Expected Value of Poisson Process

Question

"A sandwich shop has n different types of sandwiches and is visited by a number K of customers in a given period of time, where K is a Poisson random variable with mean λ. Assume each customer orders a single sandwich, with all types of sandwiches being equally likely, independent of the number of other sandwiches and the types of sandwich they order. Find the expected number of different types of sandwiches ordered."

So I know that the total number of sandwiches purchased will be equal to K. If I define the random variable X as the total number of unique sandwich types purchased, then X must be less than or equal to both K and n.

I'm having difficulty quantifying the PMF of X, and so can't figure out how to find E[X]. How can I proceed here?

Answer 1

I probably made a mistake somewhere; feel free to berate me and my decaying brain.

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Indicator random variables may be helpful. Let $I_i$ equal $1$ if [at least] one of the $K$ people ordered the $i$th type of sandwich, and $0$ otherwise. Then $X = I_1 + \cdots + I_n$.

Hint 1: If instead $K$ is a fixed number (say, $5$), can you then answer the question?

Let $K$ be fixed. Note $$E[X_i] = P(\text{$i$th sandwich was ordered}) = 1 - \left(1 - \frac{1}{n}\right)^K$$ for each $i$, so $E[X] = \sum_{i=1}^n E[X_i] = n(1 - (1 - n^{-1})^K)$.

Hint 2: Use the tower rule to use the above work to handle the case where $K$ is random.

The above work implies $E[X \mid K] = n(1-(1-n^{-1})^K)$, so $$E[X] = E[E[X \mid K]] = n(1 - E[(1-n^{-1})^K]).$$ You can compute or look up the PGF for a Poisson random variable: $E[z^K] = e^{\lambda(z-1)}$. Plugging in $z=1-n^{-1}$ yields $E[X] = n(1-e^{-\lambda/n})$.