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Section 7.4 Negative Binomial

Consider the situation where one can observe a sequence of independent trials where the likelihood of a success on each individual trial stays constant from trial to trial. Call this likelihood the probably of "success" and denote its value by \(p\) where \(0 \lt p \lt 1 \text{.}\) If we let the variable \(X\) measure the number of trials needed in order to obtain the rth success, \(r \ge 1\text{,}\) with \(R = \{r, r+1, r+2, ... \}\) then the resulting distribution of probabilities is called a Geometric Distribution.

Note that r=1 gives the Geometric Distribution.

First, convert the problem to a slightly different form: \(\frac{1}{(a+b)^n} = \frac{1}{b^n} \frac{1}{(\frac{a}{b}+1)^n} = \frac{1}{b^n} \sum_{k=0}^{\infty} {(-1)^k \binom{n + k - 1}{k} \left ( \frac{a}{b} \right ) ^k}\)

So, let's replace \(\frac{a}{b} = x\) and ignore for a while the term factored out. Then, we only need to show

\begin{gather*} \sum_{k=0}^{\infty} {(-1)^k \binom{n + k - 1}{k} x^k} = \left ( \frac{1}{1+x} \right )^n \end{gather*}

However

\begin{align*} \left ( \frac{1}{1+x} \right )^n & = \left ( \frac{1}{1 - (-x)} \right )^n \\ & = \left ( \sum_{k=0}^{\infty} {(-1)^k x^k} \right )^n \end{align*}

This infinite sum raised to a power can be expanded by distributing terms in the standard way. In doing so, the various powers of x multiplied together will create a series in powers of x involving \(x^0, x^1, x^2, ...\text{.}\) To detemine the final coefficients notice that the number of time \(m^k\) will appear in this product depends upon the number of ways one can write k as a sum of nonnegative integers.

For example, the coefficient of \(x^3\) will come from the n ways of multiplying the coefficients \(x^3, x^0, ..., x^0\) and \(x^2, x^1, x^0, ..., x^0\) and \(x^1, x^1, x^1, x^0,..., x^0\text{.}\) This is equivalent to finding the number of ways to write the number k as a sum of nonnegative integers. The possible set of nonnegative integers is {0,1,2,...,k} and one way to count the combinations is to separate k *'s by n-1 |'s. For example, if k = 3 then *||** means \(x^1 x^0 x^2 = x^3\text{.}\) Similarly for k = 5 and |**|*|**| implies \(x^0 x^2 x^1 x^2 x^0 = x^5\text{.}\) The number of ways to interchange the identical *'s among the idential |'s is \(\binom{n+k-1}{k}\text{.}\)

Furthermore, to obtain an even power of x will require an even number of odd powers and an odd power of x will require an odd number of odd powers. So, the coefficient of the odd terms stays odd and the coefficient of the even terms remains even. Therefore,

\begin{gather*} \left ( \frac{1}{1+x} \right )^n = \sum_{k=0}^{\infty} {(-1)^k \binom{n + k - 1}{k} x^k} \end{gather*}

Similarly, \(\left ( \frac{1}{1-x} \right )^n = \left ( \sum_{k=0}^{\infty} {x^k} \right )^n = \sum_{k=0}^{\infty} {\binom{n + k - 1}{k} x^k}\)

Consider the situation where one can observe a sequence of independent trials with the likelihood of a success on each individual trial \(p\) where \(0 \lt p \lt 1 \text{.}\) For a positive integer r, let the variable \(X\) measure the number of trials needed in order to obtain the rth success. Then the resulting distribution of probabilities is called a Negative Binomial Distribution.

Since successive trials are independent, then the probability of the rth success occurring on the x-th trial presumes that in the previous x-1 trials were r-1 successes and x-r failures. You can arrange these indistinguishable successes (and failures) in \(\binom{x-1}{r-1}\) unique ways. Therefore the desired probability is given by

\begin{gather*} P(X=x) = \binom{x - 1}{r-1}(1-p)^{x-r}p^r \end{gather*}
\begin{equation*} \sum_{x=r}^{\infty} {\binom{x - 1}{r-1}(1-p)^{x-r}p^r} = p^r \sum_{x=r}^{\infty} {\binom{x - 1}{r-1}(1-p)^{x-r}} \end{equation*}

and by using \(k = x-r\)

\begin{align*} & = p^r \sum_{k=0}^{\infty} {\binom{r + k - 1}{k}(1-p)^k}\\ & = p^r \frac{1}{(1-(1-p))^r}\\ & = 1 \end{align*}

Utilize the interactive cell below to compute f(x) and F(x) for the negative binomial distribution.