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16.3 Problems on Conditional Independence, Given a Random Vector

Random samples whose common distribution depends on an unobserved parameter, updating beliefs about that parameter from the observed outcomes.

Applied Probability, 16.3 Problems on Conditional Independence, Given a Random Vector

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6 exercises

16.3 Exercise 1

The pair {X,Y} ci |H\left\{X , Y\right\} ci \left|\right. H. X∼X \sim exponential (u/3)\left(u / 3\right), given H=uH = u; Y∼Y \sim exponential (u/5)\left(u / 5\right), given H=uH = u; and H∼H \sim uniform [1,2]\left[1 , 2\right]. Determine a general formula for P(X>r,Y>s)P \left(X > r , Y > s\right), then evaluate for r=3,s=10r = 3 , s = 10.

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16.3 Exercise 2

A small random sample of size n=12n = 12 is taken to determine the proportion of the student body which favors a proposal to expand the student Honor Council by adding two additional members “at large.” Prior information indicates that this proportion is about 0.6 = 3/5. From a Bayesian point of view, the population proportion is taken to be the value of a random variable H. It seems reasonable to assume a…

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16.3 Exercise 3

Let {Xi:1≤i≤n}\left\{X_{i} : 1 \leq i \leq n\right\} be a random sample, given H. Set W=(X1,X2,⋯,Xn)W = \left(X_{1} , X_{2} , \hdots , X_{n}\right). Suppose X conditionally geometric (u)\left(u\right), given H=u; H = u; i.e., suppose P(X=k|H=u)=u(1-u)kP \left(\right. X = k \left|\right. H = u \left.\right) = u \left(1 - u\right)^{k} for all k≥0k \geq 0. If H∼H \sim uniform
on [0, 1], determine the best mean square estimator for H, given W.

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16.3 Exercise 4

Let {Xi:1≤i≤n}\left\{X_{i} : 1 \leq i \leq n\right\} be a random sample, given H. Set W=(X1,X2,⋯,Xn)W = \left(X_{1} , X_{2} , \hdots , X_{n}\right). Suppose X conditionally Poisson (u)\left(u\right), given H=u; H = u; i.e., suppose P(X=k|H=u)=e-uuk/k!P \left(\right. X = k \left|\right. H = u \left.\right) = e^{- u} u^{k} / k !. If H∼H \sim gamma (m,λ)\left(m , \lambda\right), determine the best mean square estimator for H, given W.

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16.3 Exercise 5

Suppose {N,H}\left\{N , H\right\} is independent and {N,Y} ci |H\left\{N , Y\right\} ci \left|\right. H. Use properties of conditional expectation and conditional independence to show that

E [ g ( N ) h ( Y ) | H ] = E [ g ( N ) ] E [ h ( Y ) | H ] a . s . E \left[\right. g \left(N\right) h \left(Y\right) \left|\right. H \left]\right. = E \left[g \left(N\right)\right] E \left[\right. h \left(Y\right) \left|\right. H \left]\right. a . s .
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16.3 Exercise 6

Consider the composite demand D introduced in the section on Random Sums in "Random Selecton"

D = ∑ n = 0 ∞ I { k } ( N ) X n where X n = ∑ k = 0 n Y k , Y 0 = 0 D = \sum_{n = 0}^{\infty} I_{\left\{k\right\}} \left(N\right) X_{n} \text{where} X_{n} = \sum_{k = 0}^{n} Y_{k} , Y_{0} = 0

Suppose {N,H}\left\{N , H\right\} is independent, {N,Yi} ci |H\left\{N , Y_{i}\right\} ci \left|\right. H for all i, and E[Yi|H]=e(H)E \left[\right. Y_{i} \left|\right. H \left]\right. = e \left(H\right), invariant with i. Show that E[D|H]=E[N]E[Y|H]a.s.E \left[\right. D \left|\right. H \left]\right. = E \left[N\right] E \left[\right. Y \left|\right. H \left]\right. a . s ..

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Adapted from Applied Probability © 2009 Paul E. Pfeiffer. Licensed under CC BY 3.0. Source: Applied Probability. Imported at revision d7a0420.