(I asked this question on MSE 10 days ago, but I got no answer.)

Let $X$ and $Y$ be two independent identically distributed binomial random variables with parameters $n \in \mathbb{N}$ and $p \in (0,1)$. Let $Z := XY$ be their product.

Is it true or false that $\tilde{Z} := (Z - \mathbf{E}[Z]) / \sqrt{\mathbf{Var}[Z]}$ converges in distribution to a standard normal random variable (as $n \to \infty$) ?

At a first glance, I would be tempted to write $X = \sum_{i=1}^n A_i$ and $Y = \sum_{i=1}^n B_i$, where $A_i$ and $B_i$ are Bernoulli random variables, and then to apply the central limit theorem to $Z = \sum_{i=1}^n \sum_{j = 1} A_i B_j$... but $A_i B_j$ are not independent...

Thanks for any help

P.S.1 It is easy to check that $\mathbf{E}[Z] = n^2 / 4$ and $\mathbf{Var}[Z] = n^3 / 8 + n^2 / 16$. Moreover, expanding $(XY - n^2/4)^k$ with the binomial theorem and using the formula for the moments of the binomial distribution, I got that

$$\mathbf{E}\left[\tilde{Z}^k\right] = \frac1{(n^3 / 8 + n^2 / 16)^{k/2}} \sum_{j=0}^k \binom{k}{j} \left(\sum_{i=0}^j {j \brace i} (n)_{i} (1/2)^i\right)^2 (-n^2/4)^{k-j}$$

where ${j \brace i}$ are Stirling numbers of second kind and $(n)_{i}$ is a falling factorial. With this formula, I verified that the first 20 moments of $\tilde{Z}$ tend to the moments of a standard normal variable. However, I still do not know how to prove this for all moments.

P.S.2 I think that one cannot prove the claim only using the fact that the (normalized) $X$ and $Y$ converge in distribution to normal variables. In fact, it is known that the product of two independent normal variables is not a normal variable.

(I asked this question on MSE 10 days ago, but I got no answer.)

Let $X$ and $Y$ be two independent identically distributed binomial random variables with parameters $n \in \mathbb{N}$ and $p \in (0,1)$. Let $Z := XY$ be their product.

Is it true or false that $\tilde{Z} := (Z - \mathbf{E}[Z]) / \sqrt{\mathbf{Var}[Z]}$ converges in distribution to a standard normal random variable (as $n \to \infty$) ?

At a first glance, I would be tempted to write $X = \sum_{i=1}^n A_i$ and $Y = \sum_{i=1}^n B_i$, where $A_i$ and $B_i$ are Bernoulli random variables, and then to apply the central limit theorem to $Z = \sum_{i=1}^n \sum_{j = 1} A_i B_j$... but $A_i B_j$ are not independent...

Thanks for any help

P.S.1 For $p=1/2$, it is easy to check that $\mathbf{E}[Z] = n^2 / 4$ and $\mathbf{Var}[Z] = n^3 / 8 + n^2 / 16$. Moreover, expanding $(XY - n^2/4)^k$ with the binomial theorem and using the formula for the moments of the binomial distribution, I got that

$$\mathbf{E}\left[\tilde{Z}^k\right] = \frac1{(n^3 / 8 + n^2 / 16)^{k/2}} \sum_{j=0}^k \binom{k}{j} \left(\sum_{i=0}^j {j \brace i} (n)_{i} (1/2)^i\right)^2 (-n^2/4)^{k-j}$$

where ${j \brace i}$ are Stirling numbers of second kind and $(n)_{i}$ is a falling factorial. With this formula, I verified that the first 20 moments of $\tilde{Z}$ tend to the moments of a standard normal variable. However, I still do not know how to prove this for all moments.

P.S.2 I think that one cannot prove the claim only using the fact that the (normalized) $X$ and $Y$ converge in distribution to normal variables. In fact, it is known that the product of two independent normal variables is not a normal variable.