2
$\begingroup$

Given a random variable $X$ with probability density function $$f(x)=\frac5{x^6}\mathbf 1_{\{x>1\}}$$ I am trying to find the mean of $1/X$. After making the transformation I find that $f(1/x)=5x^4$. So far I've come up with only whole numbers but the answer is supposed to be a fraction.

Any ideas?

$\endgroup$
  • $\begingroup$ You mean that $f(x)=5/x^6$ is the pdf? $\endgroup$ – Jimmy R. Feb 21 '16 at 10:12
  • $\begingroup$ Your transformation is fine, but what happened to the indicator function? $\endgroup$ – Will Orrick Feb 21 '16 at 10:29
3
$\begingroup$

$$ <1/x>=\int_1^\infty dx (1/x)5/x^6=5\int_1^\infty dx\frac{1}{x^7}=\frac{5}{6}\ . $$

$\endgroup$
2
$\begingroup$

\begin{align}\Bbb E\left[\frac1X\right]&=\int_{-\infty}^{+\infty}\frac1xf(x)d x=\int_{-\infty}^{+\infty}\frac1x\cdot \frac5{x^6}\mathbf 1_{\{x>1\}}dx=\int_{1}^{+\infty}\frac{5}{x^7}dx\\[0.2cm]&=\left[\frac5{-6x^6}\right]^{+\infty}_1=0-\frac{5}{-6}=\frac56\end{align}

$\endgroup$

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.