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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

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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
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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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