The number of students per day has the distribution N ∼ Poisson(10). The students of CSUEB withdraw money from a cash machine according to the following probability function (X):
X | 50 | 100 | 200
P(X = x) | 0.3| 0.5 | 0.2
Let $T_N = X_1 +X_2 +···+X_N$ be the total amount of money withdrawn in a day, where each $X_i$ has the probability function in the table, and $X_1,X_2,...$ are independent of each other and of N. Here $T_N$ is a randomly stopped sum, stopped by the random number of N customers.
I've found E[X] = 105 and Var[X] = 2725
My attempt at solving this is:
$Var[T_N | N] = Var[X_1] + \cdot\cdot\cdot+Var[X_N] = 2725N = 27250$
I thought this was pretty straight forward but I'm not too confident. Is it correct to make the assumption that $Var[T_N] = 2725N$?
If I use the property
$V[X] = V[E[X|Y]]+E[V[X|Y]$
$E[T_N|N] = 105N$, $V[T_N|N] = 2725N$, $V[N] = 10$, $E[N] = 10$
$V[T_N] = V[105N] + E[2725N] = 105^2N + 2725N = 137500$