Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to solve the following question: Given an exponential R.V. X with rate parameter $\lambda > 0$, find the PDF of $V=|X-\lambda|$.

In order to find the PDF, I would like to use the CDF method (i.e. finding the CDF and then taking the derivative to obtain the PDF). I realize this function is not one to one on the range between 0 and $2\lambda$, so the CDF should be broken into three parts: $0>w$, $0<w<2\lambda$ and $2\lambda<w$. For $0<w<2\lambda$, $Pr(|X-\lambda| < w)$ = $Pr(-w+\lambda < x < w+\lambda)$, and then I think I want to do the double integral of $\int_0^\infty\int_{-w+\lambda}^{w+\lambda}\lambda e^{-\lambda x}dx$, with the function being integrated there being the exponential distribution pdf. After getting this, I should be able to take the derivative and get the PDF for $0<w<2\lambda$, right? (For the record, I get $\lambda e^{-\lambda ^2 - \lambda w} - \lambda e^{-\lambda ^2+\lambda w}$ when i try this; its possible i'm wrong though!)

For the last part, I believe the bounds are $2\lambda <w<\infty$, so i want to do this integral: $\int_0^\infty\int_{2\lambda}^{\infty}\lambda e^{-\lambda x}dx$, but i'm not really sure about this at all.

I understand there are probably more efficient ways of solving this, but I'm specifically trying to do it using the CDF method!

share|cite|improve this question
up vote 3 down vote accepted

You have done most of the analysis, so I will be to a great extent repeating what you know. We want to find an expression for $\Pr(V\le w)$.

In general, $V\le w$ iff $|X-\lambda| \le w$ iff $X-\lambda\le w$ and $X-\lambda\ge -w$, that is, iff $$\lambda-w \le X\le \lambda+w.$$

There are three cases to consider, (i) $w\le 0$; (ii) $0\lt w\le \lambda$; and (ii) $w \gt \lambda$.

Case (i): This is trivial: if $w\le 0$ then $\Pr(V\le w)=0$.

Case (ii): We want $\Pr(X\le \lambda+w)-\Pr(X\lt \lambda -w)$. This is $$(1-e^{-\lambda(\lambda+w)})-(1-e^{-\lambda(\lambda-w)}).$$ There is some immediate simplification, to $e^{-\lambda(\lambda-w)}-e^{-\lambda(\lambda+w)}$, and there are various alternate ways to rewrite things, by introducing the hyperbolic sine.

Case (iii): This one is easier. We simply want $\Pr(X\le \lambda+w)$. For $w\ge -lambda$, this is $$1-e^{-\lambda(\lambda+w)}.$$

We could have set up the calculations using integrals, but since we already know that $F_X(x)=1-e^{-\lambda x}$ (when $x\gt 0$) there is no need to do that.

Now that we have the cdf of $V$, it is straightforward to find the density. For $w\le 0$, we have $f_V(w)=0$. For $0\lt w\lt \lambda$, we have $f_V(w)=\lambda e^{-\lambda(\lambda-w)}+\lambda e^{-\lambda(\lambda+w)}$. Finally, for $w\gt \lambda$ we have $f_V(w)=\lambda e^{-\lambda(\lambda+w)}$.

Remark: Suppose that we did not have a nice expression for the cdf of $X$. That happens, for example, with the normal, and a number of other distributions. We could still find the density function by setting up our probabilities as integrals, and differentiating under the integral sign.

share|cite|improve this answer
Thank you so much! I really appreciate the help - you made everything much clearer! – JenK Nov 29 '12 at 5:51

You are on the right track, but there are a few things that need corrected. First, you do not need to be taking double integrals here. Lets look at each interval one at a time. Let $W=|X-\lambda|$. To find the CDF of $W$ we divide the positive real line into three segments. To find the right intervals, it helps to draw a graph of the function $f(X)=|X-\lambda|$.

1) Since $|X-\lambda|\geq0$ it is clear that $P(|X-\lambda|\leq w)=0$ for $w<0$.

2) For $0\leq w< \lambda$, $P(|X-\lambda|\leq w)=P(-w+\lambda\leq X\leq w+\lambda)$ as you correctly stated. But this probability is given by the integral

$$ \int_{-w+\lambda}^{w+\lambda}\lambda e^{-\lambda x}dx=e^{\lambda w-\lambda^2}-e^{-\lambda w - w\lambda^2} $$

3) If $\lambda\leq w$, then $-w+\lambda\leq 0$ and $P(-w+\lambda\leq X\leq w+\lambda)=P(0\leq X\leq w+\lambda)$ since $X\geq 0$. This probability is given by the integral

$$ \int_{0}^{w+\lambda}\lambda e^{-\lambda x}dx=1-e^{-\lambda w - \lambda^2} $$

Edit: Now we can correctly combine cases 1-3 to get the CDF, which we can differentiate to get the PDF.

$$ F_{W}(w)= \begin{cases} 0 &\mbox{if } w<0 \\e^{\lambda w-\lambda^2}-e^{-\lambda w - w\lambda^2} &\mbox{if } 0\leq w < \lambda \\1-e^{-\lambda w - \lambda^2} &\mbox{if } \lambda\leq w \end{cases} $$

share|cite|improve this answer
I think I lost you at the end, where you get your result for $\lambda < w $... should it not be the result you got in step 3.)? The way it is currently set up, if we consider $\lambda = 1$ and $w = \infty$, we get $2-e^2-0$, which is not one! – JenK Nov 29 '12 at 5:50
Oops! Fixed. Thinking about an issue that doesn't apply here. Thanks @JenK. – caburke Nov 29 '12 at 15:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.