Multiplying Binomial N and P for Expectation?

Question

The problem given is X ~ Bin(5,.5) and the E(X) was calculated by just multiplying 5 and .5 together to get 2.5. I thought to calculate the expectation we would take values X=0 (p 0) + X=1 (p 1) etc. Can you calculate it this way? It seems like they both give different answers. Can np be used to find the expectation of any binomial distribution?

Answer 1

The expected value of the Binomial distribution is $np$ where the random variable $X$ is the expected number of successes in $n$ independent trials and $p$ is the probability of success of each trial. We could form random variables $T_1, T_2, ..., T_n$ where $T_i$ is the result of the $i$th trial. Then $P(T_i=1)=p$, $P(T_i=0)=1-p$, and $$ E[X] = E[\sum_{i=1}^n T_i] = \sum_{i=1}^n E[T_i] = \sum_{i=1}^n p = n p. $$ You can also use the probability mass functions of $X$
$$P(X=k) = {n\choose k} p^k(1-p)^{n-k}.$$
Then $$ E(X) = \sum_{k=0}^n k\,P(X=k) = \sum_{k=1}^n k {n\choose k}p^k(1-p)^{n-k} $$ $$ = \sum_{k=1}^n n p {n-1\choose k-1}p^{k-1} (1-p)^{n-k} =np ( p + (1-p))^{n-1}=n p. $$

Feel free to put any questions in the comments below.