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Conditional Expectations of 2 Variables
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[QUOTE="transmini, post: 6058264, member: 556213"] [h2]Homework Statement [/h2] Suppose that the number of eggs laid by a certain insect has a Poisson distribution with mean ##\lambda##. The probability that anyone egg hatches is ##p##. Assume that the eggs hatch independently of one another. Find the expected value of ##Y##, the total number of eggs that hatch. [h2]Homework Equations[/h2] ##E(Y_1) = E(E(Y_1|Y_2))## [h2]The Attempt at a Solution[/h2] Call ##Y_1 = Y## and ##Y_2 = N## with ##N## being the number of eggs laid. Then ##E(Y|N=n) = \sum_{y=0}^{n} y(\frac{n!}{y!(n-y)!})p^y (1-p)^{n-y} = np## since given that ##n## eggs are laid, the number of eggs hatching has a constant probability ##p## and thus the number of eggs that hatch would be a binomial distribution. so ##E(Y) = E(E(Y|N=n)) = E(np) = pE(n) = p\sum_{n=0}^\infty ne^{-\lambda}\frac{\lambda^n}{n!}## since the number of eggs laid is modeled by a Poisson distribution. and then ##E(Y) = p\lambda## The book itself doesn't have an answer listed in the back of the book and I don't have a solutions manual, so I was relying on Chegg to check my answer. However, Chegg lists the answer as ##(1-p)e+p## and models ##\lambda## as a Bernoulli random variable. I don't see how making ##\lambda## a Bernoulli random variable makes sense, as nowhere in the problem do they mention ##\lambda## varies; it's always the same mean. Is my answer incorrect and Chegg's correct? If so, why would we follow Chegg's solution, rather than my own? Or is this just one of the many problem's Chegg has a wrong answer to? [/QUOTE]
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