I tried to multiply two independent binomials, each distributed with $n=2$ and probability p. So using the piecewise function for $P(XY),$ I have $$P(Z=z) = P(XY = xy) = \sum_{xy=z} z \cdot P(X=x, Y=y)$$ leading to $Z \sim \text {Binom}\ (4,p)$. But using the piecewise function to compute $E[XY]$...I got $4p^2.$ But if $Z \sim \text {Binom}\ (4,p),$ the expectation should be $4p.$ Could someone let me know where I went wrong?
Thank you in advance.
On
We have
$$\begin{align*} \mathsf P(Z=0) &=\mathsf P(X=0,Y=0)+2\cdot \mathsf P(X=0,Y=1)+2\cdot \mathsf P(X=0,Y=2)\\\\ &=(1-p)^4+4p(1-p)^3+2p^2(1-p)^2 \end{align*}$$
and
$$\begin{align*} \mathsf P(Z=1) &=\mathsf P(X=1,Y=1)\\\\ &=\mathsf 4p^2(1-p)^2 \end{align*}$$
and
$$\begin{align*} \mathsf P(Z=2) &=2\cdot\left(\mathsf P(X=1,Y=2)\right)\\\\ &=4p^3(1-p) \end{align*}$$
and
$$\begin{align*} \mathsf P(Z=4) &=\mathsf P(X=2,Y=2)\\\\ &=p^4 \end{align*}$$
Hence we get the pmf
$$ p_{Z}(z)= \begin{cases} (1-p)^4+4p(1-p)^3+2p^2(1-p)^2 & z=0 \\ 4p^2(1-p)^2 & z=1 \\ 4p^3(1-p) & z=2 \\ p^4 & z=4 \\ \end{cases} $$
It would behoove you to check that the respective probabilities sum to $1$.
If $$X \sim \operatorname{Binomial}(2,p), \quad Y \sim \operatorname{Binomial}(2,p), \\ Z = XY, \quad W \sim \operatorname{Binomial}(4,p),$$ and $X$ and $Y$ are independent, then $$\begin{align*} \Pr[Z = 1] &= \Pr[XY = 1] \\ &= \Pr[(X = 1) \cap (Y = 1)] \\ &= \Pr[X = 1]\Pr[Y = 1] \\ &= \left(\binom{2}{1}p^1 (1-p)^{2-1}\right)^2 \\ &= 4p^2(1-p)^2. \end{align*}$$ However, $$\Pr[W = 1] = \binom{4}{1} p^1 (1-p)^{4-1} = 4p(1-p)^3 \ne \Pr[Z = 1],$$ demonstrating $W$ and $Z$ are not identically distributed.
In fact, it is the sum of independent and identically distributed binomial random variables that are also binomially distributed, not the product.