If $X\sim\mathcal{N}(\mu,\sigma)$, what's the CDF of $X u(X)$?

Question

Let $X \sim \mathcal{N}(\mu, \sigma^2)$ be a gaussian random variable. What is the CDF (Cumulative Distribution Function) and (if it exists) the PDF (Probability Density Function) of the variable $$ Y = X u(X) $$ where $u(\cdot)$ is the unit step function, i.e. $$ u(x) = \begin{cases} 1 & x \geq 0 \\ 0 & x < 0 \end{cases} $$ In the following, I will denote the CDF of $Y$ by $F_Y(\cdot)$ and the standard gaussian CDF by $\Phi(\cdot)$.

I tried $$ F_Y(t) = P(Y < t) = P(Xu(X) < t) $$ and now I'm not sure what to do, as $u(X)$ depends on whether $X<0$ or not.

score 3 · Accepted Answer · answered Mar 24 '17 at 12:03

3

For $t \geq 0$, we have \begin{align} F_Y(t) &= \Pr(Xu(X) \leq t) \\ &= \Pr(Xu(X) \leq t, X < 0) + \Pr(Xu(X)\leq t, X \geq 0) \\ &= \Pr(X < 0) + \Pr(0 \leq X \leq t)\\ &= F_X(t) \end{align} Therefore, \begin{align} F_Y(t) = \begin{cases} F_X(t) & \text{if } t \geq 0 \\ 0 & \text{if } t < 0 \end{cases} \end{align} and it does not have a PDF.

answered Mar 24 '17 at 12:03

PSPACEhard

10,283

Thank you, only one thing is not clear: why doesn't it have a PDF? The CDF is derivable everywhere except in $t=0$, which is a zero-measure set. – PseudoRandom Mar 24 '17 at 13:28
@PseudoRandom $\Pr(Y = 0) \neq 0$. – PSPACEhard Mar 24 '17 at 13:57

If $X\sim\mathcal{N}(\mu,\sigma)$, what's the CDF of $X u(X)$?

1 Answers1