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I have the following problem:

Let $X$ be the number of distinct birthdays in a group of 110 people (i.e., the number of days in a year such that at least one person in the group has that birthday). Under the usual assumptions (no February 29, all the other 365 days of the year are equally likely, and the day when one person is born is independent of the days when the other people are born), find the mean and variance of $X$.

The solution proceeds as follows:

Let $I_j$ be the indicator r.v. for the event that at least one of the people was born on the $j$th day of the year, so $X = \sum_{j = 1}^{365}$ with $I_j \sim \text{Bern}(p)$, where $p = 1−(364/365)^{110}$. The $I_j$'s are dependent but by linearity, we still have $\mathbb{E}(X) = 365p \approx 95.083$. By symmetry, the variance is $\text{Var}(X) = 365 \text{Var}(I_1) + 2 {365 \choose 2} \text{Cov}(I_1, I_2)$.

I don't understand how the author got the following fact:

By symmetry, the variance is $\text{Var}(X) = 365 \text{Var}(I_1) + 2 {365 \choose 2} \text{Cov}(I_1, I_2)$.

I also searched for this "variance symmetry" formula, but I was unable to find it.

I would greatly appreciate it if people could please take the time to explain this.

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You are looking for $\text{Var}(X)=\text{Cov}(X, X) = \text{Cov}\left(\sum\limits_{i=1}^{365} I_i, \sum\limits_{j=1}^{365} I_j\right) = \sum\limits_{i=1}^{365}\sum\limits_{j=1}^{365}\text{Cov}\left( I_i, I_j\right)$

There are $365^2$ terms in the double sum

  • $365$ of the terms are $\text{Cov}\left( I_i, I_j\right)$ where $i=j$, and are equal to each other by symmetry, being $\text{Cov}\left( I_1, I_1\right)=\text{Var}(I_1)$

  • $365^2-365= 2{365 \choose 2}$ of the terms are $\text{Cov}\left( I_i, I_j\right)$ where $i\not=j$, and are equal to each other by symmetry, being $\text{Cov}\left( I_1, I_2\right)$

so $\text{Var}(X)=365\text{Var}(I_1)+2{365 \choose 2}\text{Cov}\left( I_1, I_2\right)$

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  • $\begingroup$ Thanks for the answer. It seems very odd that someone would somehow know that $365^2 - 365 = 2{365 \choose 2}$. Am I missing something here? $\endgroup$
    – The Pointer
    Commented Oct 27, 2019 at 16:36
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    $\begingroup$ (a) Choose two days in order $\left(365^2\right)$ so they are distinct $(-365)$ or (b) choose two distinct days ${365 \choose 2}$ where order matters $(\times 2)$. I think like (a) but they may have thought like (b); it makes no difference $\endgroup$
    – Henry
    Commented Oct 27, 2019 at 16:57

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