Finding distribution of $Y=-1/\log X$

Question

Having some difficulty with this particular question: Let the pdf of $X$ be $f(x)=\lambda*x^{\lambda-1}$, $x=[0,1]$

Let $Y=-1/\log(X)$

Solving for inverse of $Y$ we get

$ \log(X)=-1/Y$ and $ X=e^{-1/Y}$

Now, $dY/dX=1/(x\log^2(X))$ $>0$ for $x$

Now for the inverse function $dX/dY=-e^{-1/Y}$

After substituting into the general formula for transformation of variables we receive

$f(y)=\lambda*e^{-1/y*(\lambda-1)}*e^{-1/y}=\lambda*e^{-\lambda/y}$, $y=[0,1]$

However the answer should be $f(y)=\lambda/y^2*e^{-\lambda/y}$ for $y=[0,\infty]$

I have no idea at all where the $1/y^2$ comes from. Would be very grateful to receive some help.

Kind regards,

Answer 1

1. $\frac{dX}{dY}=X\log^2(X)=e^{-1/Y}\left(\frac{-1}Y\right)^2=\frac{e^{-1/Y}}{Y^2}$

Or you can do $X=e^{-1/Y}\implies \frac{dX}{dY}=\frac1{Y^2}e^{-1/Y}$.

The exponent is $-1/Y,$ not $-Y$.

2. $Y=-1/\log X$ ranges from $0^+\to\infty$ as $X$ ranges from $0^+\to1^-$.

Answer 2

I have no idea at all where the $1/y^2$ comes from.

Using fundamental transformation theorem

$$f_Y(y)=f_X[g^{-1}(y)]\Bigg|\frac{d}{dy}g^{-1}(y)\Bigg|$$

... you need also to calculate

$$\Bigg|\frac{d}{dy}g^{-1}(y)\Bigg|=e^{-\frac{1}{y}}\frac{1}{y^2}$$