Expectation of square of Gaussian random variable.

Question

Let $Z\sim \mathcal{N}(0,1)$. For $\lambda <1/2$, why does this hold?

$$ \mathbb{E}\left[\exp\left(\lambda\left(Z^2 - E[Z^2]\right)\right)\right] = \exp\left(-\frac{1}{2}\log(1-2\lambda) - \lambda\right) $$

Answer 1

In this case, $X=Z^2 \sim \chi^2_1$, so $$ f_X(x) = \frac{x^{-1/2}e^{-x/2}}{2^{1/2} \Gamma(1/2)} = \frac{x^{-1/2}e^{-x/2}}{\sqrt{2\pi}} $$ and $\mathbb{E}[x] = 1$, so your LHS is $$ \mathrm{LHS} = \int_0^\infty e^{\lambda (x-1)} \frac{x^{-1/2}e^{-x/2}}{\sqrt{2\pi}} dx = \frac{e^{-\lambda}}{\sqrt{2\pi}} \int_0^\infty x^{-1/2} e^{x(\lambda-1/2)}\ dx $$ can you take it from here?

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UPDATE

Let's substitute $a = 1/2 - \lambda,u = ax$ so $du = adx$ and since $\lambda < 1/2$, the $a>0$, so $u$ and $x$ have the same sign. Thus, as $x \to \infty$, we have $u \to \infty$ as well. Now $$ \mathrm{LHS} = \frac{e^{-\lambda}}{\sqrt{2\pi}} \int_0^\infty (u/a)^{-1/2}e^{-u} \frac{du}{a} $$ Could you express that as a Gamma($\Gamma$) function?