Question

A penny which is unbalanced so that the probability of heads is 0.4 is tossed twice. Let

Z be the number of heads obtained in the first toss. Let W be the total number of heads

obtained in the two tosses of the coin.

a)Calculate the correlation coefficient between Z and W

b)Show numerically that Cov(Z, W) = VarZ

c)Show without using numbers that Cov(Z, W) = VarZ

Answer #1

The joint distribution of Z, W here is obtained as:

W = 0 | W = 1 | W = 2 | |

Z = 0 | TT, Prob = 0.6*0.6 = 0.36 | TH, Prob = 0.6*0.4 = 0.24 | 0 |

Z = 1 | 0 | HT, Prob = 0.4*0.6 = 0.24 | HH, Prob = 0.4*0.4 = 0.16 |

a) The probability distribution for W and Z here are obtained
as:

P(W = 0) = 0.36, P(W = 1) = 0.48, and P(W = 2) = 0.16

P(Z = 0) = 0.36 + 0.24 = 0.6, and P(Z = 1) = 0.4

Therefore, E(Z) = 0*0.6 + 1*0.4 = 0.4, E(Z^{2}) = 0*0.6
+ 1*0.4 = 0.4

Therefore, Var(Z) = E(Z^{2}) - [E(Z)]^{2} = 0.4 -
0.4^{2} = 0.24

Now for W, E(W) = 0*0.36 + 1*0.48 + 2*0.16 = 0.8,
E(W^{2}) = 0*0.36 + 1*0.48 + 2^{2}*0.16 =
1.12

Therefore, Var(W) = E(W^{2}) - [E(W)]^{2} = 1.12 -
0.8^{2} = 0.48

E(ZW) = 1*1*0.24 + 1*2*0.16 = 0.56

Therefore, Cov(Z, W) =E(ZW) - E(Z)E(W) = 0.56 - 0.4*0.8 = 0.24

Now the correlation coefficient here is computed as:

**Therefore 0.7071 is the correlation coefficient
here.**

b) We have already proved above that Var(Z) = Cov(W, Z) = 0.24

c) Now we know that W is nothing but the sum of two random variables Z, as Z is the number of heads in the first toss and W is the number of heads in total two tosses.

Therefore the covariance of Z and W would be nothing but the covariance between Z and Z which is the variance of Z which is same as Cov(W, Z) = Var(Z)

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