Need the answer to Part (a). Thank you! Let X, Y, Z be discrete random such that

Question

Need the answer to Part (a). Thank you!

Let X, Y, Z be discrete random such that, with probability 1, X, Y, and Z are all unequal. Let a = P(X > Y), b = P(Y > Z), and c = P(Z > X). (a) Show that min {a, b, c} lessthanorequalto 2/3 [This part is related to the observation that in an election, it is possible for more than half the voters to prefer candidate A to candidate B, more than half B to C, and more than half C to A.] (b) Show that if X, Y, and Z are independent and identically distributed then a = b = c = 1/2. (c) Assume P(X = 0) = 1, Y and Z are independent, P(Z = 1) P(Y = 1) = p, and P(Z = -2) P(Y = 2) = 1 - p. Compute max {a, b, c} as a function of p, and maximize with respect to p (d) Construct an example where a = b = c = 2/3 Applications of statistics along these lines are used to model "mathematical games which are used in political decision making all the time. Arguably, game theory, not nuclear weapons, won WWII.

Explanation / Answer

Given that,

a=P(X>Y)

b=P(Y>Z)

c=P(Z>X)

Let us assume that a,b 2/3

According to the laws of probability,

P(Z>X) = 1 -

P(X>Y Y>Z)

i.e. c = 1 – (ab)

Using contradiction method, let us assume that

Min (a,b,c) 2/3                 ...(1)

i.e. all numbers are greater than 2/3

This implies that,

a + b + c 2                       ...(2)

(Since, a,b,c 2/3)

Since any equality has probability zero, thus the possible cases are:-

P1 = P(X>Y>Z)

P2 = P(X>Z>Y)

P3 = P(Y>Z>X)

P4 = P(Y>X>Z)

P5 = P(Z>X>Y)

P6 = P(Z>Y>X)

And according to laws of probablity, sum of all probablities is 1. Hence,

P1+P2+P3+P4+P5+P6 = 1 ....(3)

where,

a = P1+P2+P5

b=P3+P4+P1

c=P5+P6+P3

Since a,b,c are sum of probabilities, the result can't be negative.

Hence,

a+b+c = P1+P2+P5+P3+P4+P1+P5+P6+P3

From eq (3),

a+b+c = 1+P1+P5+P3 ...(4)

From eq(2) and eq(4), this suggests that a+b+c lies in the range [1,2]

This is possible only when P1=P3=P5=1/3 and P2=P4=P6=0

This contradicts our assumption that min(a,b,c) 2/3

HENCE, Min (a,b,c) 2/3

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