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2. Let X · Unif(0, 1) and Y = ₁. Find the probability density function of Y. X+1' Solution. We let g: [0, 1] → [0,1/2] be given by g(x) = 41. By writing g(x) = 1+1, see that g is strictly increasing (also note that g(0) = 0 and g(1) = 1/2). The inverse function is given by | g^¹₁ (y) = 1 ² 2 y Y y = [0,1/2] 2 - (this is obtain by solving for x in y = 11). We then have fy (u) = fx (9¯¹ (u)) · |— 9¯¹ (1) = 0. (1). 7 d Y dy 1-y fx (x) = g(x) = 1 + 1 (1 — y)² ¹ y = [0,1/2]. we 1-Y Y =

2. Let X · Unif(0, 1) and Y = ₁. Find the probability density function of Y. X+1' Solution. We let g: [0, 1] → [0,1/2] be given by g(x) = 41. By writing g(x) = 1+1, see that g is strictly increasing (also note that g(0) = 0 and g(1) = 1/2). The inverse function is given by | g^¹₁ (y) = 1 ² 2 y Y y = [0,1/2] 2 - (this is obtain by solving for x in y = 11). We then have fy (u) = fx (9¯¹ (u)) · |— 9¯¹ (1) = 0. (1). 7 d Y dy 1-y fx (x) = g(x) = 1 + 1 (1 — y)² ¹ y = [0,1/2]. we 1-Y Y =

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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could u please tell me why the part I circled is 1 rather than y? thx :)

2. Let X Unif(0, 1) and Y = X X+1* fy(y) = fx(g¯¯¹(y)) Find the probability density function of Y. X 1 By writing g(x) = x+1° Solution. We let g: [0, 1] → [0, 1/2] be given by g(x) = see that g is strictly increasing (also note that g(0) function is given by g¯¹ (y) = (this is obtain by solving for x in y = Y 1- y' 1 1+ X · | d9¯¹ (0)| dy 1 1+ 1 X = 0 and g(1) = 1/2). The inverse | =(1). Y? y = [0, 1/2] ). We then have fx (x) = g(x) = 1 + 1 d Y dy 1 - y (1 — y)² ¹ - | - - y = [0, 1/2]. we 1-Y 11 Y
Transcribed Image Text:2. Let X Unif(0, 1) and Y = X X+1* fy(y) = fx(g¯¯¹(y)) Find the probability density function of Y. X 1 By writing g(x) = x+1° Solution. We let g: [0, 1] → [0, 1/2] be given by g(x) = see that g is strictly increasing (also note that g(0) function is given by g¯¹ (y) = (this is obtain by solving for x in y = Y 1- y' 1 1+ X · | d9¯¹ (0)| dy 1 1+ 1 X = 0 and g(1) = 1/2). The inverse | =(1). Y? y = [0, 1/2] ). We then have fx (x) = g(x) = 1 + 1 d Y dy 1 - y (1 — y)² ¹ - | - - y = [0, 1/2]. we 1-Y 11 Y
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