Let X be a U[0,1] random variable, and let Y = 1/X. Find the density f(y) of Y.

Question

Attached is my solution for this problem.

Can someone explains to me why do we multiply f(x) with absolute value of dy/dx to get the f(y)?

Also, what are the steps to ace this kind of question? First, I have to find X in terms of Y. And then?

Thanks in advance! :)

Answer 1

It is basically the chain rule.   The probability density function (for an absolutely continuous random variable) is the unsigned derivative of its cummulative distribution function.   So when $X=g(Y)$ where $g:Y\mapsto X$ is bijective over the supports, this happens:

$$\begin{align}f_Y(y) &= \left\lvert \dfrac{\mathsf d F_Y(y)}{\mathsf d y~~~~~~~~} \right\rvert \\ &= \left\lvert \dfrac{\mathsf d F_X(g(y))}{\mathsf d y~~~~~~~~~~~~~} \right\rvert \\ &= \left\lvert \dfrac{\mathsf d g(y)}{\mathsf d y~~~~~} \cdot\dfrac{\mathsf d F_X(g(y))}{\mathsf d g(y)~~~~~~~~~}\right\rvert \\ &= \left\lvert \dfrac{\mathsf d g(y)}{\mathsf d y~~~~~} \right\rvert f_X(g(y)) \end{align}$$

Where $F_X, F_Y, f_X, f_Y$ are the CDF and pdf of the denoted random variables.

In this case $g(y)=y^{-1}$, so $f_Y(y)= \lvert -y^{-2}\rvert~f_X(y^{-1})$, and because $X$ is uniform over $[0;1]$ that becomes $f_Y(y) = y^{-2}\mathbf 1_{y\in [1;\infty)}$

Answer 2

It will become clear if you start from the CDF. Write \begin{align*} F_Y(y) \doteq P(Y \leq x) = P(X \geq 1/x) = 1 - 1/x. \end{align*} Now differentiate to find the density.