Math 151. Rumbos Spring Solutions to Assignment #15

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Claud Butler
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1 Math 151. Rumbos Spring 28 1 Solutions to ssignment #15 1. Let F (X,Y ) be the joint cdf of two random variables X and Y. For real constants a < b, c < d, show that Pr(a < X b, c < Y d) F (X,Y ) (b, d) F (X,Y ) (b, c) F (X,Y ) (a, d)+f (X,Y ) Use this result to show that F (, y) cdf of two random variables. 1 if + 2y 1, otherwise, cannot be the joint Solution: Let (, y) R 2 a < b, c < y d}; we then want to compute Pr[(X, Y ) ]. In addition, define the events: y d 3 c a b 2 4 Figure 1: Events, 1, 2, 3 and 4 in the y plane 1 (, y) R 2 b, y d}, 2 (, y) R 2 a, y c}, 3 (, y) R 2 a, c < y d}, and 4 (, y) R 2 a < b, y c}. Then, 1 is a disjoint union of the events, 2, 3 and 4 (see Figure 1). It then follows that Pr[(X, Y ) 1 ] Pr[(X, Y ) ]

2 Math 151. Rumbos Spring 28 2 or Pr[(X, Y ) 1 ] Pr[(X, Y ) ] + Pr[(X, Y ) 2 ]+ +Pr[(X, Y ) 3 ] + Pr[(X, Y ) 4 ]. (1) Observe that Pr[(X, Y ) 1 ] Pr(X b, Y d) F (X,Y ) (b, d) and Pr[(X, Y ) 2 ] Pr(X a, Y c) F (X,Y ) It then follows from equation (1) that Pr[(X, Y ) ] F (X,Y ) (b, d) F (X,Y ) (a, c) Pr[(X, Y ) 3 ] Pr[(X, Y ) 4 ]. (2) On the other hand, observe that and Moreover, Pr[(X, Y ) 3 2 ] Pr(X a, Y d) F (X,Y ) (a, d) (3) Pr[(X, Y ) 4 2 ] Pr(X b, Y c) F (X,Y ) (b, c). (4) Pr[(X, Y ) ( 3 2 ) ( 4 2 )] Pr[(X, Y ) 3 2 ] +Pr[(X, Y ) 4 2 ] Pr[(X, Y ) 2 ], since ( 3 2 ) ( 4 2 ) 2. It then follows from equations (3) and (4) that Pr[(X, Y ) ( 3 2 ) ( 4 2 )] F (X,Y ) (a, d) +F (X,Y ) (b, c) F (X,Y ) However, since ( 3 2 ) ( 4 2 ) 2 3 4, we also get that Pr[(X, Y ) ( 3 2 ) ( 4 2 )] Pr[(X, Y ) 2 ] +Pr[(X, Y ) 3 ] +Pr[(X, y) 4 ].

3 Math 151. Rumbos Spring 28 3 We therefore get, using Pr[(X, Y ) 2 ] F (X,Y ) (a, c), that Pr[(X, Y ) 3 ] + Pr[(X, Y ) 4 ] F (X,Y ) (a, d) + F (X,Y ) (b, c) 2F (X,Y ) Substituting this into equation (2 ) yields Pr[(X, Y ) ] F (X,Y ) (b, d) F (X,Y ) (a, c) F (X,Y ) (a, d) F (X,Y ) (b, c) + 2F (X,Y ) (a, c), from which we get Pr[(X, Y ) ] F (X,Y ) (b, d) F (X,Y ) (a, d) F (X,Y ) (b, c) + F (X,Y ) 1 if + 2y 1, Net, suppose that F (, y) otherwise, two random variables X and Y. Consider the set By what we just proved, (, y) R 2 < 1, < y 1/2}. is the joint cdf of Pr[(X, Y ) ] F (1, 1/2) F (, 1/2) F (1, ) + F (, ) <, which is impossible since Pr[(X, Y ) ]. Therefore, F cannot be a joint pdf. 2. Let g(t) denote a non negative, integrable function of a single variable with the property that Define g(t) dt 1. 2g( 2 + y 2 ) π for < <, < y <, f(, y) 2 + y 2 otherwise. Show that f(, y) is a joint pdf for two random variables X and Y.

4 Math 151. Rumbos Spring 28 4 Solution: First observe that f is non negative since g is non negative. Net, compute R 2 f(, y) ddy 2g( 2 + y 2 ) π 2 + y 2 ddy. Switching to polar coordinates we then get that R 2 f(, y) ddy π/2 2g(r) πr rdrdθ π 2 1, 2 g(r) dr π g(r) dr and therefore f(, y) is indeed a joint pdf for two random variables X and Y. 3. Let X and Y have joint pdf e y for < <, < y <, (, y) otherwise. Define Z X + Y. Compute Pr(Z ) for < < and give the pdf of Z. Solution: y y Figure 2: Event in the y plane

5 Math 151. Rumbos Spring 28 5 Compute Pr(Z ) Pr(X + Y ), for > (, y) ddy, where (, y) R 2 >, y >, + y } (see Figure 2). We then have that Pr(Z ) e y dyd e e y dyd e [ e y] d e (1 e ( ) ) d (e e ) d [ e ] e 1 e e. Thus, F Z () 1 e e for all >. It then follows that the pdf of Z is e if >, f Z () otherwise. 4. Let X and Y have joint pdf (, y) 1 for < < 1, < y < 1, otherwise. Find the cdf and pdf of the product Z XY.

6 Math 151. Rumbos Spring 28 6 Solution: Compute Pr(Z ) Pr(XY ), for < < 1 (, y) ddy, where (, y) R 2 < < 1, < y < 1, y } (see Figure 3). y 1 y 1 Figure 3: Event in the y plane Then, Pr(Z ) ddy 1 dyd + 1 / dyd + 1 ln, d for < < 1. It then follows that the cdf of Z is if, F Z () ln if < < 1, 1 if 1, ln if < < 1, and the corresponding pdf of Z is f Z () otherwise.

7 Math 151. Rumbos Spring [Eercise 11 on page 136 in the tet] Suppose that two persons make an appointment to meet between 5 PM and 6 PM at a certain location and they agree that neither person will wait more than 1 minutes for each person. If they arrive independently at random times between 5 PM and 6 PM, what is the probability that they will meet? Solution: Let X denote the arrival time of the first person and Y that of the second person. Then X and Y are independent and uniformly distributed on the interval (5 PM, 6 PM), in hours. It then follows that the joint pdf of X and Y is 1 if 5 PM < < 6 PM, 5 PM < < 6 PM, (, y) otherwise. Define W X Y ; this is the time that one person would have to wait for the other one. Then, W takes on values, w, between and 1 (in hours). The probability that that a person would have to wait more than 1 minutes is Pr(W > 1/6), since the time is being measured in hours. It then follows that the probability that the two persons will meet is 1 Pr(W > 1/6) Pr(W 1/6) F W (1/6). We will therefore first find the cdf of W. To do this, we compute Pr(W w) Pr( X Y w), for < w < 1, where is the event (, y) ddy, (, y) R 2 5 PM < < 6 PM, 5 PM < y < 6 PM, y w}. This event is pictured in Figure 4. We then have that Pr(W w) ddy area(),

8 Math 151. Rumbos Spring 28 8 y 6 PM 5 + w 5 PM 5 PM 5 + w 6 PM Figure 4: Event in the y plane where the area of can be computed by subtracting from 1 the area of the two corder triangles shown in Figure 4: Pr(W w) 1 (1 w) 2 2w w 2. Consequently, F W (w) 2w w 2 for < w < 1. Thus the probability that the two persons will meet is F W (1/6) ( ) , or about 3.56%.

Math 151. Rumbos Spring Solutions to Review Problems for Exam 1

Math 151. Rumbos Spring Solutions to Review Problems for Exam 1 Math 5. Rumbos Spring 04 Solutions to Review Problems for Exam. There are 5 red chips and 3 blue chips in a bowl. The red chips are numbered,, 3, 4, 5 respectively, and the blue chips are numbered,, 3