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.
Random Variables – How to Find the PDF of One Random Variable Given Another’s PDF
random variables
Best Answer
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.