# Binomial distribution not adding up

Suppose that there are 730 people in the class and each person’s birthday is equally likely to be any of the 365 days in the year (for this question you should ignore leap years).

Define $X1$ be the random variable corresponding to the number of people born on the first day of the year.

Similarly, for $i = 2, 3, . . . , 365$, define $Xi$ to be the number of people born on the $ith$ day of the year.

So each $Xi$ takes a value in the range {0, 1, . . . , 730} and, for example, $$P(X1 = 10) = {730\choose{10}}  \frac{1}{365}^{10} \frac{364}{365}^{720}$$

I realize that this is a Binomial distribution. Now:

Let X = X1 + X2 + . . . + X365

$$E(X) = 730$$

$$var(X) = 0$$

why is E(X) = 730? I thought that in a binomial, the $E(X) = Np$, which in this case would be $\frac{730}{365}$

• Pay attention to what $E[X]$ represents. It represents the expected total number of people born on the first day or the second day or the ... or the 365th day. In other words, $E[X]$ represents the total number of people. Since there are $730$ people regardless how the birthdays are distributed, the expected value is identically $730$. Since it is always $730$, the variance is zero. – JMoravitz Mar 9 '16 at 23:43
• $E[X_i]$ on the other hand is the expected number of people born on the $i^{th}$ day. $E[X_i]$ is not the same thing as $E[X]$. Indeed, $E[X_i]$ will be $\frac{730}{365}$ for exactly the reason you state, it is a binomial distribution. $X_i$ is a binomial distribution, $X$ is not. – JMoravitz Mar 9 '16 at 23:44

If $Y_1,\dotsc,Y_n$ are iid $\text{Bin}(n_i,p)$, then $$Y = Y_1+\dotsb+Y_n\sim\text{Bin}\left(\sum_i n_i, p\right)$$
Notice, that $Y_i$ are iid. Your $X_i$ are not iid, and $X$ is deterministic, $$X = X_1+\dotsb+X_{365} = 730$$ since each person is born on a particular day during the year. In other words, the number of people born in the year is $730$.
Hence $$E[X] = E = 730,$$ and $$\text{Var}(X) = \text{Var}(730) = 0.$$