Question

Consider the probability mass distribution PſY = 1] = c and PſY = k] = k · P[Y = k – 1] for k = 2, ...5. for some constant c. a) Calculate c. b) Calculate the probability that Y is an odd number. b) Calculate expectation and variance of Y.

Answer 1

A probability mass distribution is given by P(Y=1) = 0 P(Y=K) =K. P (Y=k-1). for K = 2,3,4,5. K = 2 → PCY= 2) = 2. P(Y=2-1) = 2xP(Y=1) So = 2XC PCY=2) =20 K=3 = P (Y= 3)= 3.PCY= 3-1) = 3x PX=2)=3X2C = 6C P (Y= 3) = 60 K=4 → PCY=4) = 4. PCY=4+) = 4xP(Y=3) = 4 x 6 C .:P(Y=4) = 24c K=5> PX-5) = 5.PV=5-1) = 5x P(X=4) = 5x24c P (8=5) = 1200 By definition of probability mass distribution, É PCY=k)=1 5 K-1 PCY=1) +P (Y=2) +PCY=3)+P (Y=4)+PCY=5) = 1

c+2ct GC + 24C+ 1200 = 1 => 153.C = 1 + c = 153 Ans b) P (8= odd number): PCY= odd) = POY=1) + PCY= 3) + PCY=5) = c +60 + 1200 127C = 127x I 153 0.8301 Ans c) Expectation of Y: E (Y) = EYPCY) Here Y; takes 1, 2, 3, 4,5 Pari) takes 153 to 52 120 respectie ..EO) = (x +33) +23) Hx162) +(4x2%) + 153 5x 120 153 1 + 4+18+96+600 153 719 153 4.699 Ans

2 Variance of Y: Var (%) = E(82) - [EM] Let's find EY2) E (Y2) = { }; ². P (Y;) = (xs) + 3) + 120 + 15 x 153 1+8 + 54+384+ 3000 153 3447 153 Var (1) = E (42) - LE CO] 2 3447 153 719 1153 52739)- 51696) 1532 = 10430 1532 = 0.4456 ans