# How to find the PDF of one random variable when the PDF of another random variable and the relationship between the two random variables are known?

The probability density function of random variable X is given as $f_x(x) = \lambda e^ {-\lambda x} , x \ge 0.$ A new random variable $Y = e^ {-\lambda X}$ is formed. Find the PDF of Y.

There are shortcuts, but we will use a basic method. The idea is to find the cumulative distribution function of $Y$, and then differentiate to find the density function. We have $$F_Y(y)=\Pr(Y\le y)=\Pr(e^{-\lambda X}\le y)=\Pr(-\lambda X \le \log y).$$ (We took the logarithm: this preserves inequalities.) Thus $$F_Y(y)=\Pr\left(X\ge -\frac{\log y}{\lambda}\right).$$ We know that $\Pr(X\ge t)=e^{-\lambda t}$, if $t$ is positive. If $t$ is negative, the probability is $1$.

Substitute for $t$. There is dramatic simplification. The $\lambda$'s cancel, and we get $e^{\log y}$, that is $y$. But note this is correct only when $-\log y$ is $\ge 0$, that is, when $y\le 1$. If $y\gt 1$, $F_Y(y)=1$.

Finally, differentiate. The density function is $1$ on the interval $(0,1)$, and $0$ elsewhere.

There is a nice "change of variables" formula for doing this. If $g$ is a monotonic function, then

$$f_Y(y)=\left\vert\frac{d}{dy}(g^{-1}(y))\right\vert f_X(g^{-1}(y))$$

So, let $g(x)=e^{-\lambda x}$; this is a monotonic function, so the formula applies. $g^{-1}(y)=-\ln(y)/\lambda$, and so

$$\left\vert\frac{d}{dy}g^{-1}(y)\right\vert=\frac{1}{\lambda y}$$

putting the pieces together, we get:

$$f_Y(y)=\frac{1}{\lambda y}\lambda \exp[-\lambda g^{-1}(y)]=\frac{1}{y}\exp[\ln(y)]=1$$

Notice that since $x\geq 0$, $0<y\leq 1$ (assuming $\lambda>0$) so this distribution makes sense.