Chapter 7: Q.7.61 (page 357)

Let $X 1, . . .$ be independent random variables with the common distribution function $F$ , and suppose they are independent of $N$ , a geometric random variable with a parameter $p$ . Let $M = m a x (X 1, . . ., X N) .$
(a) Find $P {M \dots x}$ by conditioning on $N$ .
(b) Find $P {M \dots x | N = 1}$ .
(c) Find $P {M \dots x | N > 1}$
(d) Use (b) and (c) to rederive the probability you found in (a)

Short Answer

Expert verified

a) $P {M \leq x} = \frac{N F (e)}{1 - (∣ P) F (x)}$

b) $P {M \leq x ∣ N = 1} = F (x)$

c) $P {M \leq x ∣ N > 1} = F (x) P {M \leq x}$

d) The probability of part(a)is rederived by using (b) and (c)as $P {M \leq x} = \frac{F (x) p}{1 - (1 - p) F (x)}$

Step by step solution

Step 1:Given Information(part a)

Given that the distribution function $F$ , and suppose they are independent of $N$ , a geometric random variable with parameter $p$ . Let $M = m a x (X 1, . . ., X N) .$

Step 2:Explanation(part a)

Discover by $p {M . . . x}$ conditioning on $N$

$P {M \leq x} = \sum_{n = 1}^{\infty} P {M \leq x ∣ N = n} P {N = n}$

$S n \ d i s p l a y s t y l e \ {s u m_\ {n = 1 \} * \ {\ i n f t y \} $ F^{\ w e d g e} \ {n \} (x) p (1 - p)^{*} \ {n - 1 \} s $$

$$ = ∖ frac {p F (x)} {1 - (1 - p) F (x)} $$

Step 3:Final Answer(part a)

$p {M . . . x}$ by conditioning on $N$ is

P {M \leq x} = \sum_{n = 1}^{\infty} P {M \leq x ∣ N = n} P {N = n}

Step 4:Given Information(part b)

Given that $M = m a x (X 1, . . ., X N)$ and the parameter $p$ .

Step 5:Explanation(part b)

Discover $P {M \leq x ∣ N = 1}$

We have

P {M \leq x ∣ N = 1} = F (x)

Step 6:Final Answer(part b)

$P {M \leq x ∣ N = 1} = F (x)$

Step 7:Given Information(part c)

Given that $M \leq x$ and $N > 1$ .

Step 8:Explanation(part c)

Discover $P {M \leq x ∣ N > 1}$

We have

P {M \leq x ∣ N > 1} = F (x) P {M \leq x}

Step 9:Final Answer(part c)

$P {M \leq x ∣ N > 1} = F (x) P {M \leq x}$

Step 10:Given Information(part d)

Given that $P {M \leq x ∣ N = 1}$ and $P {M \leq x ∣ N > 1}$ .

Step 11:Explanation(part d)

$P {M \leq x} \cdot P {M \leq x ∣ N = 1} P {N = 1} \cdot P {M \leq x ∣ N > 1} P {N > 1}$

$= F (x) p + F (x) P {M \leq x} (1 - p)$

Hence,

$P {M \leq x} = \frac{F (x) p}{1 - (1 - p) F (x)}$

Step 12:Final Answer(part d)

The probability of part(a)is rederived by using (b) and (c)as $P {M \leq x} = \frac{F (x) p}{1 - (1 - p) F (x)}$

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Short Answer

Step by step solution

Step 1:Given Information(part a)

Step 2:Explanation(part a)

Step 3:Final Answer(part a)

Step 4:Given Information(part b)

Step 5:Explanation(part b)

Step 6:Final Answer(part b)

Step 7:Given Information(part c)

Step 8:Explanation(part c)

Step 9:Final Answer(part c)

Step 10:Given Information(part d)

Step 11:Explanation(part d)

Step 12:Final Answer(part d)

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