Q. 4.28 - A First Course in Probability

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.

Step-by-Step Solution

Answer

We have proved that

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

Step 1 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.

Step 2 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 $X>n$.

Step 3 Final answer

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

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