1
$\begingroup$

Let $(X_i)_{i=1}^\infty$ be independent nonnegative integer valued random variables. Suppose that $X_n \succeq X_{n+1}$ (in the stochastic dominance sense). Does it follow that $X_n \overset{d}\to X$ for some random variable $X$?

It seems to me that if we let $F_n(k) = \mathbf P[ X_n \leq k]$, then because $X_n \succeq X_{n+1}$, we have $F_n(k)$ are increasing, and have limits, $F(k)$. I believe the stochastic dominance also implies that $F(k) \leq F(k+1) \leq 1$ for all $k$. So, it appears $F$ defines a unique distribution for a r.v. $X$. Is this reasoning sound?

$\endgroup$
  • $\begingroup$ Hi Matt, This looks fine, but I think people don't much like this questions of the form "please check this proof". $\endgroup$ – Anthony Quas Mar 17 '16 at 18:35
  • $\begingroup$ This is fine, and you don't need the independence condition. $\endgroup$ – Iosif Pinelis Mar 17 '16 at 19:33
  • $\begingroup$ You also don't need the random variables to be integer valued. $\endgroup$ – Iosif Pinelis Mar 17 '16 at 19:36
  • $\begingroup$ Thanks Iosif, I agree. It's a bit embarrasing, but I see now that this is one of the equivalent formulations of convergence in distribution in the Portmanteau lemma. And, thanks Anthony. I can see why "check my math" is not a popular type of post here. I do appreciate the feedback, and will be more conscientious for future posts. $\endgroup$ – Matthew Junge Mar 19 '16 at 6:01

Your Answer

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

Browse other questions tagged or ask your own question.