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Exercise 2.49

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(a)

\begin{array}{l} P (A_{2}) & = p_{1} p_{2} + q_{1} q_{2} \\ = (1 - q_{1}) (1 - q_{2}) + (b_{1} + \frac{1}{2}) (b_{2} + \frac{1}{2}) \\ = (b_{1} - \frac{1}{2}) (b_{2} - \frac{1}{2}) + (b_{1} + \frac{1}{2}) (b_{2} + \frac{1}{2}) \\ = \frac{1}{2} + 2 b_{1} b_{2} \end{array}

(b)

By strong induction,

P (A_{n}) = \frac{1}{2} + 2^{n - 1} b_{1} b_{2} . . . b_{n}

for $n \leq 2 .$

Suppose the statement holds for all $n \leq k - 1$ . Let $S_{i}$ be the event that the $i$ -th trial is a success.

\begin{array}{l} P (A_{k}) & = p_{k} P (A_{k - 1}^{c} | S_{k}) + q_{k} P (A_{k - 1} | S_{k}^{c}) \\ = p_{k} (1 - (\frac{1}{2} + 2^{k - 2} b_{1} b_{2} . . . b_{k - 1})) + q_{k} (\frac{1}{2} + 2^{k - 2} b_{1} b_{2} . . . b_{k - 1}) \\ = p_{k} (\frac{1}{2} - 2^{k - 2} b_{1} b_{2} . . . b_{k - 1}) + q_{k} (\frac{1}{2} + 2^{k - 2} b_{1} b_{2} . . . b_{k - 1}) \\ = \frac{1}{2} + (q_{k} - p_{k}) 2^{k - 2} b_{1} b_{2} . . . b_{k - 1} \\ = \frac{1}{2} + 2 b_{k} 2^{k - 2} b_{1} b_{2} . . . b_{k - 1} \\ = \frac{1}{2} + 2^{k - 1} b_{1} b_{2} . . . b_{k - 1} b_{k} \end{array}

(c)

p_{i} = \frac{1}{2}

for some

i

, then

b_{i} = 0

and

P (A_{n}) = \frac{1}{2} .

if $p_{i} = 0$ for all $i$ , then $b_{i} = \frac{1}{2}$ for all $i$ . Hence, the term $2^{k - 1} b_{1} b_{2} . . . b_{k - 1} b_{k}$ equals $\frac{1}{2}$ . Thus, $P (A_{n}) = 1 .$ This makes sense since the number of successes will be $0$ , which is an even number.

if $p_{i} = 1$ for all $i$ , then $b_{i} = - \frac{1}{2}$ for all $i$ . Hence, the term $2^{k - 1} b_{1} b_{2} . . . b_{k - 1} b_{k}$ will either equal to $\frac{1}{2}$ or $- \frac{1}{2}$ depending on the parity of the number of trials. Thus, $P (A_{n})$ is either $0$ or $1$ depending on the parity of the number of trials.

This makes sense since, if every trial is a success, the number of successes will be even if the number of trials is even. The number of successes will be odd otherwise.

fifthist

2021-12-05 00:00

Exercise 2.49

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