Problem is this. Given probability density function of $ f(x)=1 , \phantom2 0<x<1 $ when variable $X$ is transformed into $ Y=-\log(X) $, I have to find the probability density function of $Y$ and its mean and variance. I managed to find $Y$'s pdf, which is $ f_{Y}(y)=e^{-y}$ and calculated the mean, but to calculate variance, I needed 2nd moment of $Y$ but $$ E\left[Y^{2}\right] = \int_{\infty}^{0} y^{2}e^{-y}dy = [-e^{-y}y^{2}]_{\infty}^{0} -\int_{\infty}^{0}-2ye^{-y}dy .$$ For the latter part, I can reuse $E[Y]$ that i got. The problem is the first part which is $[-e^{-y}y^{2}]_{\infty}^{0}$ I'm not sure whether this part converges or diverges, I heard you need analysis to do this but I haven't studied analysis at all... Can somebody tell me how to solve this?

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    $\begingroup$ Can you show that $y^2/2^y$ converges? $\endgroup$
    – Glen_b
    Commented Apr 25, 2016 at 8:57

1 Answer 1


Since the random variable $Y$ can take values from $0$ to $\infty$. The pdf $f_Y(y)=e^{-y}$ is an exponential distribution with parameter $\lambda=1$. You can get mean and variance from wikipedia

If you want to calculate the variance yourself, you may need to use integration by parts, and have a look here.

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    $\begingroup$ I don't understand your answer. I am already using integration by parts. The problem is in doing so, I can't figure out whether $[-e^{-y} y^{2}]_\infty^{0}$ converges or diverges $\endgroup$
    – Kyoon
    Commented Apr 25, 2016 at 12:44
  • $\begingroup$ It converges to $0$. Note that what you have is $[ - \frac{y^2}{e^{y}}]$. This is equal to $0$ for $y=0$. For the case that $y \rightarrow \infty$ it leads to a limit of the style $\frac{\infty}{\infty}$. To solve this limit you have two options: 1) exponential functions grow much faster than plynomial functions, hence limit is $0$ for $y\rightarrow \infty$, and 2) you can apply the Hopital rule link. $\endgroup$
    – PolBM
    Commented Apr 25, 2016 at 13:01

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