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Let $g(x)$ be an integrable function, i.e. $\int_{-\infty}^{\infty}|g(x)|dx < \infty$. Let $f(x)$ be a probability density function with support $S_X$. Show that the variance of $g(X)/f(X)$, where $X ∼ f(x)$, is finite if $|g(x)/f(x)| < C < \infty$ for some $C \in R$ and all $x ∈ S_X$

would using taylor series work here?

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    Wait, you know that the random variable $Y=g(X)/f(X)$ is such that $|Y|$E(Y^2)$ is finite? – 2017-02-08

1 Answers 1

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For the mean of $\frac{g(X)}{f(X)}$ we have

$$E\left[\frac{g(X)}{f(X)}\right]=\int_{S_X}\frac{g(x)}{f(x)}f(x)\ dx\le \int_{S_X}\mid g(x)\mid \ dx<\infty.$$

For the second momentum of the same random variable we have

$$E\left[\frac{g^2(X)}{f^2(X)}\right]=\int_{S_X}\frac{g(x)}{f(x)}g(x)\ dx\le C\int_{S_X}\mid g(x) \mid \ dx< \infty.$$

And, finally, for the variance

$$\operatorname {Var}\left(\frac{g^2(X)}{f^2(X)}\right)= E\left[\frac{g^2(X)}{f^2(X)}\right]-E^2\left[\frac{g(X)}{f(X)}\right]< \infty,$$