The problem
A clinic is equally likely to have $2$, $3$, or $4$ doctors volunteer for service on a given day. No matter how may volunteer doctors there are on a given day, the numbers of patients seen by these doctors are independent Poisson random variables with mean $30$. Let $X$ denote the total number of patients seen in the clinic by all doctors on a given day. (a) Find $E[X]$ (b) Find $Var(X)$
My question
The answers are $E[X] = 90$ and $\operatorname{Var}(X) = 690$. The book gets the $\operatorname{Var}(X)$ using the random sums formula which does make sense to me. But I don't know why my alternative approach is wrong. I am getting $890$ instead of $690$.
My approach
Let $Y$ be the number of patients that one doctor sees then $Y \sim Poisson(\lambda = 30)$.
Let $N$ be the number of doctors then $N$ is a discrete uniform RV that can take values $2,3,4$ with probability $1/3$. Therefore $X = NY$.
$$E[X] = E[E[X \mid N]] = \frac{1}{3} ( 2E[Y] + 3E[Y] + 4E[Y]) = \frac{1}{3}(9 \cdot 30) = 90$$
Now I want to find the variance of $X$ using $Var(X) = E[X^2] - (E[X])^2$.
$$E[X^2] = E[E[X^2 \mid N]] = \frac{1}{3}(E[(NY)^2 \mid N = 2] + E[(NY)^2 \mid N = 3] + E[(NY)^2 \mid N = 4] $$
$$= \frac{1}{3} (4E[Y^2] + 9E[Y^2] + 16E[Y^2])$$
and since $\operatorname{Var}(Y) = 30 = E[Y^2] - 30^2$ this means that $E[Y^2] = 930$ and so
$$E[X^2] = \frac{1}{3} (4\cdot 930 + 9\cdot 930 + 16\cdot 930) = 8990$$
and so $\operatorname{Var}(X) = 8990 - 90^2 = 890$ but this approach is wrong because the answer is $690$. Thanks for your help!
You have $Y \sim \mathrm{Poisson}(\lambda = 30)$, all right, where $Y$ is the number of patients one particular doctor sees, but $X \neq NY.$
If $Y$ were the number of patients the first doctor sees on a particular day, and you are guaranteed that no matter what happens and no matter which doctors show up on that particular day, every doctor will see the exact same number of patients as the first doctor, then indeed $X$ would be $NY.$ But you have been given that the numbers of patients seen by the various doctors are independent random variables. So if there are four doctors on a given day, they may see (respectively) $Y_1,$ $Y_2,$ $Y_3,$ and $Y_4$ patients.
You could write $X = Y_1 + Y_2 + 1_{N\geq3} Y_3 + 1_{N\geq4} Y_4.$ But that won't fit your calculations.