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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 $XY$?

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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.

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Let $X$ be a discrete random variable with probability mass function $p_X : \mathcal{X} \to [0,1]$, where $\mathcal{X}$ is a discrete set (possibly countably infinite). Random variable $X$ can be thought of as a continuous random variable with the following probability density function

$$f_X (x) = \sum_{x_k \in \mathcal{X}} p_X (x_k) \, \delta (x - x_k)$$

where $\delta$ is the Dirac delta function.

If $Y$ is a continuous random variable, then $Z := X+Y$ is a hybrid random variable. As we know the probability density functions of $X$ and $Y$, we can compute the probability density function of $Z$. Assuming that $X$ and $Y$ are independent, the probability density function of $Z$ is given by the convolution of the probability density functions $f_X$ and $f_Y$

$$f_Z (z) = \sum_{x_k \in \mathcal{X}} p_X (x_k) \, f_Y (z - x_k)$$

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  • $\begingroup$ Why the downvote? $\endgroup$ – Rodrigo de Azevedo Jun 8 '16 at 12:56
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    $\begingroup$ Yeah, I'm also curious about the downvote $\endgroup$ – Yair Daon Jun 8 '16 at 13:58
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    $\begingroup$ @Yair I haven't downvoted it, but it looks like a misleading and incomplete answer. Simply writing a distribution in terms of a Dirac delta doesn't make it continuous! This answer is also limited in that (a) it does not consider general discrete distributions, which may have a countable infinity of atoms and (b) it implicitly assumes $X$ and $Y$ are independent. $\endgroup$ – whuber Jun 8 '16 at 14:07
  • $\begingroup$ @whuber I agree with (b). However, it is said that a discrete RV "can be thought of as...", so I think it adds an interesting view. $\endgroup$ – Yair Daon Jun 8 '16 at 14:15
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    $\begingroup$ This is why I wrote that your answer is misleading. Because the question concerns the distinction between discrete and continuous distributions--and that distinction is a matter of mathematical definition, not "taste"--your efforts to confuse the two are likely to be less than helpful. $\endgroup$ – whuber Jun 8 '16 at 14:24
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This answer assumes that $X$ and $Y$ are independent. Here is a solution which does not need that assumption.

Edit: I am assuming that "continuous" means "having a pdf." If continuous is instead intended to mean atomless, the proof is similar; simply replace "Lebesgue null set" with "singleton set" in what follows.


Let the support of $X$ be the countable set $\{x_1,x_2,x_3\dots\}$. I will use

Lemma: A random variable $Z$ is continuous if and only $P(Z\in E)=0$ for all Borel measurable sets $E$ with Lebesgue measure zero.

Proof: Use the Lebesgue-Radon-Nikodym theorem. $ \square$

To prove $X+Y$ is continuous, take any null set $E$, and note that $$ P(X+Y\in E)=\sum_k P(\{Y+x_k\in E\}\cap \{X=x_k\})\le \sum_k P(Y+x_k\in E) $$ But $Y+x_k\in E$ if and only if $Y\in E-x_k$. The shifted set $E-x_k$ is still Lebesgue null. Since $Y$ is continuous, this means $P(Y+x_k\in E)=0$, so the above summation is zero, proving $X+Y$ is continuous.

For the question of products, the same logic applies as long as $P(X=0)=0$. If $P(X=0)=1$, then $XY$ is discrete with $P(XY=0)=1$. Otherwise, $XY$ is a nontrivial mixture.

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