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My question is very simple: why does the index in the third step go from $0$ to $1$? No change of variable seems to happen, the only thing that happens is a rewriting and cancellation of $k$.. can you help me understand it? $$ \begin{align} \mu &= \sum_{k=0}^nk\binom nkp^k(1-p)^{n-k}\\ &=np\sum_{k=0}^nk\frac{(n-1)!}{(n-k)!k!}p^{k-1}(1-p)^{(n-1)-(k-1)}\\ &=np\sum_{k=1}^n\frac{(n-1)!}{((n-1)-(k-1))!(k-1)!}p^{k-1}(1-p)^{(n-1)-(k-1)}\\ &=np\sum_{k=1}^n\binom{n-1}{k-1}p^{k-1}(1-p)^{(n-1)-(k-1)}\\ &=np\sum_{l=0}^{n-1}\binom{n-1}lp^l(1-p)^{(n-1)-l}\\ &=np\sum_{l=0}^{m}\binom{m}lp^l(1-p)^{m-l}\\ &=np(p+(1-p))^m\\ &=np \end{align} $$

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    One might ask why they waited until the third line to decide the $k=0$ term is zero and can be ignored. The $k$ is already shown as a factor of the general term on the first line. – 2017-01-16
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    @DavidK exactly! Then why do we even define the binomial distribution from $0$ to $n$ and not from $1$ to $n$ at all? – 2017-01-16
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    They are calculating a mean, right? Well, in a mean (of a discrete random variable) you sum over all values that can be taken weighted by their probabilities. If you want to write that out in full, you include the value $0$ times its probability to begin with for illustrative purposes (so you can see all of the values and their probabilities), and then use the facts $0\cdot x=0$ and $0+x=x$ to simplify. – 2017-01-16
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    @arctictern okay fair enough then , thank you all – 2017-01-16
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    A remark: Do you know that this proof not the shortest one ? By far, the simplest is by considering a binomial RV as a sum $X=\sum_{i=1}^nX_i$ of Bernoulli variables Ber(p), thus $E(X)=\sum_{i=1}^nE(X_i)=\sum_{i=1}^n p=np.$ – 2017-01-16
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    @JeanMarie yeah I know thank you! But I have an exam in Statistical Distribution Theory and I need to be able to find these kind of things! – 2017-01-16

2 Answers 2

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Because the term with $k=0$ vanishes.

If $a_k\in\mathbb{R}$ and $a_0=0$ then $$\begin{align*} \sum_{k=0}^na_k &= a_0+a_1+a_2+\cdots+a_n \\ &= 0+(a_1+a_2+\cdots+a_n) \\ &=a_1+a_2+\cdots+a_n \\ &=\sum_{k=1}^na_k. \end{align*}$$

A clearer proof would have begin with this issue first: $$\begin{align*} \mu &= \sum_{k=0}^nk\binom{n}{k}p^k(1-p)^{n-k} \\ &= \sum_{k=1}^nk\binom{n}{k}p^k(1-p)^{n-k} \\ &= \cdots \end{align*} $$

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    I don't understand can you explain a bit more? I see that $k$ vanishes as it is cancelled with the factorial, but why does it mean a change of index? – 2017-01-16
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    @Euler_Salter If $a_0=0$ then $a_0+a_1+\cdots+a_n=a_1+\cdots+a_n$. – 2017-01-16
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    @Euler_Salter See the edit. – 2017-01-16
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    @arctictern true. However the term with $k=0$ was indeed $0$ since the beginning. However once you delete $k$ (by simplifying with the factorial) your term for $k=0$ is non negative, right? – 2017-01-16
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    @Euler_Salter What? I don't understand. What part of $0+x=x$ confuses you? – 2017-01-16
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    @arctictern nothing about it. Just that I don't see why it matters at that point of the proof, since clearly it didn't matter before, when we still had $0+x = 0$ – 2017-01-16
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    @Euler_Salter If an expression that was equal to zero becomes non-zero when you cancel factors within the expression, you did the cancellation wrong. But I share your wonder that they didn't just delete the $k=0$ term right away when it was more obvious. – 2017-01-16
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$$\frac{k}{k!} = \begin{cases} 0 & k=0 \\ \frac{1}{(k-1)!} & k>0 \end{cases}.$$

Thus they have removed the $k=0$ term and simplified the remaining terms.

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    why is $\frac{1}{(k-1)!}$ defined only for strictly positive $k$s? I am pretty sure I saw a question here regarding negative factorials and they are defined! – 2017-01-16
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    @Euler_Salter $k!$ by definition is defined for nonnegative integer $k$. On the other hand, the Gamma function has poles at the nonpositive integers. This can be somewhat loosely interpreted as saying that $|k!|=\infty$ when $k$ is a negative integer, so that $\frac{1}{k!}$ can be understood as $0$ when $k$ is a negative integer. This is consistent with the manipulation that we did here, which didn't require any Gamma function trickery. – 2017-01-16
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    @Euler_Salter Ian did not say $\frac{1}{(k-1)!}$ wasn't defined for $k=0$. In fact, using the gamma function, $(-1)!=\infty$ is a thing we could do, so $0/(-1)!$ would be $0$, exactly as Ian said. Of course when $k=0$ we could just evaluate $0/0!$ directly without having to extend the scope of the formula $k/k!=1/(k-1)!$. – 2017-01-16
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    Negative factorials are not defined. The Gamma function is defined for negative values, but not negative integers. Also, the equality is simply not true if $k=0$, even if $(k-1)!$ was defined as some value. – 2017-01-16
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    But when we had $$\sum_{k=0}^{n}\binom{n}{k}p^k(1-p)^{n-k}$$ then the $k=0$ term, i.e. the first term was zero. So 1. why did we include it since the beginning ? and 2. when we delete $k$ we have $$np\sum_{k=0}^{n}\frac{(n-1)}{(n-1)-(k-1))!(k-1)!}p^{k-1}(1-p)^{(n-1)-(k-1)}$$ which doesn't vanishes for $k=0$, or does it? – 2017-01-16
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    We didn't have that, we had $$\sum_{k=0}^n k\binom{n}{k}p^k(1-p)^{n-k}$$ to start with. $\binom{n}{k}\neq 0$ for $k=0,1,\dots,n$. @Euler_Salter – 2017-01-16
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    @ThomasAndrews yeah but $\binom{n}{k}$ is multiplying by $k$.. – 2017-01-16
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    So what? $k\binom{n}{k}=0$ if $k=0$. The first term of your sequence is not $0$, the first term of the actual sequence is zero. @Euler_Salter – 2017-01-16
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    yes that is what I am saying. The first term, corresponding to the index $k=0$ is equal to $0$, so it could have been neglected since the beginning – 2017-01-16
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    @Euler_Salter Except that term initially arose out of the definition of the expectation: $X$ is supported on $\{ 0,1,\dots,n \}$ so $E[X]=\sum_{k=0}^n k P[X=k]$. – 2017-01-16