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Let $T_N = X_1 +X_2 +···+X_N$ be the total amount of money withdrawn in a day, where each $X_i$ has the probability function in the table, and $X_1,X_2,...$ are independent of each other and of N. Here $T_N$ is a randomly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.

Find $Var(T_N)$.


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

$V[X] = V[E[X|Y]]+E[V[X|Y]$


$E[T_N|N] = 105N$, $V[T_N|N] = 2725N$, $V[N] = 10$, $E[N] = 10$


$V[T_N] = V[105N] + E[2725N] = 105^2N + 2725N = 137500$


No, it is not correct.

The variability in $T_N$ arises not only due to $X_i$ bit also due to random number of summands $N$.

Hint: To calculate the required quantities, use the law of iterated expectations which results in following identities:

\begin{align} E(X) &= E(E(X\mid Y)) \\V(X) &= V(E(X\mid Y)) + E(V(X\mid Y)) \end{align}

Note: What you have calculated is in fact $V(T_N\mid N)$.

  • $\begingroup$ I've updated my question using the properties you've provided. It is very close to the simulation of the variance that I've done in R. Thank you. $\endgroup$ – ThrmsNPrfs May 13 '19 at 18:30
  • $\begingroup$ @ThrmsNPrfs My pleasure $\endgroup$ – Sanket Agrawal May 14 '19 at 11:06

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