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

$P\{X>n\}=P\{Y<r\}$

Hint: Either one could attempt an analytical proof of the preceding equation, which is equivalent to proving the identity

$\begin{array}{r}\sum _{i=n+1}^{\mathrm{\infty }} \left(\begin{array}{c}i-1\\ r-1\end{array}\right)p^r(1-p{)}^{i-r}=\sum _{i=0}^{r-1} \left(\begin{array}{c}n\\ i\end{array}\right)\\ \times p^i(1-p{)}^{ni}\end{array}$

or one could attempt a proof that uses the probabilistic interpretation of these random variables. That is, in the latter case, start by considering a sequence of independent trials having a common probability p of success. Then try to express the events to express the events $\{X>n\}$and $\{Y<r\}$in terms of the outcomes of this sequence.

Short answer

We have proved that

$P(X>n)=P(Y<r)$

Step-by-step solution

Given information

We are going to prove that events $X>n$ and $Y<r$ are equivalent. As a consequence, these events will have the same probabilistic measure.

Explanation

If $X>n$, that means that we needed more than $n$attempts to reach $r$successes that happens with probability $p$. That implies that in $n$attempts we made strictly less that $r$successes, which is exactly $Y<r$.

On the other hand, if $Y<r$, that means that in $n$attempts we made strictly less that $r$successes. So, in order to reach $r$successes, we have to go on with our trials. Hence, the total number of trials until we reach $r$successes will be strictly greater that $n$. That is exactly $X>n$.

Final answer

So, we have proved that $\{X>n\}=\{Y<r\}$which implies

$P(X>n)=P(Y<r)$