 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability-resit/7/b/ii/solution

Put

D = 1 - (1 - p) (1 - q) = p + q - pq

P (Y = k) = n = k \sum \infty p (1 - p)^{n} (k n) q^{k} (1 - q)^{n - k} = p [(1 - p) q]^{k} n = k \sum \infty (k n) [(1 - p) (1 - q)]^{n - k} .

For

0 \leq t < 1

\sum_{n = k}^{\infty} (k n) t^{n - k} = \frac{1}{k !} \frac{d ^{k}}{d t ^{k}} \frac{1}{1 - t} = \frac{1}{( 1 - t ) ^{k + 1}} .

Taking

t = (1 - p) (1 - q)

therefore yields

P (Y = k) = \frac{p}{D} (\frac{( 1 - p ) q}{D})^{k}, k \geq 0.

Since

D - p = (1 - p) q

p_{*} = \frac{p}{p + q - pq}, P (Y = k) = p_{*} (1 - p_{*})^{k} .

0 < p_{*} < 1

0 < p, q < 1

n \geq k

P (X = n ∣ Y = k) = \frac{p ( 1 - p ) ^{n} ( k n ) q ^{k} ( 1 - q ) ^{n - k}}{p [( 1 - p ) q ] ^{k} / D ^{k + 1}} = (k n) [(1 - p) (1 - q)]^{n - k} D^{k + 1} .

Thus

P (X = n ∣ Y = k) = h_{n, k} (r), r = D = p + q - pq .

For

n < k

X - k

is the failure-count form of a negative binomial distribution, with

k + 1

successes and success parameter

r

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