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I am looking for some help with how to derive the distribution of a random variable that uses the max function. What I have is the following information:

$$x_1\sim N(\mu_1,\sigma_1^2)$$ $$x_2\sim N(\mu_2,\sigma_2^2)$$ where $x_1$ and $x_2$ are independent and $$y=\max\{0,x_1+x_2\}$$

And so I would like to figure out what the distribution of $y$ is. I know that if, say, $z=x_1+x_2$ then $z\sim N(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2)$ but the $\max$ part is what is throwing me off. My guess is that it would be a truncated normal distribution (truncated at 0) but I am not 100% sure about that? Any suggestion (or solutions) are greatly appreciated!

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    $\begingroup$ An easy way to check your guess would be to imagine sampling from y. You would first sample $z\sim\textsf{Norm}(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2)$, then check to see if $z\leq 0$. If it is, you return $0$, otherwise you return $z$. So you can imagine this shifts all of the mass of the distribution left of $0$ to $0$. Now consider a truncated normal. This spreads out all of the mass left of $0$ $evenly$ over the portion of the distribution right of $0$. Thus it's not a truncated normal. $\endgroup$
    – aleshing
    Commented Jan 11, 2018 at 8:29
  • $\begingroup$ Answer is contained here: stats.stackexchange.com/questions/82495/… $\endgroup$
    – kjetil b halvorsen ♦
    Commented Jan 11, 2018 at 9:38

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