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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!

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

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