While reading wikipedia article on Exponential distribution, I found the statement on scaling the random variable. Let $Exp(\lambda)$ be the distribution of the exponential random variable with paramter $\lambda > 0$ whose probability density function is $\lambda e^{-\lambda x}$ for $x\geq 0$.

The family of exponential distribution is closed under scaling by a positive factor; that is, if $X \sim Exp(\lambda)$ then $kX \sim Exp(\lambda/k)$ for $k>0$.

How can I prove it?

  • $\begingroup$ It is the family of exponential distributions that is closed under scaling. I wouldn't speak of simply a single distribution being closed under scaling. $\endgroup$ Nov 25, 2011 at 16:18

2 Answers 2


We give two solutions. The first is a formal calculation. The second is informal, but more informative.

Formal verification: Let $X$ have exponential distribution with parameter $\lambda$. Let $k$ be a positive constant, and let $Y=kX$. The cumulative distribution function $F_Y(y)$ of $Y$, for $y \ge 0$, is given by $$F_Y(y)=P(Y \le y)=P(X \le y/k)=F_X(y/k),$$ where $F_X$ is the cumulative distribution function of $X$.

Now there are several ways to find the density function $f_Y(y)$ of $Y$: (i) We could find $F_X(x)$ by integrating, substitute $y/k$ for $x$, and differentiate with respect to $y$; (ii) We could look up $F_X(x)$ and then proceed as in (i); (iii) Or else we can use the Fundamental Theorem of Calculus. Let $x=y/k$. By the Chain Rule, $$\frac{dF_X(y/k)}{dy}=\frac{dx}{dy}\frac{dF_X(x)}{dx}=\frac{1}{k}f_X(x)=\frac{1}{k}\lambda e^{-\lambda x}=\frac{\lambda}{k}e^{-\lambda y/k}=\lambda'e^{-\lambda' y},$$ where $\lambda'=\lambda/k$. That is what we wanted to show.

Note: The approach (iii) is in general the easiest, so that's what we used. It may look more difficult than approach (i), but that's because $\lambda e^{-\lambda x}$ is exceptionally easy to integrate. If we are working with the normal instead of the exponential, we can't even start on (i), while (iii) goes through just as easily as for the exponential.

Informal justification: The random variable $X$ might describe the lifetime, in days, of an atom of a certain radioactive substance $S$. Recall that the mean lifetime is $1/\lambda$.

On Planet $K$ far far away, the length of the day is one-third the length of an Earth day. So, on Planet $K$, and measuring time in $K$-days, substance $S$ will have mean lifetime $3/\lambda$ $K$-days. It follows that the lifetime $Y$ of an atom of substance $S$, as measured in $K$-days, has exponential distribution with parameter $\lambda/3$.

Similarly, if the length of the day on Planet $K$ is $1/k$ times an Earth day, then on Planet $K$, measuring time in $K$-days, substance $S$ has mean lifetime $k/\lambda$. So the lifetime $Y$ of an atom of $S$, measured in $K$-days, has exponential distribution with parameter $\lambda/k$.

  • $\begingroup$ nice! but how did you know you have to start from the CDF and calculate the PDF? how did you know that you cannot do the substitution in the PDF directly $\lambda e^{-\lambda cx}$ ? $\endgroup$
    – ihadanny
    May 30, 2020 at 8:16

$P(kX \ge t) = P(X \ge t/k) = \exp(-\lambda t / k)$. This is the survival function of a Exp$(\lambda / k)$ random variable. $t \ge 0$, of course; if $t < 0$ the probability is $1$.

Something that is often more useful in more general situations is the following. If $g$ is 1-1 and smooth and $Y = g(X)$ with $X$ being continuous with a density $f$ then the density of $Y$ is $f(g^{-1} (y)) |\frac d {dy} \ g^{-1}(y)|$. This generalizes to multivariate transformations, with the derivative term becoming the absolute value of the jacobian of $g$.


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