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If I define $Z = X+ Y$ where $X$ is discrete and $Y$ is continous random variable.

How can I show that $Z$ is a continuous random variable ?

I have known that $F_Z(z) = \sum F_Y(z-k). P(X=k)$ but don't know how to proceed from here. Or maybe there is another solution.

Thank you very much for your help!

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    $\begingroup$ There are conflicting definitions of continuous r.v's. Please give your definition of a continuous r.v. $\endgroup$ Commented Jan 21, 2022 at 9:02
  • $\begingroup$ $F_Z(z) = \sum F_Y(z-k). P(X=k)$ is wrong without independence. $\endgroup$ Commented Jan 21, 2022 at 9:18
  • $\begingroup$ @KaviRamaMurthy Are there any definitions of continuous random variable where $Z$ might not be a continuous random variable? $\endgroup$ Commented Jan 21, 2022 at 9:21

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Note: It is important to answer this question without assuming independence of $X$ and $Y$ so your approach is not valid.

One definition of a continuous r.v. is one whose CDF is a continuous function. With this definition here is a proof:

$P(Z=z) = \sum P(Y=z-k, X=k) \leq \sum P(Y=z-k)=\sum 0=0$.

Another definition: A r.v. is continuous if it has a density function.

With this definition here is a proof: Suppose $E$ has Lebesegue measure $0$. Then $P(Z \in E) =\sum P(X=k, Y\in E-k)\leq \sum P(Y\in E-k)$. But $E-k$ also has Lebesgue measure $0$ so we get $P (Z\in E)=0$. Hence, $X$ has a density function.

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  • $\begingroup$ Hi, thanks for your help! Do you use the independence assumption in your proof please ? $\endgroup$ Commented Jan 21, 2022 at 9:40
  • $\begingroup$ @InTheSearchForKnowledge No, the conclusions hold without independence. $\endgroup$ Commented Jan 21, 2022 at 9:43
  • $\begingroup$ Hi, thanks for your reply. In fact I think it is interesting and complete to keep your proof with 2 definitions. However, for the part you prove CDF is continuous, I think that we should prove CDF of $Z$ is left-continuous right ? Why the fact that $P(Z=z)=0$ is equivalent to the continuity of the CDF ? could you please add a little bit more detail in your proof in that part ? Thanks for your help!!! $\endgroup$ Commented Jan 21, 2022 at 9:48
  • $\begingroup$ @InTheSearchForKnowledge If $F$ is the CDF of a r.v. $Z$ then $P(Z=z)=F(z)-F(z-)$ (the jump in value of $F$ at the point $z$). So $F$ is continuous at $z$ iff $F(z)=F(z-)$ iff $P(Z=z)=0$. $\endgroup$ Commented Jan 21, 2022 at 9:51
  • $\begingroup$ Hi, can you explain a little more detail about the first inequality in your proof ? I think it should be an equality, i.e. $P(Z=z) = \sum P(Y=z-k, X=k)$, right ? $\endgroup$ Commented Jan 21, 2022 at 10:13

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