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X conforms to a normal distribution with mean a and variance b.

Every time I randomly generates a value of X first, say x0. If x0 is greater than preset threshold Q, then I generates Y from another normal distribution with mean c and variance d. If x0 is less or equal to threshold Q, then stop.

Let Z=X+Y

I did below calculation, but that only gives me the expectation. How can I derive the distribution of Z?

E(Z)=E(X)+E(Y)=a+E(Y|X>Q)+E(Y|X<=Q)=a+c+0=a+c

Var(Z)=?

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    What is $Y$ if $X$ is less or equal to threshold $Q$? Does not defined at all? – 2017-02-14
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    @NCh sorry didn't make it clear. In that case Y = 0. – 2017-02-14
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    Use Law of total expectation for $E(Y)$. Both conditional expectations need to be multiplied by probabilities of $X>Q$ (respectively $X\leq Q$). See also https://en.wikipedia.org/wiki/Law_of_total_variance – 2017-02-14
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    @nch thanks will look into that. That only gives me the variance right? It seems Z doesn't have a typical distribution – 2017-02-14
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    The alternative simple way is to take $Y\sim N(c,d)$ independent of $X\sim N(a,b$, and $Z=X+Y\cdot I(X\geq Q)$, where $I(X\geq Q)$ means indicator function which is equal to 1 if $X\geq Q$ or 0 else. Then calculate $Var(Z)$ using properties of variance or simply through the second moment. – 2017-02-14

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