$X_1,X_2,\ldots,X_n$ are i.i.d. random variables, $X_1>0$, $E[X_1]=\mu$, $E[X_1^k]<\infty$ for $1<k \leq2$. Prove: $$ E\left[\left(\frac{1}{n}\sum_{i=1}^nX_i\right)^k\right]\leq E\left[X_1\left(\frac{X_1+(n-1)\mu}{n}\right)^{k-1}\right]. $$

I have problem with showing that this inequality really does hold. I know that Jensen inequality should be applied, but I do not know how.

Any help would be great. Thanks!



$$\left( \frac{1}{n} \sum_{i=1}^n X_i \right)^k = \left( \frac{1}{n} \sum_{j=1}^n X_j \right) \left( \frac{1}{n} \sum_{i=1}^n X_i \right)^{k-1}$$

we get

$$\mathbb{E} \left[ \left( \frac{1}{n} \sum_{i=1}^n X_i \right)^k \right] = \frac{1}{n} \sum_{j=1}^n \mathbb{E} \left[ X_j \left( \frac{1}{n} \sum_{i=1}^n X_i \right)^{k-1} \right].$$

Since the random variables are independent and identically distributed, this implies

$$\mathbb{E} \left[ \left( \frac{1}{n} \sum_{i=1}^n X_i \right)^k \right] = \mathbb{E} \left[ X_1 \left( \frac{1}{n} \sum_{i=1}^n X_i \right)^{k-1} \right]. \tag{1}$$

Using the tower property and the independence of the random variables, we find

$$\begin{align*} \mathbb{E} \left[ X_1 \left( \frac{1}{n} \sum_{i=1}^n X_i \right)^{k-1} \right] &= \mathbb{E} \left( \mathbb{E} \left[ X_1 \left( \frac{1}{n} \sum_{i=1}^n X_i \right)^{k-1} \mid X_1 \right] \right) \\ &= \mathbb{E} \left( X_1 \cdot \mathbb{E} \left[ \left( \frac{x+ \sum_{i=2}^n X_i}{n} \right)^{k-1} \right] \bigg|_{x=X_1} \right). \tag{2}\end{align*}$$

Since $0<k-1 \leq 1$ it follows from Jensen's inequality that

$$\begin{align*} \mathbb{E} \left[ \left( \frac{x+ \sum_{i=2}^n X_i}{n} \right)^{k-1} \right] &= \left(\mathbb{E} \left[ \left( \frac{x+ \sum_{i=2}^n X_i}{n} \right)^{k-1} \right] \right)^{\frac{k-1}{k-1}} \\ &\leq \left( \mathbb{E} \left[ \frac{x+\sum_{i=2}^n X_i}{n} \right] \right)^{k-1} \\ &= \left( \frac{x+(n-1) \mu}{n} \right)^{k-1}. \tag{3}\end{align*}$$

Combining $(1)$, $(2)$, $(3)$ finishes the proof.

  • $\begingroup$ Excellent! Thank you! $\endgroup$ – Kyoto Jan 31 '15 at 19:51
  • $\begingroup$ @Kyoto You are welcome. If you like the answer, you can upvote it by clicking on the arrow next to it. $\endgroup$ – saz Jan 31 '15 at 19:52
  • $\begingroup$ Could you explain how we got to equation $(1)$? I don't seem to understand why we can write it like that (ie RV being IIDs implies that).. $\endgroup$ – Tyler Hilton Feb 11 '15 at 18:43
  • $\begingroup$ @TylerHilton It holds that $$\mathbb{E} \left[ X_j \left( \frac{1}{n} \sum_i X_i \right)^{k-1} \right] = \mathbb{E} \left[ X_1 \left( \frac{1}{n} \sum_i X_i \right)^{k-1} \right]$$ for all $j \in \{1,\ldots,n\}$. This follows from the fact that the random variables are IID. [...] $\endgroup$ – saz Feb 11 '15 at 18:52
  • $\begingroup$ Indeed: Write $$\begin{align*} \mathbb{E} \left[ X_j \left( \frac{1}{n} \sum_i X_i \right)^{k-1} \right] &= \int \dots \int x_j \left( \frac{1}{n} \sum_i x_i \right)^{k-1} \, d\mu(x_1) \dots \, d\mu(x_n) \\ &= \int \dots \int x_1 \left( \frac{1}{n} \sum_i x_i \right)^{k-1} \, d\mu(x_1) \dots \, d\mu(x_n) \\ &= \mathbb{E} \left[ X_1 \left( \frac{1}{n} \sum_i X_i \right)^{k-1} \right] \end{align*}$$ where $\mu$ denotes the distribution of $X_1$. $\endgroup$ – saz Feb 11 '15 at 18:52

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