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How do I prove that $(E[X_1])^q \leq E([\frac{1}{n}\sum_{j=1}^{n}X_j])^q \leq E\bigg[X_1\bigg(\frac{X_1+(n-1)\mu}{n}\bigg)^{q-1}\bigg]$ using Jensen's inequality? I tried using $\phi(x) = x^q$ and $x^q = xx^{q-1}$ but I don't know how to use Jensen's Inequality for conditional expectations. Also, $X_1, \dots, X_n$ are independent random variables with $X_1 > 0, E[X_1]=\mu$ and $E[X_1^q]< \infty, 1<q\leq 2$

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  • $\begingroup$ You notation is confusing. By, $E[X_1]^q$, do you mean $E[X_1^q]$ or $(E[X_1])^q$ ? $\endgroup$
    – dohmatob
    Jul 17, 2020 at 14:19
  • $\begingroup$ @dohmatob it is $(E[X_1])^q$, I will edit if it is better. Do you have any idea? $\endgroup$
    – Bro Livros
    Jul 17, 2020 at 14:30
  • $\begingroup$ Welcome to SE! If you say $E[X_1]^q$ means $(E[X_1])^q$ (which I doubt), then when do you mean by the condition "$E[X_1]^q < \infty$" ? Isn't it sufficient to ask for $E[X_1] := \mu < \infty$ ? Maybe you mean the moment condition "$E[X_1^q] < \infty$" (which is stronger than $E[X_1] < \infty$). I advice, you take some time to read you question and be sure that everything makes sense. People won't pay much attention to your question if it has many notation errors / inconsistencies. $\endgroup$
    – dohmatob
    Jul 17, 2020 at 14:51
  • $\begingroup$ @dohmatob now everything is correct. This is an exercise I really need help. $\endgroup$
    – Bro Livros
    Jul 17, 2020 at 14:57
  • $\begingroup$ Just to confirm that in the last term of the inequality there isn't a missing $n^{2-q}$ factor. $\endgroup$
    – Mittens
    Jul 17, 2020 at 15:22

2 Answers 2

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It may be that there is a $n^{q-2}$ factor missing in your statement.

Here is what I got:

Jensen's inequality implies that

$$(E[X_1])^q=\left(E\Big[\tfrac{X_1+\ldots+ X_n}{n}\big]\right)^q\leq E\Big(\tfrac{X_1+\ldots+ X_n}{n}\Big)^q$$

As $\{X_1,\ldots,X_n\}$ ia an i.i.d. family $$ \begin{align} E\left[\Big(\tfrac{X_1+\ldots+ X_n}{n}\Big)\Big(\tfrac{X_1+\ldots+ X_n}{n}\Big)^{q-1}\right]&=\frac{1}{n}\sum^n_{j=1}E\left[X_j\Big(\tfrac{X_1+\ldots+ X_n}{n}\Big)^{q-1}\right]\\ &= E\left[X_1\Big(\tfrac{X_1+\ldots+ X_n}{n}\Big)^{q-1}\right]\\ &=E\left[X_1 E \left[\Big(\tfrac{X_1+\ldots+ X_n}{n}\Big)^{q-1}|X_1\right]\right] \end{align} $$

Since $0<q-1\leq 1$,

$$ \Big(\frac{X_1+\ldots+X_n}{n}\Big)^{q-1}\leq\frac{X^{q-1}}{n^{q-1}}+\ldots +\frac{X^{q-1}_n}{n^{q-1}} $$

Integrating with respect the conditional probability given $X_1$ leads to

$$ \begin{align} E\left[\Big(\frac{X_1+\ldots+X_n}{n}\Big)^{q-1}|X_1\right]&\leq \frac{X^{q-1}_1}{n^{q-1}}+\frac{E[X^{q-1}_2]}{n^{q-1}}+\ldots + \frac{E[X^{q-1}_n]} {n^{q-1}}\\ &\leq \frac{X^{q-1}_1}{n^{q-1}}+\frac{(E[X_2])^{q-1}}{n^{q-1}}+\ldots + \frac{(E[X_n])^{q-1}}{n^{q-1}}\\ &= n\frac{X^{q-1}+(n-1)\mu^{q-1}}{nn^{q-1}}\leq n\Big(\frac{1}{n}\Big(\frac{X_1}{n}+\frac{\mu}{n}+\ldots+\frac{\mu}{n}\Big)\Big)^{q-1}\\ &=n^{2-q}\Big(\frac{X_1+(n-1)\mu}{n}\Big)^{q-1} \end{align} $$

Here we have twice used Jensen's inequality for the concave function $x\mapsto x^{q-1}$. Putting things together

$$\mu^q=(E[X_1])^p\leq n^{q-2}E\left[X_1\Big(\frac{X_1+(n-1)\mu}{n}\Big)^{q-1}\right] $$

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  • $\begingroup$ Thanks for the answer. Pretty clear now! $\endgroup$
    – Bro Livros
    Jul 17, 2020 at 19:07
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$$ \begin{split} (E[X_1])^q &\overset{(a)}{=} \left(E\left[\frac{1}{n}\sum_{i=1}^nX_i\right]\right)^q \overset{(b)}{\le} E\left[\left(\frac{1}{n}\sum_{i=1}^nX_i\right)^q\right]\\ & \overset{(c)}{=} E\left[\left(\frac{1}{n}\sum_{i=1}^nX_i\right)\left(\frac{1}{n}\sum_{i=1}^nX_i\right)^{q-1}\right]\\ &\overset{(d)}{=}\frac{1}{n}\sum_{j=1}E\left[X_j\left(\frac{1}{n}\sum_{i=1}^nX_i\right)^{q-1}\right], \end{split} $$ where

  • (a) is because the $X_i$'s are iid and so $E[X_i] = E[X_1]$ for all $i$.
  • (b) is Jensen's inequality for the convex function $x \mapsto x^q$.
  • (c) Just expanding things out.

Try to continue from here...

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  • $\begingroup$ Ok, perfect. Will try to solve it now. Thanks a lot! $\endgroup$
    – Bro Livros
    Jul 17, 2020 at 19:07

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