Random variable X has exponential density $\lambda e^{-\lambda x}$. Then comment on $Var(logX)$

Question

Let $\lambda>0$, Random variable X has exponential density $\lambda e^{-\lambda x}$ on $[0,\infty)$. Then $Var(logX)$:

1. is strictly increasing in $\lambda$
2. is strictly decreasing in $\lambda$
3. does not depend on $\lambda$
4. first increases and then decreases in $\lambda$

My attempt:

If $X \sim \exp (\lambda)$, then what does $\log(X)$ follow, or what is its moment generating function? If I can find that, then I can find the expression for $Var(X)$, hence find $\frac{dVar(X)}{d\lambda}$ and the second derivative and decide if it is increasing or depressing.

I am using the solution given in Let $X$ have the exponential distribution with $\lambda=2$ Find the density function of the random variable $Y=\ln(X)$

Then $Y=log(X)$ has the pdf $f(y)=\lambda e^{-\lambda e^y+y}$

Now, $Var(Y)=E(Y^2)-[E(Y)]^2$

$E(Y)=\int_{-\infty}^{\infty} y \lambda e^{-\lambda e^y+y} dy$ and $E(Y^2)=\int_{-\infty}^{\infty} y^2 \lambda e^{-\lambda e^y+y} dy$.

I am having trouble integrating these. Is there any other approach?

Answer 1

Another approach.

---

Observe that $U:=\lambda X$ has exponential distribution with parameter $1$.

Based on that we find: $$\mathsf{Var}(\ln X)=\mathsf{Var}\left(\ln\left(\frac1{\lambda}U\right)\right)=\mathsf{Var}(\ln U-\ln\lambda)=\mathsf{Var}(\ln U)\tag1$$and we are released from the annoying parameter now.

It is enough now to solve the following integrals:$$\mathbb E[\ln U]=\int_0^{\infty}e^{-u}\ln u\;du$$and:$$\mathbb E[\ln U]^2=\int_0^{\infty}e^{-u}(\ln u)^2\;du$$

I hope that they are easier to find.