Question

Suppose that a person plays a game in which he draws a ball from a box of 10 balls numbered 0 through 9. He then puts the ball back and continue to draw a ball (with replacement) until he draws another number which is equal or higher than the first draw. Let X denote the number drawn in the first draw and Y denote the number of subsequent draws needed.

(a) Find the conditional probability distribution of Y given X = x, for x = 0, 1, · · · , 9.
(b) Find the joint probability distribution of X and Y , P(X = x, Y = y).
(c) Find the probability distribution of Y , P(Y = y).
(d) Find E(Y ).

Answer 1

a) The conditional probability distribution of Y given X = x, for x = 0, 1, · · · , 9 is

10 X 10 x 1 PMx=x)(Yx) 10 10 X Y-1 10 -x PYxx)(yx) 10X0,1,2,..., 9, y 1,2,3,... 10

b)Here .

Px(X x)0x 0, 1,2,.., 9

The joint distribution is

P(X X, Y y) P(xx)(YX )px (X x) x y-1 10X P(X x, Y y) = 1, 2,3,... 100X0,1, 2,..., 9,y 10

c) The marginal PMF of is

P(Y y) P(X x, Y = y) x-0 9 x y-1 10 x 10 100 x-0 1 y-1 X X P(Y = y) = 10 10/ 10 X 0 9 1 Le y-1 P(Y y)= 10 X (10)y 1,2,3,. 10 x-0

A closed form expression cannot be obtained for .

PY y

d)The expecte value of   is

E(Υ)-ΣγP(Yy) y-1 Χ )y-1 E(Υ)-ΣΗΣ 10. 10 10 y 1 x 0 LE-eξα - 1 y-1 y Χ ΕΥ)-10ΣΣ |( ) (1 10 x-0 y 1 σ

Let , then

X Z = 10

Σν -y(10 xy-1 y 10 Χ ( 6)-Σlyw1-yr 10 y 1 y 1 Σ dzy Ζ. dz Χ )y-1 dzy Χ dz 10 y-1 y 1 dΣ-2/ 5) -y(x- Σ7 Σγ(6) y-0 Z- dz dz 10 y-1 (P Ζ dz ΣΥΗ5- Xy-1 Η dz 10 10 -Aτ

xy-1 y 10 1 X X y 10/ (1 2)2 (1-z y 1 1 -y(10 x Y-1 X 1 Z 10 y-1 y-1 1 X 1 10 10 X 10 y 1 10 Xy-1 -y (10) 10 10 X y-1

10 1 E(Y) 1010 X x-0 9 1 E(Y) 10-X x 0 E(Y) 2.928968