distribution of $Y= \exp(-X^2)$

Question

I have a question about distribution functions. Assume that $X \sim N(\mu, \sigma^2)$. I want to compute the mean and variance of $Y = \exp(-\frac{X^2}{2h^2})$ analytically. What is the solution?

Answer 1

The mean of $e^{-X^2}$ is the integral of $e^{-x^2}$ weighted by the density function of $X$ (which is another exponential function). For an analytic expression, notice that the integral now contains a product of two exponentials. Combine the product of exponentials into an exponential of a sum and re-write the resulting exponent as a single negative square (plus a constant).

The variance is the mean of the square of $e^{-X^2}$ minus the square of the mean, so its analytic computation is very similar.

Answer 2

Mean:

You integrate $y=e^{-x^2}$ times a Gaussian pdf. Hence, you get an exponential with exponent

$$-x^2-\frac{(x-\mu)^2}{2\sigma^2}=-\frac{2\sigma^2x^2+x^2-2\mu x+\mu^2}{2\sigma^2}.$$

By completing the square,

$$2\sigma^2x^2+x^2-2\mu x+\mu^2=(2\sigma^2+1)(x-\frac\mu{2\sigma^2+1})^2-\frac{\mu^2}{2\sigma^2+1}.$$

This yields the pdf of a Gaussian of $\sigma'=\frac\sigma{\sqrt{2\sigma^2+1}}$, times the coefficient $e^{\mu^2/(2\sigma^2+1)2\sigma^2}$. The integral is still $1$, provided you adjust for the change of $\sigma$, with the factor $\sigma/\sigma'$.

Hence

$$\overline{y}=\frac1{\sqrt{2\sigma^2+1}}\,e^{\mu^2/(2\sigma^2+1)2\sigma^2}.$$

Variance:

Repeat this procedure for $y^2=e^{-2x^2}$, to get

$$\overline{y^2}=\frac1{\sqrt{4\sigma^2+2}}\,e^{\mu^2/(4\sigma^2+1)2\sigma^2}.$$

The final value is

$$\sigma^2_y=\overline{y^2}-\overline y^2.$$