Calculate the variance of this random variable

Question

Let $p_1$ have $p_x(x) = 0.3*0.7^X$. Let $p_2$ have the exact same $p_x(x) = 0.3*0.7^X$. Assume this is valid for all $x\geq 0$.

Assume that $p_1,p_2$ are independent random variables.

Let $Y=(p_1)(p_2)$. Calculate $var(Y)$.

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Here is my attempt:

$var(Y)=[E(Y^2)]-[E(Y)]^2=[E(p_1p_2)^2]-[E(p_1p_2)]^2=[E(p_1^2p_2^2)]-[E(p_1p_2)]^2$

Usually my process for this is to calculate $E[Y], E[Y^2]$, and then apply the formula (squaring $E[Y]$ as a whole, of course]

So, $E[Y]=E[p_1p_2]=E[p_1]E[p_2]$. Calculating one will give me the other, as they are the exact same.

I notice that $p_1$ is $Geo(\theta=0.3)=\theta(1-\theta)^X$. The expectation is given by $(1-\theta)/\theta=0.7/0.3=2.33$

This helps me with $E[Y]=(2.33)(2.33)$, but how do I calculate $E[Y^2]$?

Answer 1

\begin{align} E[Y^2] &= E[p_1^2p_2^2]\\ &=E[p_1^2]E[p_2^2]\\ &= (Var(p_1)+E[p_1]^2)^2 \end{align}

Answer 2

Notice that if $p_1$ and $p_2$ are independent then so are $p_1 ^2$ and $p_2 ^2$. Therfore $$ E (p_1 ^2 p_2 ^2) = E (p_1 ^2)E( p_2 ^2) $$

Answer 3

$$E(Y^2){=E(p_1^2p_2^2)\\=E(p_1^2)E(p_2^2)}$$according to Geometric distribution(https://en.wikipedia.org/wiki/Geometric_distribution) we obtain $$E(p_1^2)=E(p_2^2)={0.7\times 1.7\over 0.3^2}={119\over 9}$$therefore $$E(Y^2)={14161\over 81}\approx 174.83$$