Derive the probability density function of $u$

Question

Suppose that $\tau^2\sim\operatorname{Exp}(1/2)$ and suppose that $u$ is given by $$u = (1+\tau^2)^{-1}$$

Question: How do I derive the probability density function of $u$?

I've tried to do the following:

$$P(u\leq x) = P((1+\tau^2)^{-1}\leq x) = \ldots = 1 - P(\tau^2\leq 1/x - 1)$$ Since $\tau^2\sim\operatorname{Exp}(1/2)$ that means that $$\displaystyle P(u\leq x) = 1 - \int_0^{\frac{1}{x}-1}\frac{1}{2}e^{-\frac{1}{2}x}dx =\ldots=e^{-\frac{1}{2x}+\frac{1}{2}}$$ So $u$ has the cdf $\exp^{\dfrac{1}{2x} + \dfrac{1}{2}}$. To find the pdf I now need to take the derivative:

$$\dfrac{d}{dx}\exp^{-\frac{1}{2x} + \frac{1}{2}} = \dfrac{1}{4x^2}\exp^{-\frac{1}{2x} + \frac{1}{2}}$$

I'm just not sure if this is correct as it feels a bit weird..

Answer 1

Your cdf is correct. $\checkmark$ But the differentiation failed. Let $g(x)=-\frac1{2x}+\frac1{2}=-\frac{1}{2}x^{-1}+\frac1{2}$.

Then $g^{'}(x)=(-1)\cdot \left(-\frac{1}{2} \right)\cdot x^{-2} =\frac{1}{2x^2}$. Thus the pdf is

$$f_U(x)=\begin{cases}{} \frac{1}{2x^2}\cdot e^{\frac12(1-\frac1{x})} , \text{if} \ x\in (0,1) \\ 0, \text{elsewhere} \end{cases}$$