calculating $\mathbb E\left(\exp\left(\frac{1}{2}\sum_{i=1}^n X_i^2\right)\right)$

Question

suppose $X_1,X_2,\ldots,X_n \sim \mathcal N(0,\sigma^2)$. How can I calculate $$\mathbb E\left(\exp\left(\frac{1}{2}\sum_{i=1}^n X_i^2\right)\right)$$

Answer 1

One of the comments addresses the solution. But, I will outline the logic and show the direct computation which is not immediate. You have not assumed anything about the $X_i$, $1 \leq i \leq n$ except that they are identically $N(0, \sigma^2)$. I assume you mean also to assume they are independent since you assert they come from a univariate normal distribution.

Note that by property of exponentials, $$e^{\frac{1}{2} \sum\limits_{i=1}^{n} X_i^2} = \prod\limits_{i=1}^{n} e^{\frac{1}{2}X_i^2}$$ Recall that if $X, Y$ are two independent random variables then, $$\mathbb{E}[h(X)h(Y)]= \mathbb{E}[h(X)]\mathbb{E}[h(Y)]$$ Which gives us that, $$\mathbb{E}\Big[e^{\frac{1}{2} \sum\limits_{i=1}^{n} X_i^2}\Big] = \mathbb{E}\Big[\prod\limits_{i=1}^{n} e^{\frac{1}{2}X_i^2}\Big] = \prod\limits_{i=1}^{n} \mathbb{E}\big[e^{\frac{1}{2} X_i^2}\big]$$

Therefore, we need only compute $\mathbb{E}\big[e^{\frac{1}{2} X_i^2}\big]$.

Note that if $Z \sim N(0,1)$ then $Z^2 \sim \chi_1^2$ and $\mathbb{E}[Z^2] = 1$. We can write $\frac{1}{2}X_i^2 = \big(\frac{X_i}{\sqrt{2}}\big)^2$.

Define $X^{'}_i = \frac{X_i}{\sqrt{2}}$ so that $X_i^{'} \sim N(0,\frac{\sigma^2}{ 2})$. Then, we can scale $X_i^{'}$ by $X_i^{'} = \frac{\sigma}{\sqrt{2}} Z_i$ where $Z_i \sim N(0,1)$. Therefore, $X_i^{'2} = \frac{\sigma^2}{2} Z_i^2$. We know $Z_i^2 \sim \chi_1^2$ and for $\frac{\sigma^2}{2} > 0$, $\frac{\sigma^2}{2} Z_i^2 \sim \Gamma\big(\frac{1}{2},\sigma^2\big)$ Chi-Square Dist.

Note that the moment generating function $M_X(t) := \mathbb{E}[e^{t X}]$. Since $X^{'2} \sim \Gamma\big(\frac{1}{2}, \sigma^2\big)$. We have that,

$$M_{X^{'2}}(t) = (1 - \sigma^2 t)^{-\frac{1}{2}}$$

Which can be looked up Here.

If I let $t = 1$, then,

$$M_{X^{'2}}(1) = \mathbb{E}[e^{X^{'2}}] = (1 - \sigma^2)^{-\frac{1}{2}}$$

This proves that, $$\mathbb{E}\Big[e^{\frac{1}{2} \sum\limits_{i=1}^{n} X_i^2}\Big] = \prod\limits_{i=1}^{n} (1 - \sigma^2)^{-\frac{1}{2}} = (1 - \sigma^2)^{-\frac{n}{2}}$$

Note that the Gamma MGF is only defined for $0 < \sigma^2 < 1$ for $t = 1$. In the derivation of the Gamma MGF, one realizes this restriction is necessary or else the integral is infinite. That is, this restriction gives you the range for $\sigma^2$ so that the expectation is finite.