probability - Finding PDF involving absolute value

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: $0>w$, $0think 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

For the last part, I believe the bounds are $2\lambda

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

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.