Calculate $\sum_{k=0}^n k \binom{n}{k} p^k (1-p)^{n-k}$

Question

For $p \in [0,1]$ calculate

$$S =\sum_{k=0}^n k \binom{n}{k} p^k (1-p)^{n-k}.$$

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Since $$ (1-p)^{n-k} = \sum_{j=0}^{n-k} \binom{n-k}{j} (-p)^j, $$ then $$ S =\sum_{k=0}^n \sum_{j=0}^{n-k} k \dfrac{n!}{k!j!(n-k-j)!} p^k (-p)^j. $$ If it weren't for that $k$, I would have $$ S = (1+p-p)^n = 1\ldots $$

Answer 1

Use the identity $k\binom{n}{k}=n\binom{n-1}{k-1}$ and the binomial identity to get $$ \begin{align} \sum_{k=0}^nk\binom{n}{k}p^k(1-p)^{n-k} &=\sum_{k=1}^nn\binom{n-1}{k-1}p\,p^{k-1}(1-p)^{n-k}\\ &=np\sum_{k=1}^n\binom{n-1}{k-1}p^{k-1}(1-p)^{n-k}\\ &=np\,(p+(1-p))^{n-1}\\[12pt] &=np \end{align} $$

Answer 2

We have the binomial expansion $$ (p+q)^n=\sum_{k=0}^n\binom{n}{k}p^kq^{n-k}. $$ Differentiating with respect to $p$ we get $$ n(p+q)^{n-1}=\sum_{k=0}^nk\binom{n}{k}p^{k-1}q^{n-k} $$ and hence $$ np(p+q)^{n-1}=\sum_{k=0}^nk\binom{n}{k}p^{k}q^{n-k}. $$ Setting $q=1-p$ above we finally obtain $$ np=\sum_{k=0}^nk\binom{n}{k}p^{k}(1-p)^{n-k}. $$

Answer 3

An answer from the point of view of probability theory rather than logic is as follows.

Let $X_1,\ldots,X_n$ be independent random variables, each equal to $1$ with probability $p$ and to $0$ with probability $1-p$.

The probability that exactly $k$ of them are equal to $1$ is $\dbinom n k p^k (1-p)^{n-k}$. The expected number of them that are equal to $1$ is therefore $\displaystyle\sum_{k=0}^n k\binom n k p^k (1-p)^{n-k}$.

But the expected value is $$ \mathbb E(X_1+\cdots+X_n) = \mathbb E(X_1)+\cdots+\mathbb E(X_n) = p + \cdots + p = np. $$