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If X(n) and Y(n) converges to X and Y in distribution respectively then does X/Y(n) also converge to X/Y in distribution? Prove or disprove.

I feel that this is correct but have not been able to proceed at all with the proof. If I move to distribution of X/Y then it becomes too complicated and I am not able to proceed any other way since this is the weakest sort of convergence. Moreover I have also not been able to construct any counter example. Slightly detailed explanation would be appreciated.

PS This is not an assignment question but my teacher sort of implicitly used this in stats class while doing Large Sample Theory.

Govind Saran Gautam
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As convergence in distribution only cares about the distribution of the random variables, one must we very careful when applying "pointwise" operation like division. Consider $\Omega = \{1,2\}$ with the uniform distribution, and $X \colon \Omega \to \mathbf R$ the identity embedding. Then $X$'s distribution is $\mathbf P_X = \frac 12 \delta_1 + \frac 12\delta_2$ and its the same as the distribution of $Y$ given by $Y(\omega) = 3 - \omega$. Now let $X_n = Y_n = X$. Then $X_n \to X$, $Y_n \to Y$ in distribution (as all variables have the same distribution $\frac 12 \delta_1 + \frac 12 \delta_2$). But $X_n/Y_n = 1$, hence $\mathbf P_{X_n/Y_n} = \delta_1$, but $$ \frac XY(1) = \frac 12, \quad \frac XY(2) = \frac 21 = 2 $$ hence $\mathbf P_{X/Y} = \frac 12 \delta_{\frac 12} + \frac 12 \delta_2$, then is $\frac{X_n}{Y_n} \not\to \frac XY$

martini
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  • However will this be valid if both X(n) , Y(n) were continuous type random variables? – Govind Saran Gautam Feb 04 '16 at 15:01
  • No. Almost the same argument works for $\Omega = [1,2]$ with the uniform distribution, $X$ the identity, $Y(\omega) = 3-\omega$ and $X_n = Y_n = X$ – martini Feb 04 '16 at 15:18
  • I guess I will have ro give you the complete context then... – Govind Saran Gautam Feb 04 '16 at 15:24
  • Define X(n) as the nth sample from a Normally distributed population with mean mu and variance sigma-square. By (CLT Sample mean-mu)/(sigma/under-root n) converges to Standard Normal Dist (let this convergence be #). Also I have proved sample-variance/sigma converges to 1 in probability by Chebyshev inequality and there it converges in distribution to 1(let this convergence be @), Now I have to see whether (#/@) is valid which would complete the proof of the theorem done by my professor. – Govind Saran Gautam Feb 04 '16 at 15:32