Consider the following game based on the toss of two coins: First, a fair coin o

Question

Consider the following game based on the toss of two coins: First, a fair coin of tossed. If the result is a Head, the second fair coin is tossed and the result from both coins is recorded. If the result from the first coin toss is a Tail, then an unfair coin is tossed with the probability mass function Prob{H} = 1 3 , Prob{T} = 2 3 and the result from two coins is recorded. Define random variables X1 and X2 corresponding to the first and the second coins as X1(H) = 1, X1(T) = 0, X2(H) = 1, X2(T) = 0. Also, define the vector-valued random variable X = (X1, X2) and the corresponding state space, i.e. X(HH) = (1, 1), X(HT) = (1, 0), X(T H) = (0, 1), X(T T) = (0, 0).

1. Compute the joint probability mass function pX(x) and summarize as a table.

2. Compute marginal probability mass functions pX1 (x1) and pX2 (x2).

3. Determine if variables X1 and X2 are independent.

Consider the following payoff strategy based on observing both coins:

Win $1 if the first coin is a Head,

Win $3 if the second coin is a Head,

Lose $5 if both coins are Tails.

4. Determine possible payoffs in a single game.

5. Define the random variable describing the payoff strategy described above and compute the corresponding probability mass function.

6. Determine the expected payoff in n games.

Explanation / Answer

Solution

The working is neatly presented in the following tables.

First Part

Joint Probability

X1

X2

Total

1(H)

0(T)

1(H)

( ½ )( ½ ) = ¼

( ½ )( ½ ) = ¼

½

0(T)

(1/2)(1/3) = 1/6

(1/2)(2/3) = 1/3

½

Total

5/12

7/12

1

Marginal Probabilty – X1

x1

1

0

P(x1)

½

½

Marginal Probabilty – X2

x2

1

0

P(x2)

5/12

7/12

From the above tables,

P(X1 = 1, X2 = 1) = ¼

P(X1 = 1) = 1/2, P(X2 = 1) = 5/12

Since P(X1 = 1, X2 = 1) P(X1 = 1) x P(X2 = 1), the variables are NOT independent.

Second Part

Possibility

HH

TH

HT

TT

TOTAL

Pay-off (x)

1 + 3 = 4

3

1

- 5

-

Prob p(x)

¼

1/6

¼

1/3

1

x. p(x)

¾

½

1/6

- 5/3

½

Thus, expected pay-off per game = ½ and hence for n games = n/2

X1

X2

Total

1(H)

0(T)

1(H)

( ½ )( ½ ) = ¼

( ½ )( ½ ) = ¼

½

0(T)

(1/2)(1/3) = 1/6

(1/2)(2/3) = 1/3

½

Total

5/12

7/12

1

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