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Here is something that I find hard to make sense of. Suppose $X_1, X_2, ..., X_n$ are independent draws from some distribution. By AM-GM inequality, we have:

$$ \left( X_1 X_2 .. X_n \right)^\frac{1}{n} \le \frac{X_1 + X_2 +...+X_n}{n} $$

Now $E[X_1 X_2...X_n]=E[X_1]^n$ by independence with any $n$.

Also as $n \to \infty$, $\frac{X_1 + X_2 +...+X_n}{n} \to E[X_1]$ by law of large numbers.

Plug these terms into the first equation, we have:

$$E[X_1] \le E[X_1]$$ when $n \to \infty$

Note that equality of the above is only possible when $X_1 = X_2 = ... = X_n$ which is highly unlikely and irrelevant.

Can someone tell me what is going on and what I am missing?

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The expectation of $\left(\frac{X_1 + X_2 +...+X_n}{n} \right)^n$ is not $E[X]^n$. For $n=2$ as an example, $$E[(X+Y)^2/4] = E[(X^2 + Y^2 + 2XY)/4] = (E[X^2] + E[X]^2)/2 = E[X]^2 + \frac{1}{2}{\rm Var}(X) \geq E[X]^2$$ if $X$ and and $Y$ are independent samples from the same distribution.


In the Edited version, the same problem moves to the other side of the claimed equation; $E[\left( X_1 X_2 .. X_n \right)^\frac{1}{n}] = E[X^{\frac{1}{n}}]^n$ is not $E[X]$.

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By your logic on the left-hand side of this inequality (which is not actually valid, but since @zyx already explained the problems on the left-hand side, I'm going to assume that it is valid to explain why your reasoning on the right side is invalid), the following is true for any $n$ with the situation you gave: $$E[X_1] \leq \frac{X_1+X_2+X_3+...+X_n}{n}$$

You then say that $\frac{X_1+X_2+X_3+...+X_n}{n} \to E[X_1]$ as $n \to \infty$. That's true, but using that, you can not substitute $E[X_1]$ for the average because that only becomes true as $n \to \infty$. It's not necessarily true that $\frac{X_1+X_2+X_3+...+X_n}{n}=E[X_1]$ for any $n$, but rather that, as $n$ gets larger, the first value becomes closer to the second value.

What you can say is the original inequality at the top of this answer and that as $n \to \infty$, the right-hand side becomes closer to $E[X_1]$. The right-hand side approaches the left-hand side as $n \to \infty$. That is not the same as saying that $E[X_1] \leq E[X_1]$ as $n \to \infty$ with equality only if all $X_i$ are equal, which simply does not make any sense.

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  • $\begingroup$ Sorry I edited the original question so that $n$ is no longer in the final inequality. $\endgroup$ – Tom Bennett Feb 13 '16 at 22:52
  • $\begingroup$ Your reasoning is still wrong. You simply can't manipulate expressions in limits like that. $\endgroup$ – Noble Mushtak Feb 13 '16 at 23:05
  • $\begingroup$ @TomBennett I clarified my answer above to adjust to your edits. $\endgroup$ – Noble Mushtak Feb 13 '16 at 23:11
  • $\begingroup$ The expected geometric mean of n samples from the distribution of X converges to the geometric mean of the distribution of X, when the latter exists. Hence the strict inequality E[GM] < E[AM] for finite samples also holds in the limit of large samples. On the other hand E[AM] = E[X] for finite samples and in the limit. Maybe the convergence of the geometric means to E[GM] is monotonic but other than that I don't know what else can be said about the interaction of AM-GM and expected value under finite i.i.d sampling from one distribution. $\endgroup$ – zyx Feb 14 '16 at 17:36
  • $\begingroup$ @zyx I'm a bit confused as to what you said because I'm not that familiar with statistics, but what I was trying to say in my answer was that his logic on the right-side was fallacious because he took expressions out of context with the limit. In order to make that point, I assumed that his statistical reasoning on the left-side was correct even though you showed it was wrong. I did this in order to discuss only his reasoning on the right-side since you already explained the left-side. $\endgroup$ – Noble Mushtak Feb 14 '16 at 20:05

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