Question

Exercise 10.33. Let (X,Y) be uniformly distributed on the triangleD with vertices (1,0), (2,0) and (0,1), as in Example 10.19. (a) Find the conditional probability P(X ≤ 1 2|Y =y). You might first deduce the answer from Figure 10.2 and then check your intuition with calculation. (b) Verify the averaging identity for P(X ≤ 1 2). That is, check that P(X ≤ 1 2)=:∞ −∞ P(X ≤ 1 2|Y =y)fY(y)dy.

Example 10.19. Let (X, Y) be uniformly distributed on the triangle D with vertices (1, 0), (2,0) and (0, 1), as in Example 6.20 (see Figure 10.2). Find the conditional density function fx\y(r|y) and the conditional expectation E[X|Y = y]. Verify the averaging identity (10.18) for E[X]. The area of the triangle is ^, so the joint density function of (X, Y) is given by if 0 y< 1 and 1 - y < r < 2 - 2y 2, Sх,у (r, y) if (r, y) e D 0. = otherwise 0, if (r, y) D In the second version above we expressed (r, y) e D with inequalities in a way that is suitable for the sequel. In Example 6.20 we found the marginal density function of Y as 2 2y, 0 < y < 1 fy(y) = y 0 or y 1 Уo 1-y 2-2y 1 2 Figure 10.2. Conditional on Y = yo, the point (X, Y) lies on the intersection of the blue dashed line with the shaded triangle. Since the conditional distribution of X is uniform on the interval (1 yo, 2 2yo) marked by the solid red line, its conditional expectation is the midpoint of this interval marked by the red bullet. To have fy(y) > 0 we restrict to 0 < y < 1. Then 2 1 (10.20 for 1 yr < 2 - 2y. fr(y) 2 2y 1 y

Answer 1

The conditional expectation of X given Y = y is 2-2y rfxy (ydr EXYy 1-y 2-2y dr 1-/ 2-2y 1 1 2 1-y (1 y)33y) 2(1 y) 1.5 1.5y Then the expectation of X is E[X] EEXY]] = E[1.5 - 1.5Y] = 1.5 1.5E[Y] - now EY y(2 2y) dy 1 2y3 1/3 - 3 1.5/3 = 1.5 .. E[X] 1.5 0.5 = 1 Now from the joint distribution, we get E[X] 2r dy dr (r,y)ED -1-/2 1-/2 dy dr 2 dy dr 0 J1-T 2 x(1 /2) d 2 IX- 2 0 1 2 2 d r) da a(2 0 1 2 3 3- 4/3 2/3 1/3 = 1 - 3 0 which agrees with our previous calculation