If $X$ is a discrete and $Y$ is a continuous random variable then what can we say about the distribution of $X+Y$? Is it continuous or is it mixed?

What about the product $XY$?

**Answer**

Suppose $X$ assumes values $k \in K$ with discrete distribution $(p_k)_{k \in K}$, where $K$ is a countable set, and $Y$ assumes values in $\mathbb R$ with density $f_Y$ and CDF $F_Y$.

Let $Z = X + Y$. We have

$$ \mathbb P( Z \leq z) = \mathbb P(X + Y \leq z) = \sum_{k \in K} \mathbb P(Y \leq z – X \mid X = k) \mathbb P(X = k) = \sum_{k \in K} F_Y(z-k) p_k,$$ which can be differentiated to obtain a density function for $Z$ given by

$$ f_Z(z) = \sum_{k \in K} f_Y(z-k) p_k.$$

Now let $R = X Y$ and assume $p_0 = 0$. Then

$$ \mathbb P(R \leq r) = \mathbb P(X Y \leq r) = \sum_{k \in K} \mathbb P(Y \leq r/X) \mathbb P(X= k) = \sum_{k \in K} F_Y(r/k) p_k,$$

which again can be differentiated to obtain a density function.

However if $p_0 > 0$, then $\mathbb P(X Y = 0) \geq \mathbb P(X = 0) = p_0 > 0$, which shows that in this case $XY$ has an atom at 0.

**Attribution***Source : Link , Question Author : user666 , Answer Author : Joris Bierkens*