The problem is as follows:
A pizza parlor serves n different types of pizza, and is visited by
a number $K$ of customers in a given period of time, where $K$ is a Poisson random
variable with mean $\lambda$. Each customer orders a single pizza, with all types of pizza
being equally likely, independent of the number of other customers and the types of
pizza they order. Find the expected number of different types of pizzas ordered.
Here is how I did it:
Let $X$ be the random variable for the number of different types of pizzas ordered.
Let $X_i$ be the indicator variable for when the $i^{th}$ pizza is ordered at least once.
Thus we have:
$$X = X_1 + … + X_n$$
And:
$$E[X] = nE[X_i]$$$E[X_i]$ is simply the probability that at least one person orders the $i^{th}$ pizza which is also $1-P(no\:one\:orders)$. The probability that a customer arrives and orders the $i^{th}$ pizza at any infinitesimal period $\delta$ is $\lambda\delta\frac{1}{n}$, thus, we can think of the arrival and ordering process as a Poisson random variable with parameter $\frac{\lambda}{n}$.
Thus, the probability there are no orders in some given time interval $t$ is:
$$P_P(0,t) = e^{\frac{\lambda}{n}t}$$Here, $P_P$ is the probability distribution for the number of arrivals $k$ in some interval $\tau$:
$$P_P(k, \tau) = \frac{(\lambda\tau)^ke^{-\lambda\tau}}{k!}$$Finally, we have
$$E[X] = n(1-e^{\frac{\lambda}{n}t})$$
However, the solution seems to not use any time interval, instead the answer is:
$$E[X] = n(1-e^{\frac{\lambda}{n}})$$
Does anyone know why this is the case? Don't we have to specify a time interval for such a problem? The solution seems to directly use the Poisson distribution which does not depend on an interval. If so, what is $n$ and $p$ here?
Best Answer
Implicitly you're assuming $t=1$ unit of time, so that the number of customers arriving is actually Poisson $\lambda$.
Here's another approach that might be a bit clearer. Suppose $K=k$. Then easily $$\mathbb E[X_i\mid K=k]=1-\mathbb P(\text{no one gets }i)=1-\left(1-\frac{1}{n}\right)^k.$$ Hence $\mathbb E[X\mid K=k]=n\left(1-(1-1/n)^k\right)$. So by the law of total expectation, $$\mathbb E[X]=\mathbb E[\mathbb E[X\mid K]]=\mathbb E\left[n\left(1-(1-1/n)^K\right)\right]=n\left(1-\mathbb E\left[(1-1/n)^K\right]\right).$$ Now use PGF of the Poisson distribution $\mathbb E\left[t^K\right]=e^{\lambda(t-1)}$ to get the result $$\mathbb E[X]=n\left(1-e^{-\lambda/n}\right).$$ It looks like you're missing a minus sign in the exponent -- the expression that you've written could be negative!