I'm studying my teacher's lecture and got stuck at the proof of this equation, how can I prove it? one way I think is something like this but I don't it will lead to the correct answer or not: $\sum _{k=0}^{n}Binomial(n-1,p) = \sum _{k=0}^{n}{n-1 \choose k} p^k q^(n-1-k) = \sum _{k=0}^{n}(p+q)^k because \ p=q+1 \ then \Rightarrow \sum _{k=0}^{n} 1 ^k ....$ but I don't know how to continue it!

note that p is probability and q = 1-p


closed as unclear what you're asking by uniquesolution, Alexander Gruber Apr 29 at 23:22

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  • $\begingroup$ Something looks pretty wrong here -- the index variable $k$ does not even appear in the term you're summing. $\endgroup$ – Henning Makholm Apr 29 at 12:55
  • $\begingroup$ it's a binomial distribution that has "k" inside it @HenningMakholm $\endgroup$ – Peyman Tahghighi Apr 29 at 12:56
  • $\begingroup$ And what does $\mathit{Binomial}(n-1,p)$ mean if it's not $\binom{n-1}{p}$ (since you rolled back an edit changing it to that)? $\endgroup$ – Henning Makholm Apr 29 at 12:56
  • $\begingroup$ @HenningMakholm i corrected it. $\endgroup$ – Peyman Tahghighi Apr 29 at 12:58
  • $\begingroup$ Can you try again ? We really don't get what you are saying, in the summation you don't have any $k$, you don't define $p$ and $q$. What is $Binomial(n-1,p)$ ? $\endgroup$ – P. Quinton Apr 29 at 12:59

As a quick aside, I find it quite unfortunate that people use $\text{Distribution Name}(\text{parameters})$ to refer to the PMF/PDF, hence confusion in the comments.

We have (as much as I despise the notation) $$\text{Binomial}(n, p) = \binom{n}{k}p^k(1-p)^{n-k} = \binom{n}{k}p^kq^{n-k}\text{.}$$ with $q = 1 - p$.

What you are trying to show with this question is that $\text{Binomial}(n-1, p)$ is a valid PMF, or $$\sum_{k=0}^{n-1}\text{Binomial}(n-1, p) = 1\text{.}$$ (Notice that I'm not caring about the $n$ in the end of the summation... that's just zero because that lies outside of the support of the random variable which follows this distribution.)

Now $$\sum_{k=0}^{n-1}\text{Binomial}(n-1, p) = \sum_{k=0}^{n-1}\binom{n-1}{k}p^kq^{n-1-k} = (p+q)^{n-1}$$ by the Binomial Theorem. Since $p + q = 1$, it follows that $(p+q)^{n-1} = 1$, and the proof is finished.


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