1
$\begingroup$

Trying to find the CDF and PDF for $Y=2+|X|$ where $X$ is a standard normal random variable.

Now if am not mistaken then I need to use for the CDF $\int_{-\infty}^X\frac{1}{\sqrt{2\pi}}e^{-t^2/2}\,dt$ and the PDF is just basically the derivative of that in this case.

My problem is I am not sure how do I deal exactly with the absolute value here. So, I just want to check if my steps are correct below?

My attempt for CDF:

$$P(Y\leq t) = P(2+|X|\leq t)=P(|X|\leq t-2)$$ $$P(X \leq t-2)-P(X \leq 2-t)= \Phi(X)=\int_{2-t}^{t-2}\frac{1}{\sqrt{2\pi}}e^{-\frac{\tau^2}{2}}\,d\tau =$$ $$=\frac{1}{\sqrt{2\pi}}(-2e^\frac{2-t}{2}+2e^\frac{t-2}{2})$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.