# Sum of Bernoulli random variables

I need some help with a homework assignment. The question I'm given is: "Suppose that $X_1, X_2,..., X_n, W$ are independent random variables such that $X_i\sim Bin(1,0.4)$ and $P(W=i)=1/n$ for $i=1, 2 ,.., n$. Let

$Y=\sum\limits_{i=1}^W X_i=X_1+X_2+X_3+...+X_W$

That is, $Y$ is the sum of $W$ independent Bernoulli random variables. Calculate the mean and variance of $Y$"

Since $Y$ is a sum of Bernoulli random variables it would be a binomial random variable with mean $\mu=np$ and variance $\sigma^2=np(1-p)$, but I'm not sure how to handle this problem when $n$ itself is a random variable.

Can anyone show me how to approach this?

• Hint: Break things down according to the mutually exclusive possible values of $W$. Oct 29, 2014 at 0:23
• Wald's equation may be useful, en.wikipedia.org/wiki/Wald%27s_equation Oct 29, 2014 at 0:27
• For this problem, I'd probably use the Law of Total Expectation and the Law of Total Variance, but you can do it quite easily the way that cardinal suggests. Oct 29, 2014 at 0:28

Using iterated expectations and law of total variance here is the approach I would use as recommended by @Glen_b in the comments above. To use these you need the means and variances of the individual random variables.

$$\mathbb{E}(X_1) =0.4$$ $$\mathbb{V}ar(X_1) =0.4\times0.6=0.24$$

and for $W$,

$$\mathbb{E}(W)=\frac{1}{n}\sum_{i=1}^ni =\frac{n(n+1)}{2n}=\frac{n+1}{2}$$ $$\mathbb{V}ar(W)=\frac{1}{n}\sum_{i=1}^ni^2 -\frac{(n+1)^2}{4} =\frac{n(n+1)(2n+1)}{2n}-\frac{(n+1)^2}{4} =\frac{n+1}{4}\times(4n+2-(n+1)) =\frac{(n+1)\times(3n+1)}{4}.$$

Now conditioning on $W$ and using total variance law,

$$\mathbb{V}ar(Y)=\mathbb{V}ar(\mathbb{E}(Y|W))+\mathbb{E}(\mathbb{V}ar(Y|W)) =\mathbb{V}ar(W\mathbb{E}(X_1))+\mathbb{E}(W\mathbb{V}ar(X_1)) =\mathbb{V}ar(W)\mathbb{E}(X_1)^2+\mathbb{E}(W)\mathbb{V}ar(X_1).$$

Now we can use the expressions above to substitute into the last equality to get

$$\mathbb{V}ar(Y)=\mathbb{V}ar(W)\mathbb{E}(X_1)^2+\mathbb{E}(W)\mathbb{V}ar(X_1) =\frac{(n+1)\times(3n+1)}{4} \times \frac{4}{10} + \frac{n+1}{2}\times\frac{24}{100} =\frac{(n+1)\times(3n+1)}{10} + \frac{12\times(n+1)}{100}$$