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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**Additional info for Analogies Between Analogies: The Mathematical Reports of S.M. Ulam and his Los Alamos Collaborators**

**Sample text**

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)!