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.


2 Answers 2


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.

  • $\begingroup$ I know this is a decade old, but this is such an interesting transformation! Can you please supply a proof of this?(preferably one that does not go back to CDFs) $\endgroup$
    – DatBoi
    Oct 8, 2023 at 17:22

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