1
$\begingroup$

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.

Thanks.

$\endgroup$
2
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.