Question

Problem 2. (6 pts) Independence and Conditional Probability (a) (2 pts) An urn contains 3 red and 5 green balls. At each step of this game, we pick one ball at random, note its color and return the ball to the urn together with anoter ball of the same color. Prove by induction that the probability that the ball we pick a red ball at the n-th step is 3/8. (b) (2pts) Consider any two random variables X, Y of any distirbution and not necesarily independent. Given that P(X = a) = P1, P(max(X,Y) = a) = P2, and P(min(X,Y) = a) = P3, find P(Y = a) in terms of P1, P2 and p3. Hint: Use the first Bayes theorem. Choose a suitable event-complement pair to partition the probability space by comparing X and Y. (c) (2 pts) Show that, if X and Y are two uncorrelated (i.e. EXY = EXEY) Bernoulli (Indicator) random variables then they are independent.

Answer 1

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a)

Problem 2. @ let zh Ra event a that ball drawn at n draw is red. & have to we G = event that event that ball drawn at th draw is green prove P (Rn) 3 PC Rn) = 318 P(Rn) = P(A) tn. statement g(n) Now n=1 ا لیا have l(R) no. of balls before pt draw that are red Istal no. of ball in the even before 1st draw 3 -PCR,) 8 nal. be true your statement f (n) is true your dowl to stay statement mtt. Then Pl Ry PC кг): 8 -fth)

Now be true n=k let g(n.) P( Rp ) = 3 8 Now let no Total of balls in the before kth draw be Tk (ie Thee no. of balls after the (u-gth draw urn Now no. of red - balls draw is after with TuxP(Rp) if Ith draw green is

of no. of med balls after with draw il wth draw g ered is : Tea PLRR) + 1 & Total k th draw no. of balls after the is Tetl. Tets find P(Ret). ! Plku+1) KAU Plkti | R6) - P(RK) + P(R TR - P(R) P(Run Ru). Z + P(RKHI Gre): # of red balls after seth 5 8 lis red) drawan 3 8 a Total balls ften with draw red balls after kth draw & # of one is green Total balls after yth. a raw As

= T & P (fr) + 1 3 of lo Tik ti + Tha PC Rx) you Tett 8. Tr l(fx)+3 stati [ Ta eta K 8 Tv E 3+3 8 +3] I 8 Tut!) * 3 [Tw+] 8 (Trt 3 8 7 P (RR) = 33

8(") is true your n=ktl. give & (K) is turve. Thus we show 9112 is true & g(n) is true whenever I (n-1) is true. This by induction of 8 (n) is true for all nEN. Thus, P (Rn) = 3 &n=1,2,. 8

b)

こ P (Y= a) P(Y=a/xsy). PCX5Y) +P [ Y=0 / x70Y) -P(x>Y) P(Y Man(X,Y)= ) P( mon(x,y)= a) = P(x=a ,Y<a) + P(Y=a, x < a) +1(x=a, Y=0 ☺ Y=a). 6. & P ( min (X,Y)= a P(x=9, Y>a) + P (Y=a, Xa) + P(x=9, Y=a)

(1L we Adding 0 s get, Р{~~[9, 1)- а) + (-m(x,)) (а, ч<b) + ( Yen, xce ) +P (у-а,154) +Р (Y-а , X}a +24 Р Х = 4, Y = 4 а). х

7 P(x=a, Y>a) + P(x=a Y=a) + P(x=a, y <a pl L (un Noe Z IV . > PLY= a) P (Y= a, xya) + P (Y=a. X<a a) + P ( 4 = a, X=a) from 9 8 Thias, P(x=a + PLY=a) =) p1 + P(za) = {p(x29,779 + Pt x=9, YLa + folyza , X7a) + f (tza, Xtaj + PLY=a, X29) ::b1 + P(Y=a) = (mon an(X, Y) = 1 a) + P(min. (x, y) = a L Cfrom ] Y a) + P(x=9, Y =9)

7. PC PC 7- а Ya ) = P2 + F3 21.

c)

© b, x=1 let pmy of x be =- P(^). X (1-0, x=0 > y = 1 & pay of Y be (0) is 215 Is, y zo. naf given that Its Now, El XY) = E(X) E(9) : E(XX) =_į my PCX=w, Y=y) x=0 y=0 1.L.PLX = 1, Y=1) solong P_Xz1,42)

Now, E(x)- E: OK 2D)+ («p = p) & E(Y) = S Thus, E (X) E(Y) ps. E (XY) = E(X) E(Y) , P(x = 1, Y= 1) Now ps. axt Now P ( X = 1, Y = 1) P(X= 1) a P (Y=1 (1) to show peo X & Y are independent we need to show P(x>m, Y=y) = f(x=x) {[Yy) for all pains of (ay). for pain (1) shown in 6 Now, P{ x= 1, Yz0) = P(x=1) – P(x= 1, Y=)) 6. [[Xzd).2 I. (X.31,820.R(=1_4821.) is

2 p - ps (1-3) P(x=1) P(Y=0) !. Pl_Xz), Y=0) PLX=))(Y=0) P(x=0 , Y = 1) P(Y= 1) - P(x=01, Y = 1) [= P(X=1) PLX= 1, Y=)) + f(x=0, x=1] . Now z (1) (3 Now 2 . P(X=0 , Y = 1 s - ps I from 7. = (1-P) P{ X=0, Y= 1) P(X=0) « P(Y=1) PLX=0, Y=0) = f(x=0_) - PL x 20, Y= 1) (1-P) – 5(1-5) [- forom @] (1-2) (1-5) P(X=0,Y=0)= PCX - 0) P(Y=0) This from , & 6 P(x=2,7=4-) = P(X=x.) P(Y=y) -=0; yeo.t- on X & Y are independent て 14 .