Blitzstein Page 174, derivation of binomial variance

Question

**This from Blitzstein, page 174. I don't understand the following:

1. What is $E\binom{X}{2}$ [the lines highlighted in blue in the text]? How can one take combinations of random variables, how is it defined?
2. How do we calculate $E\binom{X}{2} = \binom{n}{2} p^2$?
3. What is the intuition of defining our random variable this way?**

Answer 1

This is my first time seeing this argument for finding $\mathbb{E}X^2$. Nonetheless I will try to answer your doubts.

1. $\mathbb{E} \binom{X}{2}$ is just the expectation of a function acting on $X$. Say $$ f(k) = \binom{k}{2} = \frac{k(k - 1)}{2}, $$ and $$ \mathbb{E} \binom{X}{2} = \mathbb{E} f(X) = \mathbb{E} \left[ \frac{X(X - 1)}{2} \right]. $$

2. First, $\binom{X}{2}$ can be understood as after the trial, $X = n$ which is deterministic, then $\binom{X}{2} = \binom{n}{2}$ is just the number of pair of trials that succeeded. This is what the italic words mean. This is the same as pairing up each trial, do the experiment and keep only the pairs that both succeeded, then count the number of such pairs. Following this line of thought, we can define an indicator for every pair of experiment, where each pair have a chance of $p^2$ to succeed due to independence. $\mathbb{E} \binom{X}{2}$ is just the sum of each indicator multiply by its chances of success, which $$ \mathbb{E} \binom{X}{2} = \binom{n}{2} p^2, $$ since each pair having the same success rate $p^2$, and there are a total of $\binom{n}{2}$ of such trials.

3. From the formula $$ \mathbb{E} \binom{X}{2} = \mathbb{E} \left[ \frac{X(X - 1)}{2} \right], $$ we can generate $\mathbb{E} X^2$. The motivation here is to use a counting argument instead of an algebraic one to derive the variance formula. Sometimes a counting argument is more direct but for this example I would prefer the computaions.

Answer 2

1. We can think about taking the expectation of combinations of random variables like this: imagine you simulate your set of $n$ iid Bernoulli rvs, resulting in an integer number $k$ of successes. Then it is straightforward to compute ${k \choose 2}$ for this specific $k$.

2. Generalizing the logic above we can write: $$ E{x \choose 2} = \sum_{k=0}^n p(k) {k \choose 2} = \sum_{k=0}^n p^k (1-p)^{n-k} {n \choose k} {k \choose 2} = \frac12 \sum_{k=0}^n p^k (1-p)^{n-k} {n \choose k} (k-1)k = \frac12 n(n-1)p^2 $$

3. The intuition here is a little complicated and I'm not sure this example computation is really that helpful. Whats going on is that we are computing the expected number of pairs of successful trials. We know algebraically that this is equal to $E[X(X-1)] = E[X^2 - X] = E[X^2] - E[X]$. Since $E[X] = np$ is easy to compute its easy to jump from $E[X^2] - E[X]$ to the variance $Var(X) = E[X^2] - E[X]^2$. I think Blitzstein means this example to highlight how we can use the fact that $Var(X) = E[X^2] - E[X]^2$ to compute the variance even if we did not know that the Binomial was a sum of iid rvs.