Let $X \sim N(0,1)$ and $Y \sim N(0,1)$ be independent RVs and let $f$ be their density function. I'd like to compute the expectation of the density ratio \begin{align} \mathbb{E}\left[\frac{f(X)}{f(Y)}\right] & = \mathbb{E} \left[e^{\frac{-X^2}{2}} e^{\frac{Y^2}{2}}\right] \\ & = \mathbb{E} \left[e^{\frac{-X^2}{2}} \right]\mathbb{E} \left[e^{\frac{Y^2}{2}}\right]. \hspace{5mm} (1) \end{align} My initial thought was that we would simply have $\mathbb{E}\left[\frac{f(X)}{f(Y)}\right] = 1$ whenever $X,Y$ are i.i.d. But my Monte Carlo simulations are not converging.

Seeing that, for example, $Y^2 \equiv Z \sim \chi^2_1$, I tried using the fact that $\mathbb{E}\left[g(Z)\right] = \int_{-\infty}^\infty g(z)f_Z(z) dz$ (Law of the Unconscious Statistician) with, $g(Z) = e^\frac{Z} {2}$ and $f_Z(z)$ being the Chi-squared density with 1 df, to evaluate the factors in $(1)$. However, I getting the integral: \begin{align} \mathbb{E}\left[e^{\frac{Z}{2}}\right] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{1}{\sqrt{z}}dz \end{align}

Am I going in the right direction?


  • 3
    $\begingroup$ Hint: By the law of the unconscious statistician, $$E[e^{Y^2/2}] = \int_{-\infty}^{\infty} e^{y^2/2}\frac{1}{\sqrt{2\pi}}e^{-y^2/2}\,\mathrm dy = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}\,\mathrm dy = \cdots$$ $\endgroup$ – Dilip Sarwate Feb 11 '14 at 0:05
  • $\begingroup$ Wow, I didn't know that even had a name! Very interesting. $\endgroup$ – user31668 Feb 11 '14 at 1:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.