Expected value of a multinomial distribution

Question

A multinomial distribution can be given as

$ M(m_1,\dots,m_K|N,P) = {N \choose m_1\dots m_K}\prod_k p_k^{m_k} $

The expected value is $Np_k$.

How can I prove it?

Answer 1

A demonstration using "equations" was requested in a comment. Here is a short, simple one that is practically painless.

Notation and definitions

Let the random $K$-vector $X$ have a multinomial distribution with parameters $\mathbb p = (p_1, p_2, \ldots, p_K)$. This means that $p_1 + p_2 + \cdots + p_K=1$, $0 \le p_i$ for $i=1, 2, \ldots, K$, and the probability that $X = (m_1, m_2, \ldots, m_K) = \mathbb m$ is given by

$$\Pr(X=\mathbb m) =\binom{N}{\mathbb m}\mathbb p^\mathbb m$$

In this shorthand notation $\binom{N}{\mathbb m} = N!/(m_1! m_2! \ldots m_K!)$ is a multinomial coefficient (which is nonzero only when all the $m_i$ are natural numbers and sum to $N \ge 1$) and $\mathbb p ^ \mathbb m = p_1^{m_1}p_2^{m_2}\cdots p_K^{m_k}.$

By definition, the expectation of $X$ is the vector

$$\mathbb E[X] = \sum_{\mathbb m} \Pr(X = \mathbb m)\mathbb m =\sum_{\mathbb m} \binom{N}{\mathbb m}\mathbb p^\mathbb m \mathbb m$$

where the sum extends over the (finite number of) values of $\mathbb m$ for which the probability is nonzero.

Solution

By expanding the sum using the definition of the multinomial coefficients, notice that

$$1 = 1^N = (p_1 + p_2 + \cdots + p_K)^N = \sum_{\mathbb m}\binom{N}{\mathbb m}\mathbb p^\mathbb m.$$

Viewing the $p_i$ as variables, we can recognize the component terms $\binom{N}{\mathbb m}\mathbb p^\mathbb m m_i$ in the expectation as the result of applying the differential operator $p_i\frac{\partial}{\partial p_i}$ to the right hand side, because $p_i\frac{\partial}{\partial p_i} \left(p_i^{m_i}\right) = m_i p_i^{m_i}.$ Another way to compute the same thing is to use the Chain Rule to differentiate the penultimate term in the preceding multinomial expansion:

$$p_i\frac{\partial}{\partial p_i}(p_1 + p_2 + \cdots + p_K)^N = p_iN(p_1 + p_2 + \cdots + p_K)^{N-1}\frac{\partial p_i}{\partial p_i} = Np_i(1)^{N-1} = Np_i.$$

Therefore

$$\mathbb E[X] = (Np_1, Np_2, \ldots, Np_K),$$

QED.

Answer 2

I think that you mean that you take $N$ draws from a multinomial distribution and the expected value of getting object $k$ is $Np_k$. The easiest way to show this is to reduce the problem to $N$ draws from a binomial distribution, with the options "not get object $k$" and "get object $k$." Consider $K$ of these separate binomial problems and you get the answer.