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Given a random variable $X$ that is exponentially distributed, what is the variance of $\frac{1}{X}$, i.e. $\operatorname{Var}\frac{1}{X}?$

Since $\operatorname{Var} X = \frac{1}{\lambda^2}$, is it $\operatorname{Var}\frac{1}{X}=\lambda^2$?

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1 Answer

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If $X\sim \mbox{Exp}(\lambda)$, then $X^{-1}$ has distribution function $F(z)=e^{-\lambda/z}$. In particular it's expected value does not exist and hence neither does it's variance.

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