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!
Best Answer
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.