Distribution of the Inverse of a Random Variable

Question

I am trying to figure out how to find the distribution of the inverse of a random variable. Say, $Y=X^{-1}$ where X can take negative values.

The two ways I know to find the distribution of a random variable Y=g(X) are:

1. $F_y(y) = P(Y < y) = P(X^{-1} < y) = f(x) = \left\{ \begin{array}{lr} P(1/y < X)&: X> 0,\; y>0\\ P(1/y > X)&: X>0,\; y<0 \\ P(1/y < X)&: X<0,\; y<0 \\ P(1/y > X)&: X<0,\; y>0 \end{array} \right. \;\;\;\;\;\;\;\;\;\;$ Then take derivative.
2. Transformation theorem: $f_Y(y) = f_x(g^{-1}(y))|\frac{\partial g^{-1}(y)}{\partial y}|$ when g is monotone.

If I do it way #1, I have indicator functions all over the place which are not differentable.

I can't do way #2 because my function is not monotone on the entire real line.

Answer 1

Go back to the definitions...

Assume that $P[X=0]=0$, otherwise $Y$ is not well defined.

• For every $y\lt0$, $[Y\leqslant y]=[1/y\leqslant X\lt0]$ hence $F_Y(y)=F_X(0)-F_X((1/y)^-)$.
• For every $y\gt0$, $[Y\gt y]=[0\lt X\lt1/y]$ hence $F_Y(y)=1-F_X((1/y)^-)+F_X(0)$.
• Finally, $[Y\leqslant0]=[Y\lt0]=[X\lt0]=[X\leqslant0]$ up to null events hence $F_Y(0)=F_X(0)$.