Originally I had asked whether two continuous random variables can sum to a discrete random variable. More specifically, I am wonder whether, if we Let $X_n \sim \text{iid } N(0,\sigma_x^2)$ and $Y_n \sim \text{iid } N(0,\sigma_y^2)$, $cov(Y_n,Y_{n+1}) = 0 \forall n \in \{1,2,\dots, \infty\}$. and define $$ Z_n = \alpha + \beta Z_{n-1} + X_n $$ and lastly, let $$ V_n = Z_n + Y_n $$ Does $V_n$ need to be a continuous random variable? Also, assume that $Z_0 =1$.

I say yes, because intuitively it makes sense, and if $P_X, P_Y$ both absolutely continuous w.r.t. $\mu$, then when $\mu(A) =0$, some set $A)$, $P_x = P_y = 0 \implies P_x + P_y = 0$.

That argument assumes though that I can add $P_x$ and $P_y$, and also that the probability measure induced by $X+Y$ is $P_x +P_y$, which may in itself require something like independence to be true.

Also, I don't see a way to rule out the possibility that $Y_t$ isn't somehow related to $X_n$ or $Z_n$ such that $Z_n + Y_n$ is discrete, even if the probability of it being so is very small.



Apologies if I'm misunderstanding your question, but isn't a counter-example just say, $Y=-X$, where $X$ is some continuous random variable? Then the sum is just zero. Another example could be something like $X=-Y if Y>5, X=10-Y if Y<5$ Then both are continuous random variables but their sum is a discrete random variable (taking on the values 0 and 10).

  • $\begingroup$ It is indeed a counter-example. I am going to change my question because I'm looking for something a little different. Thank you though. $\endgroup$ – majmun Jan 29 '16 at 6:17

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