# Calculate the variance of this random variable

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 $$var(Y)$$.

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 $$E[Y^2]$$?

\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}
Notice that if $$p_1$$ and $$p_2$$ are independent then so are $$p_1 ^2$$ and $$p_2 ^2$$. Therfore $$E (p_1 ^2 p_2 ^2) = E (p_1 ^2)E( p_2 ^2)$$
$$E(Y^2){=E(p_1^2p_2^2)\\=E(p_1^2)E(p_2^2)}$$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 $$E(Y^2)={14161\over 81}\approx 174.83$$
• How did you calculate $E(p_1^2)$? The formula on wikipedia is for $E(X)$, not $E(X^2)$? Nov 24, 2018 at 16:19
• I calculated $E(X^2)$ from the variance and the mean using $$E(X^2)=\mu^2+\sigma^2$$. Also i spotted a mistake in my answer that i will fix it soon........ Nov 24, 2018 at 16:27