I have the following question about the limiting distribution of the sum of two random variables say $Z_n = X_n+Y_n.$ I know the following:

  • Conditioned on $X_n,$ $Y_n$ has a CLT i.e.,

$$\mathbb P (Y_n \le z | X_n) \to \phi(z)$$

where $\phi(z)$ is the cdf of a standard gaussian independent of $X_n.$

  • Also, $$\mathbb P (X_n \le z) \to \phi(z)$$

From these two facts can I conclude $Z_n$ converges to $\mathcal{N}(0,2)$ in distribution?


Use characteristic functions. $Ee^{it(X_n+Y_n)} =E e^{itX_n}E(e^{itY_n}|X_n)$. Note that $E(e^{itY_n}|X_n) \to \phi (t)$ uniformly and $E e^{it(X_n)} \to \phi (t)$. It follows easily from these that $Ee^{it(X_n+Y_n)} \to \phi (t)^{2}$.

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  • $\begingroup$ Thanks. I thought about this but the outer expectation is for both the first and the second exponentials. If the two rvs were independent they would nicely factor out, but it's not so clear to me what happens when there is a dependence. There are two functions within the integral that both converge but does that imply that the integral converges say using bounded convergence? $\endgroup$ – Arun Oct 19 '18 at 12:05
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    $\begingroup$ Independence not required here. Bounded convergence works fine. $\endgroup$ – Kavi Rama Murthy Oct 19 '18 at 12:33

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