Suppose the chance of rain in Chicago and Minneapolis are correlated. Define X =

Question

Suppose the chance of rain in Chicago and Minneapolis are correlated. Define X = 1 if it is raining in Chicago some day and X = 0 if not; Similarly, define Y = 1 if it is raining in Minneapolis and Y = 0 otherwise. Now we know the probability of raining in Chicago and Minneapolis are both 0.35, i.e., P(X = 1) = 0.35 and P(Y = 1) = 0.35. Now

(a) If given it rains in Chicago, the probability of rain in Minneapolis is 0.7. What is the correlation between X and Y ?

(b) If we know the correlation between X and Y is 0.5, what is the conditional probability that it rains in Minneapolis rains given it rains in Chicago?

Explanation / Answer

Part (a)

P(M|C) = P(C and M)/P(C) (Bayes' theorem)

P(C and M) = P(M|C)*P(C) = 0.7X0.35 = .245

Cov(C,M) = E[CM]-E[C]E[M] = P(C and M)-P(C)P(M) = 0.245-0.35X0.35 = 0.1225

Var(C) = Var(M) = E[C^2]-E[C]^2 = P(C)-P(C)^2 = 0.35-.35^2 = 0.2275

Cor(C,M) = Cov(C,M)/sqrt(C)sqrt(M) =0.1225/0.2275 = 0.53846..

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Part(b)

P(M|C) = P(C and M)/P(C) (Bayes' theorem)

P(C and M) = P(M|C)*0.35

Cov(C,M) = E[CM]-E[C]E[M] = P(C and M)-P(C)P(M) = (P(M|C)*0.35 )-0.35X0.35

Var(C) = Var(M) = E[C^2]-E[C]^2 = P(C)-P(C)^2 = 0.35-.35^2 = 0.2275

Cor(C,M) = Cov(C,M)/sqrt(C)sqrt(M) = [(P(M|C)*0.35 )-(0.35X0.35 )]/0.2275

Now given Corr(C,M) = 0.5

Solve

we get

P(M|C) = 0.675

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