(5) A bivariate density isx² y?, if z € (-1, 1), 0 < y < æ?f(2, y) :=otherwise.The marginal densities of X and Y are9fx(x) = 5=x € (-1,1);fy(1) = 9y²(1 – y³/2),0 < y < 1.Obtain the conditional density fx\y=y (x).The following answers are proposed.(a) For y € (0, 1), the conditional density of X given Y = y is3if y < x? < 1,fx)Y= (x) = { 2(1–g8/2)otherwise.(b) For y E (x?, 1), the conditional density of X given Y = y is32(1-y3/2)if 0 < y < 1,fx}Y=y(x) =otherwise.(c) For y € (r?, 1), the conditional density of X given Y = y isif 0 < y < 1,fxyY=y (2) = { 2(1+y3/2)otherwise..kent.edu/-kazim/login.phpHome Work(d) For y e (0, 1), the conditional density of X given Y = y is3y?if y < a? < 1,fx\Y=y(x) = { 2(1+y/2)otherwise.(e) None of the aboveThe correct answer is(a)(b)(c)(d)(e)N/A

The conditional density fX|Y = y(x) is defined as fX|Y = y(x) = f(x, y)/fY(y)

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Obtain the conditional density fxY=y(2).

A First Course in Probability (10th Edition)

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Question

(5) A bivariate density is
x² y?, if z € (-1, 1), 0 < y < æ?
f(2, y) :=
otherwise.
The marginal densities of X and Y are
9
fx(x) = 5=
x € (-1,1);
fy(1) = 9y²(1 – y³/2),
0 < y < 1.
Obtain the conditional density fx\y=y (x).
The following answers are proposed.
(a) For y € (0, 1), the conditional density of X given Y = y is
3
if y < x? < 1,
fx)Y= (x) = { 2(1–g8/2)
otherwise.
(b) For y E (x?, 1), the conditional density of X given Y = y is
3
2(1-y3/2)
if 0 < y < 1,
fx}Y=y(x) =
otherwise.
(c) For y € (r?, 1), the conditional density of X given Y = y is
if 0 < y < 1,
fxyY=y (2) = { 2(1+y3/2)
otherwise.
.kent.edu/-kazim/login.php
Home Work
(d) For y e (0, 1), the conditional density of X given Y = y is
3y?
if y < a? < 1,
fx\Y=y(x) = { 2(1+y/2)
otherwise.
(e) None of the above
The correct answer is
(a)
(b)
(c)
(d)
(e)
N/A

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