Snapsolve any problem by taking a picture.
Try it in the Numerade app?
(d) Sanity check: Verify that the above density is a proper probability density function. Compute the expectation and variance (called conditional expectation and variance of Y given Z = 2 for Z > 0) and check that: 2 E(Y|Z) = 2 = 2 + 1 2 Var(Y|Z = 2) = (Z + 1)^2 (e) Find the probability distribution function of U. [Hint] You may distinguish the cases U ≥ 0 and U < 0 and show: 1 - 1+0, U ≥ 0 e^(u/a), U < 0 F_U(u) (f) Compute the joint density of (Y,U) for non-negative values of U. (g) Show that the conditional density P(Y|U) for non-negative values of U is given by: D_Y|U(y|u) = (1 + a)e^(-u/a) * 1(y ≥ 0) (h) Explain why the events {Z = a} and {X = a} are the same. Are the conditional densities P(Y|Z)(y|a) and P(Y|U=a)(y|a) equal? What do you think of that result?
Submitted by Kathryn G. Sep. 24, 2021 04:55 a.m.
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