Let $X \sim exp(\lambda)$. How is $\bar{X} = |X-\mu|$, where $\mu=E[X]$ distributed?

Question

Let $X \sim exp(\lambda)$, where $\lambda$ is a positive integer strictly greater than $1$. How is $\bar{X} = |X-\mu|$, where $\mu=E[X]$ distributed?

We know that: $$F_X(x) = 1-e^{-\lambda x}, x \geq 0 \text{ and } E[X] = \frac{1}{\lambda}.$$ Thus, $\mu = \frac{1}{\lambda}$. I was wondering if naturally $\bar{X} = |X-\mu|$ is distributed as $\bar{X} \sim \exp(|\lambda-u|)$, so that $$F_{\bar{X}}(x) = 1-e^{-|\lambda-u|x} \ \ x \geq 0 \text{ and } E[\bar{X}] = \frac{1}{|\lambda-u|}.$$

Answer 1

I think the natural way you can compute $F_{\bar X}(x)$ via the definition of distribution function. $$F_{\bar X}(x)=P(\bar X \leqslant x)=P(|X-\mu|\leqslant x)$$ $$=P(\mu -x \leqslant X\leqslant \mu+x)=F_{X}(x+1/\lambda)-F_X(-x+1/\lambda).$$ For the expected value, $$E|X-\mu|=\int_{0}^\infty |x-1/\lambda|\lambda e^{-\lambda x}dx=\int_0^{1/\lambda}(1/\lambda -x)\lambda e^{-\lambda x}dx+\int_{1/\lambda}^\infty (x-1/\lambda)\lambda e^{-\lambda x}dx.$$ Hope this helps.