Assume that $Y$ ~ $Exp(Ω)$. Find the cdf and pdf of $Z$ = |$Y$ - $δ$|. In order to solve this question so far, for $Y$, I am thinking about using the pdf equation for the exponential distribution i.e. $f(Y)$ = $Ωexp(-Ωx)$ for $x$ > $0$ and $0$ for $x$ < 0. But that is only $Y$, whereas furthermore $Z$ it is shifted by $δ$ to the left and is the absolute value function. Please help me on how to work out the cdf and pdf for this.
Asked
Active
Viewed 40 times
1 Answers
1
In my opinion, it is easier to start by the cdf. Let $z\in\mathbb{R}$.
$P(Z\leq z)=P(|Y-\delta |\leq z)=P(\delta - z\leq Y\leq \delta+z)=F_Y(\delta+z)-F_Y(\delta -z)$
where $F_Y(y)=(1-\exp(-\Omega y))\mathbf{1}_{y\geq 0}$ is the cdf of $Y$.
Augustin
- 8,446
-
Can you please also tell how to get the pdf? is it just differentiating the CDF? – Tera Peo Aug 26 '15 at 03:32
-
Yes, where it is differentiable. – Augustin Aug 26 '15 at 07:20