# Binomial Expectation Proof

This example is from the textbook "Introduction to Probability (2e) - Blitztsein & Hwang."

I am currently studying about expectation and had a question regarding the derivation of the Binomial expectation.

More specifically, the author shows a step-by-step derivation of the equation:

$$E(X)\ =\ np$$

as follows:

\begin{align} \sum_{k=0}^nk\binom{n}{k}p^k(1-p)^{n-k} & = n\sum_{k=0}^n\binom{n-1}{k-1}p^k(1-p)^{n-k} \\ & = np\sum_{k=0}^n\binom{n-1}{k-1}p^{k-1}(1-p)^{n-k} \\ & = np\sum_{j=0}^{n-1}\binom{n-1}{j}p^j(1-p)^{n-1-j} \\ & = np \end{align}

Due to the binomial theorem, we can conclude that the summation in the third line equals $$1$$, leaving us with $$np$$.

My question is: "Is the third line of the derivation necessary?"

To me it seems like we can use the binomial theorem from the second line without the need to substitute $$k-1$$ with $$j$$. Would my observation be correct?

Any feedback is appreciated. Thank you!

• k has to begins from 1 to n or else at the first "=" you'll have combination with negative inside – papasmurfete Oct 17 '18 at 19:08