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Suppose there are two i.i.d. sequences $\left\{X_n\right\}$ and $\left\{Y_t\right\}$ that \begin{align*} \sqrt{N}\left(\bar{X}_N-\mathrm{E}\left[X_n\right]\right)\overset{d}{\rightarrow}&\mathcal{N}\left(0,\mathrm{Var}\left[X_n\right]\right)\\ \sqrt{T}\bar{Y}_T\overset{d}{\rightarrow}&\mathcal{N}\left(0,\mathrm{E}\left[Y_t^2\right]\right) \end{align*} and they are independent. Can I infer anything about \begin{align*} \frac{1}{\sqrt{NT}}\sum_{i=1}^{N}{\sum_{t=1}^{T}{X_nY_t}}, \end{align*} i.e. the product of two sequences that converge in distribution? I think I cannot apply Slutsky's Theorem since both converge not in probability, but in distribution.

Here I am assuming the existence of the moments $\mathrm{E}\left[X_n\right]$, $\mathrm{E}\left[X_n^2\right]$, $\mathrm{E}\left[Y_t\right]$ and $\mathrm{E}\left[Y_t^2\right]$, but, if it is necessary, then I can add the assumption that both $X$ and $Y$ are normally distributed.

By the way, I tried Monte Carlo Simulation and it seems the product follows a leptokurtic distribution that is similar to Laplace Distribution.

Junyong Kim
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1 Answers1

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Neglecting centerings and so on, if $U_m$ converges in distribution to $U$, and if $V_n$ converges in distribution to $V$, and it the $U$s are independent of the $V$s, then I think it's true that the vectors $(U_m,V_n)$ converge in distribution to $(U,V)$ as $(m,n)\to(\infty,\infty)$, and hence $U_nV_n$ converges in distribution to $UV$. One can, for instance, pretend that the $U_m\to U$ almost surely (on some other probability space, for which all the marginal distributions are unchanged, say), and similarly for $V_n\to V$ a.s., and so $U_mV_n\to UV$ a.s. and hence in distribution.

Taking centerings and scalings into account: Let $\sigma^2 = \mathrm{Var}(X_1)$, let $\tau^2=\mathrm{Var}(Y_1)$.

Write $\overline X_N = \mu + \sigma Z_N/\sqrt N$ and $\overline Y_T = \tau W_T/\sqrt{T}$ where $(Z_N,W_T)$ converges in distribution to a bivariate Gaussian with unit covariance matrix and zero mean. Then $S=\sqrt{NT}\overline X_N \overline Y_T $, which the original question is about, equals $\sqrt N \mu \tau W_T + \sigma\tau Z_N W_T$. If $\mu\ne0$, $S/\sqrt N$ is asymptotically $N(0,\mu^2\tau^2)$. If $\mu=0$, $S$ is asymptotically distributed as the product of an $N(0,\sigma^2)$ times an independent $N(0,\tau^2)$. Note that the correct scaling on $S$ depends on whether $\mu=0$ or not. Does this agree with your Monte Carlo experiments?

kimchi lover
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  • I think $\left(U_m,V_n\right)\overset{d}{\rightarrow}(U,V)$ as $(m,n)\rightarrow(\infty,\infty)$ and the continuous mapping theorem can be applied since the multiplication is continuous. Can I say $U_mV_n\overset{d}{\rightarrow}\mathcal{N}\times\mathcal{N}$ if $X$ is centered as well? – Junyong Kim Nov 29 '17 at 20:01
  • Continuous mapping theorem: yes. See my edits to my answer. – kimchi lover Nov 29 '17 at 21:30
  • It seems what you add is consistent with the simulation. It seems $S$ converges to $\mathcal{N}\times\mathcal{N}$ when $\mu=0$ and $S/\sqrt{N}$ converges to $\mathcal{N}$ when $\mu\ne0$. Click here for the SAS code and its outcomes. Now the problem is the ugly distribution $\mathcal{N}\times\mathcal{N}$, which is by and large unknown unfortunately. – Junyong Kim Nov 30 '17 at 05:16
  • Let $G=XY$ where $X$ and $Y$ are iid $N(0,1)$. I think the characteristic function of $G$ is $\varphi_G(t)=1/\sqrt{1+t^2}$ (which can be seen by writing $G=(U+V)(U-V)/2$ where $X=(U+V)/\sqrt 2$ and $Y=(U-V)/\sqrt2$ for $U,V$ iid $N(0,1)$); the moments of $G$ are easy to write ($EG^k=(EX^k)^2$), and many other calculus facts about $G$ are easy to derive. So don't give up hope just yes. – kimchi lover Nov 30 '17 at 13:18
  • See https://math.stackexchange.com/questions/101062/is-the-product-of-two-gaussian-random-variables-also-a-gaussian . The last answer to this question has pointers to the literature that might be useful to you. The short answer: the pdf of $XY$ is the Bessel function $K_0(x)/\pi$. – kimchi lover Nov 30 '17 at 22:28