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 $\frac{730}{365}$
Answer
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
$$E[X] = E[730] = 730,$$
and $$\text{Var}(X) = \text{Var}(730) = 0.$$
No comments:
Post a Comment