# Prove inequality on expectation

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

• You notation is confusing. By, $E[X_1]^q$, do you mean $E[X_1^q]$ or $(E[X_1])^q$ ? Jul 17, 2020 at 14:19
• @dohmatob it is $(E[X_1])^q$, I will edit if it is better. Do you have any idea? Jul 17, 2020 at 14:30
• 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. Jul 17, 2020 at 14:51
• @dohmatob now everything is correct. This is an exercise I really need help. Jul 17, 2020 at 14:57
• Just to confirm that in the last term of the inequality there isn't a missing $n^{2-q}$ factor. Jul 17, 2020 at 15:22

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,

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

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

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

• Thanks for the answer. Pretty clear now! Jul 17, 2020 at 19:07

$$\begin{split} (E[X_1])^q &\overset{(a)}{=} \left(E\left[\frac{1}{n}\sum_{i=1}^nX_i\right]\right)^q \overset{(b)}{\le} E\left[\left(\frac{1}{n}\sum_{i=1}^nX_i\right)^q\right]\\ & \overset{(c)}{=} E\left[\left(\frac{1}{n}\sum_{i=1}^nX_i\right)\left(\frac{1}{n}\sum_{i=1}^nX_i\right)^{q-1}\right]\\ &\overset{(d)}{=}\frac{1}{n}\sum_{j=1}E\left[X_j\left(\frac{1}{n}\sum_{i=1}^nX_i\right)^{q-1}\right], \end{split}$$ 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...

• Ok, perfect. Will try to solve it now. Thanks a lot! Jul 17, 2020 at 19:07