Relationship between binomial and negative binomial probabilities

Question

Let $X$ be a negative binomial random variable with parameters $r$ and $p$, and let $Y$ be a binomial random variable with parameters $n$ and $p$. Show that $$ \mathbb{P}(X > n) = \mathbb{P}(Y < r). $$

I would like to get an analytic solution. Basically I want to show the following equality mathematically: $$ \sum_{i=n+1}^\infty \binom{i-1}{r-1} p^r (1-p)^r = \sum_{i=0}^{r-1} \binom{n}{i} p^i (1-p)^{n-i} $$

Answer 1

$X$ is the minimum number of trials needed to get $r$ successes and $Y$ is the number of successes in $n$ trials. Then it is clear that $Y<r\iff X>n$ from which the result follows.

Answer 2

We want to show that $$ 1-\sum_{j=r}^n \binom{j-1}{r-1} p^r (1-p)^{j-r} = \sum_{j=0}^{r-1} \binom{n}{j} p^j (1-p)^{n-j} $$ Equivalently $$ \sum_{j=r}^n \binom{j-1}{r-1} (1-p)^{j-r} = \sum_{j=r}^{n} \binom{n}{j} p^{j-r} (1-p)^{n-j} $$ For $n=r$, LHS = 1 = RHS. Assume the result is true for $n = r + m$, we prove it is true for $n = r + m + 1$.
LHS = $$ \sum_{j=r}^{r+m+1} \binom{j-1}{r-1} (1-p)^{j-r} = \sum_{j=r}^{r+m} \binom{j-1}{r-1} (1-p)^{j-r} + \binom{r+m}{r-1} (1-p)^{m+1}\\ =^{(Assumption)} \sum_{j=r}^{r+m} \binom{r+m}{j} p^{j-r} (1-p)^{r+m-j} + \binom{r+m}{r-1} (1-p)^{m+1} \\ =(*) \sum_{j=r}^{r+m} \binom{r+m+1}{j} p^{j-r} (1-p)^{r+m-j} - \sum_{j=r}^{r+m} \binom{r+m}{j-1} p^{j-r} (1-p)^{r+m-j} + \binom{r+m}{r-1} (1-p)^{m+1} + p \sum_{j=r}^{r+m} \binom{r+m+1}{j} p^{j-r} (1-p)^{r+m-j} - p\sum_{j=r}^{r+m} \binom{r+m+1}{j} p^{j-r} (1-p)^{r+m-j} \\ = \sum_{j=r}^{r+m} \binom{r+m+1}{j} p^{j-r} (1-p)^{r+m+1-j} + \sum_{j=r}^{r+m} \binom{r+m+1}{j} p^{1+j-r} (1-p)^{r+m-j} - \sum_{j=r}^{r+m} \binom{r+m}{j-1} p^{j-r} (1-p)^{r+m-j} + \binom{r+m}{r-1} (1-p)^{m+1} + p^{m+1} - p^{m+1}\\ = \sum_{j=r}^{r+m+1} \binom{r+m+1}{j} p^{j-r} (1-p)^{r+m+1-j} + A. $$ Here $$A = (*) \sum_{j=r}^{r+m} \Bigg\{ \binom{r+m}{j} + \binom{r+m}{j-1} \Bigg\} p^{1+j-r} (1-p)^{r+m-j} - \sum_{j=r}^{r+m} \binom{r+m}{j-1} p^{j-r} (1-p)^{r+m-j} + \binom{r+m}{r-1} (1-p)^{m+1} - p^{m+1}\\ = \sum_{j=r}^{r+m} \binom{r+m}{j} p^{1+j-r} (1-p)^{r+m-j} - \sum_{j=r}^{r+m} \binom{r+m}{j-1} p^{j-r} (1-p)^{r+m+1-j} + \binom{r+m}{r-1} (1-p)^{m+1} - p^{m+1}\\ = \sum_{j=r}^{r+m} \binom{r+m}{j} p^{1+j-r} (1-p)^{r+m-j} - \sum_{j'=r-1}^{r+m-1} \binom{r+m}{j'} p^{j'-r+1} (1-p)^{r+m-j'} + \binom{r+m}{r-1} (1-p)^{m+1} - p^{m+1} = 0. $$ The equalities with $(*)$ make use of [Pascal rule]https://en.wikipedia.org/wiki/Pascal%27s_rule#:~:text=In%20mathematics%2C%20Pascal's%20rule%20(or,natural%20numbers%20n%20and%20k%2C&text=is%20a%20binomial%20coefficient%3B%20one,(1%20%2B%20x)n.