Stochastic Modeling

Download Analogies Between Analogies: The Mathematical Reports of by S. M. Ulam, A. R. Bednarek, Françoise Ulam PDF

By S. M. Ulam, A. R. Bednarek, Françoise Ulam

Many of the rules provided preserve their value this day, and . . . are totally fundmental, either from a old and from a systematic viewpoint.--Gian-Carlo Rota, Massachusetts Institute of expertise

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The following list summarizes some properties of conditional expectations. In this list, with or without affixes, X and Y are jointly distributed random variables; c is a real number; g is a function for which E[|g(X)|] < ∞; h is a bounded function; and v is a function of two variables for which E[|v(X, Y)|] < ∞. The properties are 1. E[c1 g1 (X1 ) + c2 g2 (X2 )|Y = y] = c1 E[g1 (X1 )|Y = y] + c2 E[g2 (X2 )|Y = y]. 2. if g ≥ 0, then E[g(X)|Y = y] ≥ 0. 3. E[v(X, Y)|Y = y] = E[v(X, y)|Y = y]. 4. E[g(X)|Y = y] = E[g(X)] if X and Y are independent.

Obviously, Pr{N = 1} = 1 − Pr{N = 0} = λ1 /(λ0 + λ1 ). (c) Pr{U > t} = e−(λ0 +λ1 )t , t ≥ 0. Upon adding the result in (a), Pr{N = 0 and U > t} = e−(λ0 +λ1 )t λ0 , λ0 + λ1 to the corresponding quantity associated with N = 1, Pr{N = 1 and U > t} = e−(λ0 +λ1 )t λ1 , λ0 + λ1 we obtain the desired result via Pr{U > t} = Pr{N = 0, U > t} + Pr{N = 1, U > t} = e−(λ0 +λ1 )t = e−(λ0 +λ1 )t . λ1 λ0 + λ0 + λ1 λ0 + λ1 Introduction 39 At this point observe that U and N are independent random variables. This follows because (a), (b), and (c) together give Pr{N = 0 and U > t} = Pr{N = 0} × Pr{U > t}.

M. In words, X has a binomial distribution with parameters M and pq. Example Suppose X has a binomial distribution with parameters p and N, where N has a Poisson distribution with mean λ. What is the marginal distribution for X? Proceeding as in the previous example but now using pN (n) = λn e−λ , n! n = 0, 1, . . , we obtain ∞ pX|N (k|n)pN (n) Pr{X = k} = n=0 ∞ = n=k n! λn e−λ pk (1 − p)n−k k! (n − k)! n! Conditional Probability and Conditional Expectation = λk e−λ pk k! ∞ n=k 49 [λ(1 − p)]n−k (n − k)!

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