In which cases $E(e^{t S_n S_m})$ converges to $E(e^{t X Y})$

Question

Suppose that $S_n$ and $S_m$ are two random binomial variables, which are independent and with the same distribution parameter $p$. I am wondering, in which cases $E(e^{t S_n S_m})$ converges to $E(e^{tXY})$ when $n, m$ tend to infinity, where $t$ is real and $X$ and $Y$ are normally distributed variables with parameters $N(np,\sqrt{npq})$ and $N(mp,\sqrt{mpq})$, respectively.

However, it is clear that $e^{tS_nS_m}$ converges in probability to $e^{tXY}$, according to the Moivre-Laplace theorem, but I am not sure that any of the sufficient conditions which can be found in literature are satisfied considering the expectation.

In fact, I am trying to prove $E(e^{tS_nS_m}) \sim E(e^{tXY})$ when $n, m$ tends to infinity.

Answer 1

The answer is "no", and the reason is that $E[e^{tXY}]=\infty$ when $nm$ is large enough. To see that, note that $X=\sqrt{Var[X]}N+E[X]$ and $Y=\sqrt{Var[Y]}M+E[Y]$, where $N$ and $M$ are independent standard normal random variables. If $nm$ is large enough so that $t\sqrt{Var[X]Var[Y]}\geq 1$, we have $$ E[e^{tXY}]\geq E[\mathbf 1_{\{N\geq 0\}}1_{\{N\geq 0\}}e^{NM}]\geq \tfrac 1 4 E[e^{NM}]. $$ Direct computation shows $$ E[e^{NM}]= \int\int\frac{1}{2\pi}e^{-\frac{(x-y)^2}{2}}\mathrm dx\mathrm dy=\int\frac{\mathrm dy}{\sqrt{2\pi}}=\infty. $$