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?

  • 2
    $\begingroup$ Can you show that $y^2/2^y$ converges? $\endgroup$
    – Glen_b
    Apr 25 '16 at 8:57

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.

  • 1
    $\begingroup$ 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
    Apr 25 '16 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
    Apr 25 '16 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.