2

Let $X$ be a normal random variable with mean $\mu$ and variance $\sigma^2$.

What is the $E\left [ \frac{1}{e^X-1} \right ]$ and $Var\left [ \frac{1}{e^X-1} \right ]$?

Any thoughts or references are welcome.

Jacob
  • 540
  • 2
  • 8
  • 1
    See the Law of the Unconscious Statistician – Firebug Jun 19 '23 at 15:44
  • 1
    It's not enough to know the mean and variance of X. You need to know its full pdf. If you the pdf for the random variable X, you can derive the pdf for the random variable given by f(X). Once you know the pdf for said new variable, you can calculate its mean, variance and any other quantity of interest – gazza89 Jun 19 '23 at 15:45
  • 2
    @gazza In this it's not necessary to know the distribution: it suffices that the PDF of $X$ at $0$ is nonzero and continuous there, because that implies $1/(e^X-1)$ has undefined expectation and infinite variance. – whuber Jun 19 '23 at 15:48
  • 1
    touché, that's a good challenge (albeit rather specific to this case). That said, did OP state anywhere that the original random variable has support at X=0 ? – gazza89 Jun 19 '23 at 15:59
  • 2
    @gazza All Normal variables that have densities have densities that are everywhere nonzero and continuous. That leaves, as exceptional cases, constants (where $\sigma^2=0$). – whuber Jun 19 '23 at 16:10
  • 2
    yup my bad, for some reason despite re-reading the post multiple times, my brain didn't process the word "normal" – gazza89 Jun 19 '23 at 16:26
  • 1
    @gazza89 I suspected as much, because that sort of thing has happened to me plenty of times ;-). – whuber Jun 20 '23 at 15:31

0 Answers0