1. Suppose that the number of red blood corpuscles in humans, denoted by X, follows a Poisson distribution whose parameter depends on the observed individual. For a person selected at random we may consider the parameter value Y as an that Y y, we have X Exp(a) random variable such that, given Poisson(y); namely, Exp(a) X | Y yPoisson(y), with Y Here a is a positive constant and the density function of Exp(a) is f(y)aeay, y > 0 (a) Find the conditional expectation E(X|Y = y), where y 0 (b) Compute the expectation E(X). Remark: You may use the law of total expectation. (c) Find the distribution of X. Remark: You may use the law of total probability: P(X k) = So P(X k|Y y)fy (y)dy

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Chapter1: Combinatorial Analysis
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1. Suppose that the number of red blood corpuscles in humans, denoted
by X, follows a Poisson distribution whose parameter depends on the
observed individual. For a person selected at random we may consider
the parameter value Y as an
that Y y, we have X
Exp(a) random variable such that, given
Poisson(y); namely,
Exp(a)
X | Y yPoisson(y), with Y
Here a is a
positive constant and the density function of Exp(a) is
f(y)aeay, y > 0
(a) Find the conditional expectation E(X|Y = y), where y 0
(b) Compute the expectation E(X). Remark: You may use the law
of total expectation.
(c) Find the distribution of X. Remark: You may use the law of
total probability: P(X k) = So P(X k|Y y)fy (y)dy
Transcribed Image Text:1. Suppose that the number of red blood corpuscles in humans, denoted by X, follows a Poisson distribution whose parameter depends on the observed individual. For a person selected at random we may consider the parameter value Y as an that Y y, we have X Exp(a) random variable such that, given Poisson(y); namely, Exp(a) X | Y yPoisson(y), with Y Here a is a positive constant and the density function of Exp(a) is f(y)aeay, y > 0 (a) Find the conditional expectation E(X|Y = y), where y 0 (b) Compute the expectation E(X). Remark: You may use the law of total expectation. (c) Find the distribution of X. Remark: You may use the law of total probability: P(X k) = So P(X k|Y y)fy (y)dy
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