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In order to prove this Lemma in my course about Probability :

Let $X=(X_1,\dots ,X_p)$ be a gaussian random variable such that $\mathbb{E}[X_j]=0$ for all $j=1,\dots,p$. Then $2\mathbb{E}[\sum_{j=1}^pX_j^2] \le [\mathbb{E}[e^{-\sum_{j=1}^p X_j^2}]]^{-2}$

the teacher suppose that $X$ has a normal law i.e. $X\sim N_p(0,\Lambda)$ where $\Lambda$ is a diagonal matrix with $\lambda_{jj}=\sigma_j^2$, i.e $\Lambda=\left( \begin{array}{ccc} \lambda_{11} & \dots & & 0 \\ 0 & \lambda_{22} & \dots &0\\ \vdots & \vdots & \ddots & \vdots\\ 0&0&0&\lambda_{nn} \end{array} \right) = \left( \begin{array}{ccc} \sigma_1^2 & \dots & & 0 \\ 0 & \sigma_2^2 & \dots &0\\ \vdots & \vdots & \ddots & \vdots\\ 0&0&0&\sigma_p^2 \end{array} \right)$

Then he wrote, and this is where I get lost, $\mathbb{E}[e^{-X_j^2}]=(1+2\sigma_j^2)^{-1/2}$. Where does this result come from ?

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  • $\begingroup$ I've written a brute-force derivation for which you don't need to know anything beyond the value of the Gaussian integral. See my answer below. $\endgroup$ – Michael Hardy Jan 25 '13 at 2:27
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First, $X_j=\sigma_jZ$, where $Z$ is an $\mathcal{N}(0,1)$ variable, and hence $X_j^2=\sigma_j^2Z^2$. It is well-known that $Z^2\sim\chi^2(1)$, i.e. it follows a chi-squared distribution with $1$ degrees of freedom. Now, the expression $$ E\left[e^{-X_j^2}\right]=E\left[e^{-\sigma_j^2Z^2}\right] $$ is just the moment-generating function for $Z^2$ evaluated at $-\sigma_j^2$, i.e. $$ E[e^{-X_j^2}]=M_{Z^2}(-\sigma_j^2)=\left(1+2\sigma_j^2\right)^{-1/2} $$ according to e.g. the wikipedia article. Note that $-\sigma_j^2<\frac{1}{2}$ as it should in order for the moment-generating function to be well-defined.

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  • $\begingroup$ Thank you very much, I didn't know all that ! $\endgroup$ – Alan Simonin Jan 24 '13 at 22:10
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Stefan Hansen has written one answer; now let's try a different way: just doing it from scratch. Suppose the probability density function of the random variable $X$ is $$ x\mapsto \frac{1}{\sqrt{2\pi}\ \sigma} e^{-x^2/(2\sigma^2)}. $$ Then $$ \mathbb E e^{-X^2} = \frac{1}{\sqrt{2\pi}\ \sigma} \int_{-\infty}^\infty e^{-x^2} e^{-x^2/(2\sigma^2)} \, dx = \frac{1}{\sqrt{2\pi}\ \sigma} \int_{-\infty}^\infty \exp\left( \frac{-x^2}{2\left(\frac{\sigma^2}{1+2\sigma^2}\right)} \right) \, dx $$ $$ = \frac{1}{\sqrt{2\pi}\ \sigma} \int_{-\infty}^\infty \exp\left( \frac{-x^2}{2\tau^2} \right) \, dx = \frac{1}{\sqrt{2\pi}\ \sigma} \cdot\sqrt{2\pi} \ \tau = \frac\tau\sigma $$ (where, of course, $\tau^2=\dfrac{\sigma^2}{1+2\sigma^2}$) $$ = \frac{1}{\sqrt{1+2\sigma^2}}. $$

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  • $\begingroup$ Nice answer !! I learned a lot from what you did, thanks ! $\endgroup$ – Alan Simonin Jan 25 '13 at 2:51

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