I have the pdf of $X$: $f(x) = 2x \ \ \ \text{if}\ \ \ 0<x<1 $ and $0$ elsewhere. Then I am asked to the get the expectation: $$ E\left(\frac{1}{X}\right)$$ to which I get the following: $$ E\left(\frac{1}{X}\right) = \int_0^1 \left(\frac{1}{x}\right)(2x)dx = 2 $$ From there I am asked to do the transformation of $ Y = \frac{1}{X} $ so I then use the transformation: $$f_Y(y) = f_X(g^{-1}(y)) \mid J \mid $$ to which I get that: $$f_Y(y) = \frac{2}{y^3} $$ The only problem is that the integral from 0 to 1 does not exist for this... This is where I am stuck and do not know what I did wrong. I figured this would work considering that $f_X(x)$ is an open interval which does not contain 0. Which is why I don't know what to do to change this any suggestions would be greatly appreciated

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    $\begingroup$ But $Y$ takes values from $1$ to $\infty$! $\endgroup$ – RideTheWavelet Oct 13 '17 at 3:41
  • $\begingroup$ I see my mistake now thank you $\endgroup$ – Robert Oct 13 '17 at 3:44

$Y$ has all it's probability in the interval $[1,+\infty)$, since $$P(Y\geq 1)=P\left(\tfrac1X \geq 1>0\right)=P(X\leq 1)=1.$$

So the function that you found is indeed a pdf cause its integral from $1$ to $+\infty$ adds up to $1$ (check it!).


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