Stat150_Spring08_homework1

Stat150_Spring08_homework1 - with mean 1 and X 2 is an...

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Stat 150 (Stochastic Processes) Spring 2008 Homework # 1 Let X be a random variable taking values in the nonnegative inte- gers. The probability generating function (pgf) of X is the function φ : [0 , 1] 7→ [0 , 1] defined by φ ( s ) = E [ s X ] = X k =0 s k P { X = k } , s [0 , 1] . 1) Show that the pgf of a Poisson random variable with mean λ is s 7→ exp( λ ( s - 1)). 2) Show that the pgf of a random variable X with the geometric distri- bution P { X = k } = (1 - p ) k p, k = 0 , 1 , 2 , . . . , is s 7→ p/ (1 - (1 - p ) s ). 3) Explain why the pgf of a random variable uniquely determines the distribution of the random variable. 4) Show that E [ X ] = φ 0 (1), where φ 0 is the first derivative of φ . 5) Show that if X 1 , X 2 , . . . , X m are independent nonnegative integer valued random variables with pgf’s φ 1 , φ 2 , . . . , φ m , respectively, then the random variable X 1 + X 2 + ··· + X m has pgf φ ( s ) = φ 1 ( s ) φ 2 ( s ) . . . φ m ( s ) . 6) Show using, (1), (3) and (5), that if X 1 is a Poisson random variable
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Unformatted text preview: with mean λ 1 and X 2 is an independent Poisson random variable with mean λ 2 , then X 1 + X 2 is a Poisson random variable with mean λ 1 + λ 2 . 7) Suppose that X 1 , X 2 , . . . is a sequence of independent, identically distributed nonnegative integer valued random variables with common pgf φ , and suppose that N is an independent nonnegative integer valued random variable with pgf ψ . Show that the random variable ∑ N i =1 X i has pgf γ ( s ) = ψ ( φ ( s )). (Hint: First condition on the value of N and use (5).) 8) In the set-up of (7), suppose that X 1 , X 2 , . . . are Bernoulli with success probability p and N is Poisson with mean λ . Show that ∑ N i =1 X i is Poisson with mean λp . 1...
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