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Solution: Given that X = (0,1) and Y = (0, 1), we have U € (0,2). Then for u € (0,2), the equality U = u implies that X = u-Y. Hence, since we want 0 Uu, we observe that we need 0≤x≤u-Y, which is possible if and only if Y

Solution: Given that X = (0,1) and Y = (0, 1), we have U € (0,2). Then for u € (0,2), the equality U = u implies that X = u-Y. Hence, since we want 0 Uu, we observe that we need 0≤x≤u-Y, which is possible if and only if Y

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.8: Probability
Problem 22E
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Can you please explain how to find the bounds of the integrals for X and Y and also explain how to find the inequalites that satisfy X and Y. I've looked at the solutions but its not clear to me on how the inequalities and bounds of the integral were obtained. If possible could you explain how to find the bounds of the integrals by sketching a graph with the region of integration. Thanks

Solution: Given that X = (0,1) and Y = (0, 1), we have U € (0,2). Then for u € (0,2), the equality U = u implies that X = u-Y. Hence, since we want 0 Uu, we observe that we need 0≤x≤u-Y, which is possible if and only if Y<u. However, given that 0 < x < 1 and 0 < y < 1, we need to split into two cases: For u € (0,1), we have 0 ≤ X ≤u-Y for any 0 ≤Y<u, thus Fu (u) = P(U≤u) = P(X + Y≤u) = P(X ≤u-Y) 26 ru-y = "fx.x(x, y) drdy CH = u-y dxdy = L" (u− y) dy 1 = = u² y-0 For u (1,2), we either have 0 ≤ X ≤u-Y for any u - 1 < Y<1, or 0≤x≤1 for any 0 ≤Y≤u - 1, thus Fu(u) = P(U≤u) = P(X + Y≤u) = P(X ≤u-Y) =fxx(x, y) dxdy + =LC pu-y = = cu-1 ddy+ T (u - y) dy + 1 Cu-1 fx,y(x, y) dxdy dxdy cu-1 dy Jo 1 +u-12u- -1. So, we have y-u-1 fu (u) = = dF(u) du -E- = if 0<u<1, 2-u, if 1 < u<2, 0 otherwise.
Transcribed Image Text:Solution: Given that X = (0,1) and Y = (0, 1), we have U € (0,2). Then for u € (0,2), the equality U = u implies that X = u-Y. Hence, since we want 0 Uu, we observe that we need 0≤x≤u-Y, which is possible if and only if Y<u. However, given that 0 < x < 1 and 0 < y < 1, we need to split into two cases: For u € (0,1), we have 0 ≤ X ≤u-Y for any 0 ≤Y<u, thus Fu (u) = P(U≤u) = P(X + Y≤u) = P(X ≤u-Y) 26 ru-y = "fx.x(x, y) drdy CH = u-y dxdy = L" (u− y) dy 1 = = u² y-0 For u (1,2), we either have 0 ≤ X ≤u-Y for any u - 1 < Y<1, or 0≤x≤1 for any 0 ≤Y≤u - 1, thus Fu(u) = P(U≤u) = P(X + Y≤u) = P(X ≤u-Y) =fxx(x, y) dxdy + =LC pu-y = = cu-1 ddy+ T (u - y) dy + 1 Cu-1 fx,y(x, y) dxdy dxdy cu-1 dy Jo 1 +u-12u- -1. So, we have y-u-1 fu (u) = = dF(u) du -E- = if 0<u<1, 2-u, if 1 < u<2, 0 otherwise.
5. Suppose X and Y are two independent Uniform(0, 1) random variables. Use the cumulative distribution function method to find the probability density function of their sum U = X+Y.
Transcribed Image Text:5. Suppose X and Y are two independent Uniform(0, 1) random variables. Use the cumulative distribution function method to find the probability density function of their sum U = X+Y.
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