n f left(a I right) eq dfrac n f left(a I right) eq dfrac n f left(a I right) eq dfrac n f left(a I right) eq dfrac n f left(a I right) eq dfrac n f left(a I right Rightarrow dfrac n a i dfrac n a i dfrac n a i dfrac n a i dfrac n a i dfrac n a i dfrac n a i dfrac n a i dfrac n a i dfrac n a i eq dfrac n log a I right eq dfrac n log a I right eq dfrac n log a I right eq dfrac n log a I right eq dfrac n log a I right eq dfrac n log a I right eq dfrac n log a I right eq dfrac n In accordance with the property of logarithms, we have the following: logax+logay=logabloga = log ab and bloga=logab log x + log y = log log xy.

In order to simplify the terms on the RHS, we can write them as log(i=1nain)i=1nlogain=log(i=1nain)i=1/n.beginlog left(dfrac n a i)geq dfrac n a i = dfrac n = log(i=1nain)1/n.endlog (i=1nain)i=1nlogain=log Another demonstration, made famous by a mathematician named George Pólya, does not rely on inductive reasoning to establish its validity.

Assume that given a sequence of positive realsaka kwith1kn1leq k nand a sequence of positive realspkp kwith1kn1leq k nsuch thatk=1npk=1sum_ p k = 1, a1p1a2p2a3p3anpna1p1a2p2a3p3+anpn.a 1p1a2p2a3p First, Pólya makes the observation that 1+xex,1+xleq ex, which can be readily proven visually by noting that equality occurs only atx=0x=0.

  • This idea, according to legend, came to Pólya in a dream, in which he exclaimed that it was “the greatest mathematics he had ever imagined.” If we make the adjustment in variablesxx1xmapsto x-1, our initial observation becomesxex1xleq e, and our final observation becomesxex1xleq e.
  • As a result, we may sayA,Gexp(k=1nakpk1).
  • We have connected AA and GG via inequality, but we have not distinguished between them.
  • The concept of “normalization” comes to mind at this point.

Create a new sequence k alpha k with 1 k n, 1 leq k n, where we have k=akAA=a1p1+a2p2+a3p3+anpn.begin alpha k= frac A =a 1p 1+a2p2+a3p3+cdots+a np n.end alpha k= frac A =a 1p 1+ Using our previously established bound forGGfor our new variableskalpha k, we obtain k=1nkpkexp(k=1nkpk1)k=1n(akA)pkexp(k=1nakApk1)k=1n(akA)pkexp(k=1nakApk1)k=1n(akA)pkexp(k=1nakApk1)k=1n(akA) beginprod_ alpha kleq exp left(sum_ alpha kp k-1 right) beginprod_ alpha kleq exp left(sum_ alpha kp k-1 right) beginprod_ alpha kleq exp left(sum_ alpha kp k-1 right) beginprod_ alpha kleq exp left(sum_ alpha kp k-1 right) beginprod_ alpha kle PRODUCTION LEVEL RIGHT (FRACTURE) LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL So we have a 1p1, a 2p2, a 3p3, +anpnA=AA=1, sum_ fracp k = frac= frac= frac= 1, and it follows that exp(k=1nakApk1) = 1.exp left(sum fracp k-1right) = 1.

  • In this case, we are driven to the conclusion that k=1n(akA)pk1, k=1nakpk1, k=1nApk1, etc., etc., etc., etc., etc., etc.
  • Let SS represent a collection of positive real numbers.
  • (a 1+a 2+cdots+a n) = underbrace The existence of a word that is the sum of all the terms of SS is well known to us at this point.
  • The fact that every term is positive means that anynthntextpower expansion is bigger than its summand, and as a result, we obtain (i=1nai)Nn!
  • It is written as left(sum_ ai right) and left(prod_ ai right) in the following format: Now all we have to do is recall the definition of geometric mean, which is the thenthntextroot of the product between all terms on a set, followed by some algebraic operations to arrive at the result.
  • Now, we can establish by induction that for any positive integernn,nn!ngeq n!, and this proof will assist us in eliminating the factorial on the denominator of the previous inequality, which was previously mentioned.
  • is formed as 11!1geq 1!, which is correct.

Then, here’s what we have: (k+1)k+1=kk+1+(k+1)⋅kk⋅11+k⋅(k+1)2⋅kk−1⋅12+⋯+1k+1kk+1+(k+1)⋅kkkk⋅(2k+1).

(2k+1)≥k!.

.kk⋅(2k+1)≥k⋅k!+(k+1)!(k+1)!

.begink^geq k!

\ k^ cdot (2k+1)geq kcdot k!+(k+1)!

\ Rightarrow k^ cdot(2k+1)(k+1)!

end With(k+1)k+1 kk⋅(2k+1)(k+1)^k^ cdot (2k+1)andkk⋅(2k+1)(k+1)!

.

Now, we may apply the inequalitynn≥n!n^geq n!, which is valid for any positive integer.

1nn≤1n!

\ frac}leq frac\ fracleq frac}} \ frac ^n a i}leq frac ^n a i}}} \ m aleq frac ^n a i}}}=k, end wheremam ais the arithmetic mean.

begink m g 0implies m g in (0, k) \ kgeq m a 0implies m a in (0, k].

Otherwise, they are on an equal footing.

square mgma.m g m a.

square mgma.m g m a.

square Application for AM-GM (Main Article) More information on the applications of AM-GM inequality may be found in the paper referenced above, as well as in the section below.

The minimal value of (1+1x)(1+1y)left(1+fracright)left(1+fracright) is found if and only if x,y=8x,y=8 and x,y=8x,y=8.

From AM-GMx+y2xy4xy16xy116xy1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1 As a result, the smallest possible value of(1+1x)(1+1y)left(1+fracright)left(1+fracright)left(1+fracright)left(1+fracright)is2516frac.

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It should be noted that the equality is true whenx=y=4.x=y=4 They may be calculated by noting that the smallest value is attained whenxy=16,xy=16, together with the specified restriction thatx+y=8, and then multiplying that number by eight.

In this case, we use the 3-variable version of AM-GM withx1=a3, x2=a3, x 1=a3, and x 2 = a3.

Afterwards, we multiply both sides by three to get b2ba2 + a3 + b3geq 3a2ba3 + a3 + b3geq 3a2ba3 □_square Find all of the true answers to 2x+x2=212x2x + x2 = 2 – frac.2x+x2=2 – frac.2x+x2=2 – frac.2x+x2=2 – frac.2x+x2=2 – frac.2x+x2=2 – frac.2x+x2=2 – frac.2x+x2=2 – frac.

We have2x2+12x=2×22 leq 2x + frac= 2 – x2, so0x20 geq x2.

is the only value that can be used.

□_square Look for all of the true positive answers to 4x+18y=14,2y+9z=15,9z+16x=17.

end4x + frac=14,2y + frac=15,9z + frac=17.

Summation of the above three inequalities yields the following result: 416x, 2Y, 18Y, 9Z, 9Z16, 12 and 18=46.

4x + frac+ 2y + frac+ 9z + fracgeq 16 + 12 + 18 = 46.

Furthermore, by summing the three given equations, we obtain4x+16x+2y+18y+9z+9z=46.

With these values entered into the original equations, we observe that (x, y, and z)=(2, 3, 1)(x, y, and Z)=(2, 3, 1)(x, y, and Z) = (2, 3, 1)is in fact a solution.

Solution 1: By expanding both sides, we may cancel termsa2c2a2c2andb2d2b2d2, which means we must prove thata2d2+b2c2acbda is equal to a2d2+b2c2acbda.

2d2 + B2 = 2aCBd.

2.big(a2+b2big)big(c2+d2big) = (ac+bd)2 + (ad-bc)2 = (ac+bd)2 + (ad-bc)2 = (ac+bd)2 + (ad-bc)2 = (ac+bd)2 + (ad-bc)2 = (ac+bd)2 + (ad-bc)2 = (ac The right-hand side is bigger than or equal to (ac+bd)2, which is non-negative since squares are non-negative.

abc(a+b+c) = a4 + b4 + c4 = abc(a+b+c).

Think about how we can acquire words on the right hand side of the screen by using AM-GM.

This provides a hint to trya4+a4+b4+c4a2bc.a4 + a4 + b4 + c4 eq 4 a2 b c.a4 + a4 + b4 + c4 eq 4 a2 b c.a4 + a4 + b4 + c4 eq 4 a2 b c.a4 + a4 + b4 + Similarly, we have a4+b4+b4+c4+4ab2ca4 +b4 +b4 +c4 geq 4ab2 c and a4+b4+c4+c4abc2.a4 + b4 + c4 + c4 geq 4abc2.a4 + b4 + c4 + c4 geq 4abc2.a4 + b4 + 2.

  • In the case of abc(a+b+c), the equation is a4 + b4 + c4 = a2c + ab2c + ab2 = abc(a+b+c).
  • frac+frac+frac+frac What is the smallest possible value ofP2A?frac for all rectangles with perimeterPP and areaAA?
  • Examples of problems: Figure out what the smallest possible value of4+9x2sin2xxsinxfor0x.fractext 0xpi is.
  • We can also write the expression 4+9x2SINX2XSINXfrac sin as 4xSINX.
  • squareForx0x0, the maximum value is maximizef(x)=(1+x)(1+x)(1x), and the minimum value is 1212.
  • Give your answer with as many decimal places as you choose.
  • This time, we’ll apply the identical method to Jensen’s inequality that we used to show the AM-GM inequality above to demonstrate the weighted AM-GM inequality.
  • It is true that allaka kare are equal if they are all equal.
  • It follows that if at least one of the aka kis is zero (but not all), then the weighted geometric mean is zero, but the weighted arithmetic mean is positive, and that the stringent inequality applies in this case.

sayln⁡(ω1a1+⋯+ωnanω)w1wln⁡a1+⋯+wnwln⁡an=ln⁡(a1ω1×2ω2⋯anωn∑ i).beginlnBigl(fracomega Bigr)frac wln a 1+cdots+frac wln a n =ln left(sqrtx 2cdots a n)ln right(sqrtx 2cdots a n)ln right(sqrtx 2cdots a

Arithmetic Mean is grater than geometric mean (2)

Arithmetic Mean is greater than geometric mean (2)
Arithmetic mean (A.M.) is greater than geometric mean (G.M.) for three valuables is as follows:Equality sign holds if and only ifx = y = z.This article is dedicated to the proof of the above theorem using different perspectives.
1. It is assumed that arithmetic mean is greater than geometric mean for two variables is proved:Equalitysign holds if and only if x = y.2. The reader may find it interesting to investigate theequality case in the inequalities in each of the following proofs. (We don��t discuss here)3.If we putthen (inequality 1) is equivalent to ,(inequality 2) Therefore we need to show either(inequality 1)or(inequality 2)in the following proofs.
Method1The easiest proof is: (*)since a, b, c≥ 0and the sum of complete squares in the last factor of (*) must be non-negative. ,.The weakness of this proof is ��how to get (*)?��. We can get (*) bymultiplication. Can we get (*) byfactorization? This is shown at the appendix of this article.

Method2

UsingA.M≥G.M. for two variables in the preliminary (1) in the above, we get: (i) (ii) A.M. of the L.H.S. of (i) and (ii) ≥ G.M. of R.H.S. of (i) and (ii) ,.

Method3

From, we get Similarly,Adding up the inequalities in the above and divide by 2, we haveMultiply both sides by(a + b + c)which is non-negative, the following diagram is helpful in expanding and canceling similar terms:
After eliminate the terms, only the red squares remain, we have the inequality: ,.
Method4It is interesting that A.M.≥G.M. for four variables can be proved easily: where the proof in fact uses A.M.≥G.M. for two variables twice. Now, we move back to the case for three variables. This can be done in any one of the following methods:
(Method 4a)Putin (Method 4b)Putin
Method5Ifa, b ≥ 0, thenanda �V bmust be of the same sign. Expanding and moving terms, we have: Similarly, we have And Adding the three inequalities, we have:
Method6For those who are familiar with Calculus the proof below is helpful: Let(guess how f(x) is constructed after the proof) For turning point, and When, When, f(x) is a minimum point when. Put x = a, we get,.
Method7Exercise:Show that by considering(1) (2)
Appendix
To show that:
First we note that: Now,

Art of Problem Solving

Known technically as theInequality of Arithmetic and Geometric Meansor colloquially as AM-GM in mathematics, theAM-GM Inequality says that the arithmetic mean of any list of nonnegative reals is either greater than or equal to the geometric mean of that list. Furthermore, the two means are identical if and only if every number in the list is the same for each of the two groups of numbers. In symbols, the inequality asserts that for any real numbers, withequalityif and only if is true for any real numbers.

Applications may be found at the beginner, intermediate, and olympiad levels of problems, with AM-GM being particularly important in proof-based contests, among other places.

Proofs

Induction and other more sophisticated inequalities are used in all known proofs of AM-GM, as described in the main article. They are also all more sophisticated than their application in the majority of beginning and intermediate contests, which is a good thing. AM-most GM’s simple proof makes use ofCauchy Induction, a type of induction in which one proves a result for, then uses induction to extend this to all powers of, and finally argues that assuming the result forimplies that it holds for all other powers of.

Generalizations

It has proven possible to generalize the AM-GM Inequality into a number of different inequalities. Other generalizations of AM-GM include theMinkowski Inequality andMuirhead’s Inequality, which are in addition to those already mentioned.

Weighted AM-GM Inequality

The Weighted Arithmetic-Geometric Means Inequality is a relationship between the weighted arithmetic and geometric means According to the rule, given any list of weights such that, with equality if and only if When the weighted form is reduced to the AM-GM Inequality, the result is a positive value. The proofs of the AM-GMarticle contain a number of proofs of the Weighted AM-GM Inequality.

Mean Inequality Chain

The Mean Inequality Chain is the main article. The Root Mean Square, Arithmetic Mean, Geometric Mean, and Harmonic Mean of a list of nonnegative reals are all related by the Root Mean Square, Arithmetic Mean, Geometric Mean, and Harmonic Mean Inequality Chain. It specifically specifies that with equality if and only if the following conditions are met The Mean Inequality Chain, like the AM-GM, has a weighted variant that is similar to the AM-GM.

Power Mean Inequality

Power Mean Inequality is the main article. The Power Mean Inequality is a mathematical relationship between all of the distinct power means of a list of nonnegative real numbers. The following is how the power mean is defined: It follows from this that ifb$” width=”42″ height=”13″> and equality holds only ifb$” width=”42″ height=”13″>, then plugginginto this inequality reduces it to AM-GM, and produces the Mean Inequality Chain. A weighted version of the Power Mean Inequality exists, just as there is a weighted version of the AM-GM.

Problems

  • Inequality
  • Mean Inequality Chains
  • Power Mean Inequality
  • Cauchy-Schwarz Inequality
  • Proofs of AM-GM

Geometric mean never exceeds arithmetic mean

As a result of JavaMan’s suggestion, I came up with the following: Suppose the following assertion is true:$:P(eta)equiv big frac sum limits eta: eta: eta: eta: geq: large frac sum limits eta: eta: £££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££ ^ j,::jinmathbb $ Now, $:P(1)equivlargefrac$$:geq:sqrt:longleftrightarrow :a 1-:a 2 :a 1-:a 2 $:P(1)equivlargefrac$$:geq:sqrt:longleftrightarrow $:P(1)equivlargefrac$$:geq:sqrt:longleftrightarrow $:P(1)equivlargefrac$$:geq $$sqrt+a 2:geq:0:longleftrightarrow :left $$sqrt+a 2:geq:0:longleftrightarrow $$sqrt+a 2:geq:0:longleftrightarrow $$sqrt+a 2:geq:0:longleftrightarrow $$sqrt+a 2:geq:0:longleft (sqrt-sqrtright) ^ 2:geq:0 Which is unquestionably correct.

Suppose that $:P(eta):$is true for all eta:$and that we can prove that it is also true for $:P(eta+1):$then we may go on to the next step.

Large Frac Sum Limits – to a frac sum limit of a given size – to a given size – to a given size – to a given size – to a given size – to a given size – to a given size – to a given size – to a given size – to a given size – to a given size – to a given size – to a given size – to a given size – ” $ ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” If $:a jleqlargefrac is true, then $$(a j):for all j:$then, $$(a j):$then, $:large fracsum limits eta:normalsize $:large fracsum limits $$:leq:largefracsum limits eta:frac(normalsize a j):$and$:leq:largefracprod limits eta:frac(normalsize a j):$and$:leq:largefracprod limits eta:frac(normalsize a j):$and$:leq:largefracprod limits eta:frac(normalsize a j) $$:leq:large prod limits $$:etafrac(sqrt):$ $$:leq:large prod limits $$:leq:large prod limits $$:leq:large prod limits $$:leq:large prod limits $$:leq:large prod limits $$:leq:large prod limits $$:leq:large As a result, by setting $:largePsi_ :$$=$:largefracsum limits a j:$, we get $:largePsi_ :$$=$:largefracsum limits a j:$ Phi = production limits eta sqrt:$we have $large:$Phi_ = production limits eta sqrt $1.largePsi_ = Psi left(Psi ,Psi_ right) $1.largePsi_ $$:geq:$$:largePsileft(Phi ,Phi_ )$$:largePsiright(Phi ,Phi_ )$$:largePsiright(Phi ,Phi_ )$$:largePsiright(Phi ,Phi_ )$$:largePsiright(Phi ,Phi_ )$$:largePsi $$:geq:$$largePhileft(Phi ,Phi right)=largePhi_ $$:geq:$$largePhileft(Phi ,Phi right)=largePhi_ $$:geq:$$largePhileft(Phi ,Phi right)=largePhi_ $$:geq:$$largePhileft(Phi ,P This demonstrates that $:largePsi is correct.

When $$:geq:$$:Phi:$is a power of$:eta:$, $$:geq:$$:Phi:$is a power of$:eta:$, $$:geq:$$:Phi:$is a power of$:eta:$ Suppose $$ = because the value of the variable $large omega normalsize,$$ is a power of $2$, and the value of $large omega normalsize is a power of $2$, and the value of $large omega normalsize is a power of $2$, and the value of the variable $large omega normalsize is a power of $2$, and the value of $large omega normalsize is a power of Because$:large frac sum limits omega: =large Psi_ $, it follows that$:left(large Psi_ right)cdot left(large Psi_ right)leq:left(large Psi_ right)leq:left(large Psi_ right)leq:left(large Psi_ right)leq: $Then, by placing both sides of the final inequality at the $:(largefrac):$ position, we get the following result: :$power produces the following results: $:largeprod limits ^ Large fracsum limits for the etasqrt:leq:large fracsum limits eta: $ eta: $ $:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::Largeclubsuit$

The Difference Between the Arithmetic Mean and Geometric Mean

There are several methods for evaluating the performance of a financial portfolio and determining whether or not an investment plan is effective. The geometric average, often known as the geometric mean, is frequently used by investment experts to make decisions.

Key Takeaways:

  • In the case of series that display serial correlation, the geometric mean is the most appropriate choice. Specifically, this is true for investment portfolios because the majority of financial returns are connected, such as bond yields, stock returns, and market risk premiums, among other things. Compounding becomes increasingly crucial with increasing time horizon, and the usage of the geometric mean becomes more acceptable. Because it takes into account year-over-year compounding, the geometric average gives a significantly more accurate representation of the underlying return for volatile values.

Due to the compounding that happens from period to period, the geometric mean differs from thearithmetic mean, or arithmetic mean, in how it is determined. Investors generally believe that the geometric mean is a more accurate gauge of returns than the arithmetic mean as a result of this phenomenon.

The Formula for Arithmetic Average

The portfolio returns for periodnn are represented by the numbers a1, a2,., ann where: a1, a2,., ann=Portfolio returns for periodnn=Number of periods begin A = fracsum_ n a i = fractextbfa 1, a 2, dotso, a n=textn n=textn n=textn n=textn n=textn n=textn n=textn n=textn n=textn n=textn n=textn n=textn n=textn n=textn n=textn n=text A=n1 i=1n ai =na1 +a2 +.+an where:a1,a2,.,an =Portfolio returns for periodnn=Number of periodsnn=Number of periodsnn=Number of periodsnn=Number of periodsnn=Number of periods ​

How to Calculate the Arithmetic Average

An arithmetic average is the product of the sum of a series of numbers divided by the number of numbers in that series. If you were asked to calculate the class (arithmetic) average of test results, you would simply add up all of the students’ test scores and divide that total by the number of students in the class. For example, if five students completed an exam and had scores of 60 percent, 70 percent, 80 percent, 90 percent, and 100 percent, the average for the arithmetic class would be 80 percent.

  1. It is because each score is an independent event that we use an arithmetic average to calculate test scores instead of a simple average.
  2. It is not uncommon in the field of finance to find that the arithmetic mean is not an acceptable way for determining an average.
  3. Consider the following scenario: you have been investing your funds in the financial markets for five years.
  4. With the arithmetic average, the average return would be 12 percent, which looks to be a substantial amount at first glance—but it is not totally correct in this case.
  5. They are interdependent.

Our goal is to arrive at an accurate calculation of your actual average yearly return over a five-year period. To do so, we must compute the geometric average of your investment returns.

The Formula for Geometric Average

x1,x2,=Portfolio returns for each periodn=Number of periodsbeginleft(prod_ n x i ) = sqrttextbfx 1, x 2, the number of periods is equal to the sum of the returns on the portfolio for each period (x1, x2,.) and the number of periods is equal to the sum of the returns on the portfolio for each period (n1). ​

How to Calculate the Geometric Average

It is possible to determine the geometric mean for a series of integers by multiplying the product of these values by the inverse of the length of the series. In order to do this, we add one to each of the numbers (to avoid any problems with negative percentages). In the next step, add up all of the numbers and elevate their product to a power of one divided by the number of numbers in the series. Then we take one away from the final result. When expressed in decimal form, the formula looks somewhat like this: [(1+R1)×(1+R2)×(1+R3)…×(1+Rn)] beginning with the number 1 and ending with the number 1 where R=Returnn=Count of the numbers in the series starting with the number 1 and ending with the number 1 – 1 textbftext= textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = text 1 where: R = Returnn = Number of numbers in the series 1 where: R = Returnn = Number of numbers in the series Although the formula seems complicated, it is not as tough as it appears on paper.

Using our previous example, we can calculate the geometric average as follows: The percentages of returns we received were 90 percent, 10 percent, 20 percent, 30 percent, and -90 percent, therefore we entered them into the calculation as: (1.91.11.21.30.1).

The figure obtained by applying the geometric average is far worse than the 12 percent arithmetic average we obtained previously, and regrettably, it is also the number that best depicts reality in this particular instance as well.

Links forward – The AM–GM inequality

Exercise 11 provided a geometric argument that the arithmetic mean of two positive numbers a and b is larger than or equal to the geometric mean of the two numbers. This may also be demonstrated algebraically, as seen below. Because a and b are both positive, we may define x and y as (x = sqrt ) and (y = sqrt ) respectively. Consequently, begin(x-y)2 geq 0 quad impliesquad x2+y2-2xy geq 0quad impliesquad dfracgeq xy endand so [dfracgeq xy] [dfracgeq sqrt. ] [dfracgeq xy] [dfracgeq sqrt.] [dfracgeq xy] [dfracgeq sqrt.] This is referred to as the AM–GM inequity.

Example

Find the domain and range of the function (f(x) = x2 + dfrac) for the variable x. (xneq 0).

Solution

Using the AM–GM inequality, [f(x) = x2 + dfracgeq 2sqrt = 2] is equivalent to [f(x) = x2 + dfracgeq 2sqrt = 2]. As a result, the range of f is contained within the interval [2,infty]). Make a note of the fact that f(1) = 2 and (f(x) to infty) as well as the fact that f(1) = 2 (x to infty). Considering that (f) is continuous, it follows that for each value of y more than two, there exists an equivalent value of x greater than one with f(x) equal to y. As a result, the range of f is defined as the interval [2,infty]).

  1. Locate and write down the arithmetic, geometric, and harmonic means of the numbers (3,4,5), arranged in ascending order
  2. Make a mathematical argument to support the claim that the harmonic mean of two positive real numbers a and b is less than or equal to their geometric mean

The AM–GM disparity can be expressed in the following ways. If a 1, a 2, dots, and a n are all positive real integers with n digits, then [dfracgeq sqrt. ] is true. The following exercise serves as a demonstration of this conclusion. Exercice No. 16

  1. Calculate the greatest value of the function (f(x) = log e x – x) for the variable x. (x0). As a result, we may conclude that (log e x leq x-1), for all values of x0
  2. Assume that a 1, a 2, dots, and a n are all positive real integers, and define a n as (A = dfrac). As a result of sequentially inserting x equal to dfrac for (i=1,2,dots,n) in the inequality from section (a), and summing the results, prove that [log eBigl(dfracBigr) = 0]
  3. The generalised AM–GM inequality may be derived from component (b) by exponentiating both sides of the inequality.

The following page contains information about history and applications.

Using the Arithmetic Mean-Geometric Mean Inequality in Problem Solving.

Some of the information in this article was used to construct a presentation for the Annual Meeting of the School Mathematics and Science Association, which was held in Birmingham on November 8, 2012. Version in PDF format The Arithmetic Mean-Geometric Mean Inequality (AM-GM Inquality) is a mathematical connection that is important to many mathematical concepts. It is a valuable tool for problem solving and for establishing connections with other branches of mathematics. The use of this concept in classroom mathematics should be greater than it is today.

This paper provides a broad introduction to the theorem as well as some background information and extensions, alternate demonstrations of the proof, and instances of situations that may be investigated utilizing the AM-GM Inquality.

In my examples, I will concentrate on the theorem for two positive numbers, but I will also discuss the extensions below and occasionally use the case for three positive numbers to illustrate the point.

The link raises the question of whether it is possible to generalize the evidence in accordance with the lines of argument presented by Courant and Robbins (1942).

GEOMETRIC DEMONSTRATION OF THE RMS-AM-GM-HM INEQUALITY For two positive values a and b, in particular, Using a more sophisticated degree of analysis (maybe a more basic approach?) They are all instances of Power Methods, which are defined as follows: where the power parameter has distinct values for each of the different means These are sometimes referred to as Generalized Means.

  1. For example, it is commonly known that the maximum area of a rectangle with a set perimeter is equal to the area of a square with the same perimeter.
  2. Comment: The word “minimum” should be replaced with the word “maximum” in the second statement below.
  3. 2a + 2b = 10 is the fixed perimeter of the circle.
  4. Let brerepresent the length of one side and 5 – b represent the length of the opposite side of the equation.

points on the blue curve with Area = 6.25 are always greater than points on the red curve (that is, the area of a rectangular prism will never be greater than 6.25), and the blue curve (a horizontal line representing an invariant) is tangent to the red curve when and only when (b = (5-a) = (b), that is, when b =2.5.

The other side has a value of 5 – x.

When a =b, the area is always smaller than the constant, and when a =b, the area is equal to the constant.

Here is an example of a challenge addressed to students on my website: Unlike the first problem, in which ab is a constant rather thana + b, the second problem is straightforward and arises directly from the AM-GM Inequality: In this part, I will restrict the investigation to the most straightforward scenario: The arithmetic mean and geometric mean of two positive integers are both equal to one another.

In my Problem Solving course, I want the students to discover at least 5 demonstrations or proofs, which I refer to as an EXPLORATION.

Seeing a variety of approaches to this relationship can aid in comprehending it, recognizing its significance, and recognizing its use as a problem-solving tool in the future.

Some algebra

Construct a semicircle with a diameter equal to the sum of the two numbers a and b. The Arithmetic Mean of a and b will be used to get the radius. From the common endpoints of the segments of lengtha andb, construct a line perpendicular to the diameter of the circle. Create the red segment by intersecting this perpendicular with the semicircle and extending it from there. This segment will always have a length that is less than or equal to the radius of the circle, and it will always be equal if and only if a and b have the same length.

Another Geometry Example

It is necessary to start with the identity: Sincewe havewith equality if and only ifa = b, we can proceed to the demonstration. Another Geometry Exemplification Given two tangent circles with radii a and b, solve for a. Using the two circles, construct an exterior tangent that is common to both circles, then connect the radii of each circle to the common tangent. Calculate the length indicatedc by?along the common tanget in terms ofa and b using the formula. A right triangle with legs of length?

  • The segment along the common tangent has length twice the geometric mean of the two segments.
  • Allow the CD to pass through the midway M of the AB equation.
  • The lengths are all in the positive direction.
  • We can only be certain that x + y = 2z with equality if M is the midway of the CD.

Cost of Fencing a Field

An agriculturalist wishes to fence in 60,000 square feet of land in a rectangular plot along a straight highway, posing the following problem: The fence he intends to use along the highway will cost him $2 per foot, but the fence he intends to use on the other three sides will cost him $1 per foot When purchasing fences, how much of each type will be required to ensure that costs are kept as low as possible?

What is the bare minimum in terms of expenditure? Solution: (This is a common calculus issue, yet there is no need for calculus here!)

Minimum of

The objective is to determine the smallest possible values for this function in the range or0x. The AM-GM Inequality must be applied to this equation, which necessitates a modification in the equation’s form. To begin, though, it is beneficial to examine a graph and understand the equation in some detail. There may be two minimum values in the range0x according to the graph on the right, which proposes two possibilities.

Construct a Square with Same Area as a Given Rectangle

Construction of a square with a straightedge and a compass that has the same area as a specified rectangle is the objective of the task.

The area of the square will be banded, and the length of the side of the square will be the geometric mean of the area. Consider: The geometric mean of a and b is shown by the red section.

Maximum Area of a Sector of a Circle With Fixed Perimeter

Recognizing the nature of the situation. A sector of a circle is defined by its perimeter, which is made up of two radii, and the arc of the circle that connects the ends of the two radii. Compare the areas of three sectors, each with P = 100 and center angles of 45 degrees, 90 degrees, and 180 degrees, and each with P = 100. Eighth circle, quarter circle, and semicircle sectors are all included in this category. As the angles increase, the radii decrease in length because more of the fixed perimeter is included inside the arc as the angles increase.

There is a maximum amount of space available anywhere.

Sector as Fraction of a Circle

When it comes to dealing with radian measure, some students (as well as teachers) are uncomfortable. For them, the following technique would be appropriate: In this case, letkbe the fraction of a circle represented by the thesector 0k1. Because the perimeter is a constant, we may describe it as twice the radius multiplied by the proportion of the circumference of the circle. The area of the sector in fraction notation can be expressed as either a function of r or a function of k, depending on whether the perimeter equation is substituted for r or for k in the sector equation.

The angle is a fraction of a degree less than 120 degrees.

Sector in Radian Measure

It is interesting to note the similarity between this solution and the previous one, in which the square was the maximum area for rectangles with set perimeter.

Inequalities Problems

To complete a right triangle, construct segments from the diameter’s ends in the manner shown.

Maximumand Minimum of

Rewrite The value of the function is always less than or equal to.5 when x0 is taken into consideration, according to the arithmetic mean-geometric mean inequality. When x = 1, the value of the function is equal to.5. When applied to x0, a similar parameter leads to the discovery of the minimum of the function when x = -1. We must apply the Arithmetic Mean – Geometric Mean Inequality to – x and -1/x when x0 is positive since the inequality holds only for positive numbers. We’re aware of it. Always remember that this is for x0, thus -x and -1/x are both positive.

In order to do this, the value of the function is always greater than or equal to -0.5, and it is only equal to -0.5 when x equals -1.

Maximum off(x) = (1-x)(1+x)(1+x)

It is necessary for us to know when the function reaches its maximum in an interval since we have three elements in the function. In order to benefit from the AM-GMInequality, we must ensure that the total of these elements equals a fixed amount of money.

That is why we require 2x rather than x. We may obtain this information by doing the following: In the case of x inif, the function achieves its maximum value only if2(x) = 1 + x2 – 2x = 1 + x1 = 3x = 1 + x2 – 2x = 1 + x1 = 3x = 1 + x2 – 2x = 1 + x1 = 3x = 1 + x1 = 3x = 1 + x1 = 3x That is to say,

MinimumSurface Area of a Can of Fixed Volume

The form of the can (e.g., short and fat vs. tall and slim) that is used to package a product in a given volume may be determined by different criteria such as tradition and presumptive client preferences, among other things, in the case of a fixed volume. Take, for example, the fact that all 12 oz. Coke cans are the same form – with a height of around 5 inches and a radius of approximately 1.25 inches – Why? What if the decision was made with the goal of reducing the amount of material utilized to create the can?

What is the connection between the radius and the height when attempting to decrease the surface area of a fixed volume of material?

The volume is set at a certain level.

What is the very bare minimal S?

Maximum area of a Pen, fixed length of fencing, with Partitions Parallel to a Side

Consider the following scenario: you have 100 feet of fencing. You’re looking for a rectangular pen with a divider running parallel to one of the sides. The 100 feet of fence must encircle all four sides of the building as well as the partition. What is the form of the pen, and how much space does it have at its maximum? Demonstrate that the maximum area penis has the following dimensions: 25 ft. by 16.67 ft. with a maximum area of roughly 417.7 sq. ft.

Prove the Maximum Area of a Triangle with Fixed Perimeter is Equilateral

Find the point on the graph of that is the most closely associated with the origin. (Click on the link to get to the debate) Given a triangle with one side measuring 9 units in length and the ratio of the other two sides being 40/41, find the length of the other side. Calculate the largest possible area. (Discussion may be found at the link) Calculate the maximum volume of a box created from a 25by 25 square sheet of cardboard by removing a tiny square from each corner and folding up the edges to produce a lidless box using the Arithmetic Mean-Geometric MeanInequality.

Clearly, the proof of3implies the proofs of the previous two.

On the right is a portion of a spreadsheet displaying the results of the calculation for.

The Excel file may be downloaded by clicking HERE.

Averages – Wikiversity

An average is a single number that is calculated from a collection of actual numbers in order to provide an approximation of the overall size of the numbers in the collection.

To put it another way, it is a function from a collection of nnumbers to a single number that has the following characteristics:

  1. If all of the numbers are equal, the average of all of the numbers should be equal to this value: AV(x, x, x,.) = x
  2. AV(x, x, x,.) = x
  3. The average cannot be greater than the sum of the numbers, nor can it be less than the sum of their minimums. While it is possible to be more stringent and specify that if not all of the values are equal, it must be more than the lowest and less than the maximum, doing so would rule out themedianas as an average because the median of 1, 1, 1, 1, 2 is 1
  4. This means that if all numbers are multiplied by the same constant, then their average will be multiplied by the same amount as well. It follows that AV(kx, ky) = k. the function AV(x, y)
  5. The average must be order invariant, which means that even if we permute the numbers, their average will not change: AV(y,x) = AV(y,x) (x, y). This eliminates the possibility of just selecting the first number, or the average of the first and last numbers, or any other weighted averages. The average must be monotonic, which means that if any one number grows while the others remain unchanged, the average must not drop, and the opposite is true. Some “robust measures,” in which outliers are discarded before the average is calculated, are ruled out as a result of this. We could want to go a little tighter and declare that if any number increases, the average must increase as well. In this case, the median would be ruled out as an average, because the median of two numbers such as (1,1,1,1) and (1,1,1,2) are both 1.

It is possible to believe that the average should be translation invariant, meaning that if all numbers are raised by the same constantk, their average would grow by the same number: AV(x+k,y+k) = AV(x,y)+k. However, this is not the case. However, it can be demonstrated that there is only one average that meets both of these and the other conditions strictly: the arithmetic mean, which will be explained further down the page. It is possible that the median and other quantiles might fulfill all of the requirements if the less rigorous versions of the standards were utilized; however, this is not likely.

Quantiles would be ruled out once more in this case.

The arithmetic mean

The simplest type of average is the arithmetic mean, which is defined as the sum of the numbers divided by the number of numbers in the sample. This results in (1+2+3+4+5)/5 = 3 for the average of the numbers 1, 2, 3, and 4. Exercise: Check that this satisfies all of the requirements listed above, even the more stringent variants.

The geometric mean

Another popular average is thegeometric mean, which is calculated by multiplying all of the numbers together and, if there arennumbers, by calculating thethroot of the result. This means that (1×2, 3x4x5) 1/5=2.605 is the geometric mean of the numbers 1, 2, 3, 4, and 5. (approximately). It is important to note that the geometric mean should not be employed if any of the numbers is negative (why?) and that the geometric mean should be zero if any of the numbers is zero, regardless of how large the other numbers are (why?

The arithmetic/geometric inequality

It is always true that the geometric mean is less than the arithmetic mean, unless all of the integers are equal. This is readily demonstrated for two and three numbers when just two and three numbers are used; the following are the main points of the proof: “If a and b are two unequal non-negative integers, then0″>This may be rearranged as (A multiplied by B). Assuming a, bandcare three non-negative integers that are not all equal, then0″>So0″>This may be rewritten as (a times b times c)0″>.

Quantifying the arithmetic/geometric inequality

Ascertaining whether the geometric mean is close to the arithmetic mean or much less is a challenging task. If we consider a collection of n integers and assume that their arithmetic mean is m, then the sum of the y is zero. (Why?) We assume that the x do not deviate significantly from their mean, so the y are relatively tiny values, and we can extend log (1+y) in a power series to represent the data. (Because the value of y is zero.) As a result (and why?) To put it another way, the larger the dispersion of the numbers around their mathematical mean, the greater the disparity between the two mean values of the numbers.

Exercise: In order to arrive at this conclusion, several estimates have been used. Demonstrate, by real computations, that the conclusion is correct in all cases.

Root mean square

In mathematics, the RMS is the square root of the arithmetic mean of the squares of a collection of numbers, which is to say that this average should not be used for a mixture of positive and negative integers (for what reason?). Exercise: Verify that this satisfies all of the requirements listed above, including the more stringent versions. The root mean square (RMS) is always bigger than the arithmetic mean, unless all of the integers are equal. Considering the square of the RMS and the square of the arithmetic mean, we can easily demonstrate this using just two numbers: Any amount of numbers may be proved using a proof that is comparable but more complicated.

Harmonic mean

Generally speaking, the harmonic meanof a set of integers is equal to one-half of the arithmetic meanof the reciprocals of those numbers. As a result, for three numbers, we have It is not recommended to use this mean if any of the numbers is 0 or negative (why?). Exercise: Verify that this satisfies all of the requirements listed above, including the more stringent versions. Because the harmonic means less than the geometric mean unless all of the numbers are equal, the harmonic means less than the geometric mean Due to the fact that its reciprocal is the arithmetic mean of the reciprocals of the integers, it is more significant than the geometric mean of the reciprocals, which is the reciprocal of the geometric mean, as a result.

RMS is the arithmetic mean.

Rth power mean

For every real number r, the th power mean of a set of numbers is equal to one. For reasons that are unclear, this average should only be used for positive integers when r0 and for non-negative numbers when r0. Exercise: Verify that this satisfies all of the requirements listed above, including the more stringent versions. When r = 0, this mean is undefinable, but as r approaches zero, the limit becomes the geometric mean. The averages we’ve looked at so far are all special examples of this mean (r = -1, for harmonic; r = 0, for geometric; r = 1, for algebraic; and r = 2, for RMS).

As r approaches infinity, the mean tends to the maximum of the x I conversely, as r approaches negative infinity, the mean tends to the lowest of the x i.

Power plus 1 mean

For every real number s, the power plus one mean of a collection of numbers is In the case of s=0, this is the arithmetic mean; in the case of s=-1, this is the harmonic mean Take note in particular that if s=1, the result is RMS 2 / (arithmetic mean). In the case wheremis the arithmetic mean ands2is the variance of a collection of numbers, then the s=1 mean ism”>It can be demonstrated that this sort of mean operates in a manner similar to the rth power mean. This mean is a continuous, strictly monotonic rising function of s that may be applied to any collection of positive integers (not all of which are equal).

Essentially, these two sorts of means may be seen to be specific cases of the following: The power plus 1 mean is clearly represented by s=0, while the rth power mean is represented by r=1.

Mixed averages

By combining distinct averages, it is possible to obtain more types of averages, provided that the formula is symmetric in the variables used. For example, the average of any three numbers x, y, and z is the sum of the following three numbers:

The median and other quantiles

As previously stated, if we do not require certain severe inequalities in the definition, the median is equivalent to the mean. In fact, this is true for every quantile of the distribution, including the maximum and lowest values of the distribution. It may be counterintuitive to refer to the highest and minimum as “averages.” The essential aim of an average, on the other hand, is to provide an approximation of the order of magnitude of a collection of data. A number’s maximum and lowest can be deceptive if the range of numbers is narrow; nevertheless, when the range of numbers is broad, any single figure as the average can be misleading.

According to this argument, every quantile other than the median is “biased,” and hence unsuitable.

Take, for example, the following.

The mode

The mode is not the same as the average. It is not always clearly defined; for example, there are two modes, 1 and 3, in the case of (which are also the minimum and maximum). Second, and perhaps more importantly, it fails to adhere to the monotonicity condition. Take, for example, the set of eight integers. The mode number is two. Consider the following scenario: the seventh number grows from 2 to 3: The joint modes have been renamed to 1 and 2. Consider the following scenario: the sixth number grows from 2 to 4: The mode has been changed to 1.

Transformation means

For simplicity, consider any strictly monotonic function f(x). Define the following set of numbers given a set of numbers:

  1. Assume that Y I is the solution of the equation Y=f(x I Y is the arithmetic mean of y I X is the solution of the equation Y = f(X)

X is therefore defined as the mean of the transformation of the x I with regard to f. (x). For example, the function f(x) = x r yields the rth power mean, whereas f(x) = log(x) yields the geometric mean. Exercises

  1. Will such a mean always adhere to the requirements set out at the outset of this essay
  2. Would employing a different type of mean at step 2, such as the geometric mean, result in a different type of average, as well? Is there a reason why the equation in step 3 always has a single and unique solution? Is it important that f(x) is not a continuous function
  3. Find a function f(x) such that the mean of the transformation equals the median of the transformation

See also

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