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I'm trying to do some self-studying to upgrade my statistics knowledge, and came across this term in a section discussing the weak law of large numbers and Bernoulli's theorem:

$$\sum_{k=0}^n k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k}$$

According to the book I was reading, this term can "easily be shown to equal np".

I am at a loss as to how to do so and could use some guidance!

Edited

Lower limit $k=0$, not $1$

Community
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PizzAzzra
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3 Answers3

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We have this sum: $$ \sum_{k=0}^n k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k} \tag 1 $$ First notice that when $k=0$, the term $k\dfrac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k}$ is $0$, and next notice that when $k\ne0$ then $$ \frac k {k!} = \frac 1 {(k-1)!} $$ so that $$ k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k} = \frac{n!}{(k-1)!(n-k)!} p^k (1-p)^{n-k}. $$ The two expressions inside the parentheses in the denominator now add up to $n-1$ rather $\text{than } n.$ So we can write it like this: $$ \frac{n!}{(k-1)!(n-k)!} = n\cdot \frac{(n-1)!}{(k-1)!(n-k)!} = n\cdot \binom{n-1}{k-1}. $$ Then the sum $(1)$ becomes $$ \sum_{k=1}^n n\cdot \binom{n-1}{k-1} p^k (1-p)^{n-k}. $$ Since $n$ does not change as $k$ goes from $1$ to $n$, we can pull $n$ out, getting $$ n \sum_{k=1}^n \binom{n-1}{k-1} p^k (1-p)^{n-k}. $$ Now let $j=k-1$ and observe that as $k$ goes from $1$ to $n$ then $j$ goes from $0$ to $n-1$, and $k = j+1$, so we have $$ n \sum_{j=0}^{n-1} \binom{n-1} j p^{j+1} (1-p)^{(n-1)-j}. $$ Since $p$ does not change as $j$ goes from $0$ to $n-1$, we can pull $p$ out, getting $$ np \sum_{j=0}^{n-1} \binom{n-1} j p^j (1-p)^{(n-1)-j}. $$ Now let $m= n-1$, so we have $$ np \sum_{j=0}^m \binom m j p^j (1-p)^{m-j}. $$ This sum is $1$ since it's the sum of probabilities assigned by the $\mathrm{Binomial}(m,p)$ distribution. Hence we get $$ np\cdot 1. $$

Michael Hardy
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    Really appreciate the detail at every step – this has made the derivation crystal clear and gives me some inspiration for decoding the other summations in the overall proof.

    Thanks to William for getting me started, but ultimately, thank you Michael for getting me home!

     – PizzAzzra May 22 '16 at 02:50
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    @PizzAzzra : I'm glad it helped. $\qquad$ – Michael Hardy May 22 '16 at 02:52
  • Just wanted to clarify what you meant by "The two expressions inside the parentheses in the denominator now add up to n−1 rather than n."? – PizzAzzra May 22 '16 at 02:53
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    The two expression referred to are $k-1$ and $n-k$. Their sum is $n-1$. Initially we had $k$ and $n-k$ in those roles, and their sum is $n$. $\qquad$ – Michael Hardy May 22 '16 at 02:55
  • That's a concept that I've not come across. I'm familiar with the binomial coefficients to an extent. What's the significance of the sum of the two denominators? Are they supposed to add up to equal the term for the numerator? – PizzAzzra May 22 '16 at 03:07
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    @PizzAzzra The point is to make $\frac{n!}{(k-1)!(n-k)!}$ look like a binomial coefficient. To do that, you can either divide it by $k$ (which would bring you back to what you had before) or divide it by $n$ (which is what you do in the proof). Notice that in a binomial coefficient, the sum of the arguments of the factorials in the denominator equals the argument of the factorial in the numerator. – Ian May 22 '16 at 19:16
  • @PizzAzzra : In binomial coefficients the two numbers whose factorial is taken in the denominator add up to the number whose factorial is taken in the numerator; for example $$ \binom 9 3 = \frac{9!}{3!6!} = \binom 9 6, \qquad 3+6=9. $$More generally, $$ \binom n k = \frac{n!}{k!(n-k)!} = \binom n {n-k}, \qquad n + \Big(n-k\Big) = n. $$ – Michael Hardy May 22 '16 at 22:33
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The trick is the using the identity $k { n \choose k} = n {n-1 \choose k-1}$. $$\begin{align*} &\sum_{k=1}^n k { n \choose k } p^k (1-p)^{n-k}\\ &= \sum_{k=1}^n n { n-1 \choose k-1} p^k (1-p)^{n-k}\\ &=np \sum_{k=1}^n {n-1 \choose k-1} p^{k-1} (1-p)^{(n-1)-(k-1)}\\ &=np (p+(1-p))^n\\ &=np \end{align*}$$

Maria
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Let $X_n=B(n,p)$ be a binomially distributed random variable. Also notice that $X_n=Y_1+Y_2+\cdots+ Y_n$ where $Y_i$ are i.i.d. Bernoulli with parameter $p$.

Now observe that \begin{align} \sum_{k=0}^n k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k}&= \operatorname{E}(X_n)\\ &= \operatorname{E}( Y_1+Y_2+\cdots Y_n)\\ &=\operatorname{E}( Y_1)+\operatorname{E}(Y_2)+\cdots +\operatorname{E}(Y_n)\\ &=np \end{align}

Michael Hardy
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clark
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