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Let $X_{\lambda}$ has exponential density $\lambda e^{-\lambda x}$. Then what is the nature of Var log($X_{\lambda}$) w.r.t $\lambda$ i.e. is it increasing, decreasin, etc on $\lambda$.

My approach: Without going into calucaltion, I evaluate log($X_{\lambda}$) which then has a term of the form $-\lambda$ log($\lambda$) which I know is decreasing and hence the answer is decreasing which matches up with the correct answer. But when I want to explicitly evaluate the term of E$[X]$, I am running into problem because it's of the form $\lambda log(\lambda)\int_{0}^{\infty}- x dx$. This integral would give me $\infty$ and same problem occurs with E$[X^2]$. So can anyone help?

ManishKumar Singh
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    Since $\lambda$ is the inverse of the scale of the exponential, the negative of its log is a shift parameter for the density of $\log(X)$; it has no effect on the variance whatever. – Glen_b Feb 02 '18 at 06:41
  • @Gleb_b could you elaborate your comment, perhaps in a form of an answer? – ManishKumar Singh Feb 02 '18 at 08:10

1 Answers1

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Without carrying out the variance derivation, I can say the following. If $$X\sim \text{Exp}(\lambda)$$ then $Y=\log X$ is as follows $$\begin{align} \text{Pr}(Y\le y)&=\text{Pr}(\log X\le y)\\ &=\text{Pr}(X\le e^y)\\ &=1-e^{-\lambda e^{y}},\quad\quad\quad -\infty<y<\infty \end{align}$$

We can check that this is a valid CDF if $$F(-\infty)=0$$ and $$F(\infty)=1$$ and $$\frac{d}{dy}F(y)=f(y)\ge0,\quad\forall y$$

See here for a discussion of your problem in more detail.

epp
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