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Aysun PINARBAŞI Probabilistic Models Submitted to: Assist

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1 Aysun PINARBAŞI 17500179 Probabilistic Models Submitted to: Assist
Aysun PINARBAŞI Probabilistic Models Submitted to: Assist. Prof. Dr. Sahand DANESHVAR

2 Computing Probabilities by Conditioning
From the definition of X that E[X]= P(E), E[X|Y=y] = P(E|Y=y), for any random variable Y

3 Example: Suppose that X and Y are independent continuous random variables having densities 𝑓 𝑥 and 𝑓 𝑦 , respectively. Compute P { X<Y }.

4 Example: - An insurance company supposes that the number of accidents that each of its policyholders will have in a year is Poisson distributed, with the mean of the Poisson depending on the policyholder. If the Poisson mean of a randomly chosen policyholder has a gamma distribution with density function g(λ) = λ 𝑒 −λ , λ ≥ 0 what is the probability that a randomly chosen policyholder has exactly n accidents next year?

5 Solution: X: the number of accidents that a randomly chosen policyholder has next year Y: the Poisson mean number of accidents for this policyholder Then conditioning on Y yields:

6

7 Example: Suppose that the number of people who visit a yoga studio each day is a Poisson random variable with mean λ. Suppose that each person who visits is, independently, female with probability p or male with probability 1−p. Find the joint probability that exactly n women and m men visit the academy today. Solution: 𝑁 1 : the number of women 𝑁 2 : the number of men N = 𝑁 1 + 𝑁 2 be the total number of people who visit

8 Conditioning on N gives;
P { 𝑁 1 = n, 𝑁 2 = m | N = i } = 0 when i = n + m , the preceding equation yields that

9 Given that n + m people visit it follows, because each of these n + m is independently a woman with probability p, that the conditional probability that n of them are women (and m are men) is just the binomial probability of n successes in n+m trials. Therefore; Because the preceding joint probability mass function factors into two products, one of which depends only on n and the other only on m, it follows that 𝑁 1 and 𝑁 2 are independent.

10 Moreover, 𝑁 1 and 𝑁 2  independent Poisson random variables with respective means λp and λ( 1 − p) So; The numbers of type 1 and type 2 events are independent Poisson random variables as being type 1 with probability p or type 2 with probability 1 − p.

11 Thank You

12 References Introduction to Probabilistic Models, Sheldon M. Ross


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