Two meteorologists, Agatha and Bertrand are competing to forecast the Weather. L

Question

Two meteorologists, Agatha and Bertrand are competing to forecast the Weather. Let W be the random variable that is the actual weather on a given day, and suppose W = r (rainy) with probability 1/4 and W = s (sunny) with probability 3/4. Let A and B be the random variables for the forecasts of Agatha and Bertrand, respectively. A may be either r or s: on days where it actually rains, Agatha gets it right 4/5 of the time, and on days when it actually is sunny, Agatha gets it right 2/3 of the time. B is a random variable whose only outcome is s: Bertrand always forecasts sun. (a) Compute the probability that Agatha gets the forecast correct. (b) Do the same for Bertrand. Optional: Which forecaster is better? (c) On days when Agatha calls for rain, what is the probability that it rains? On days when Agatha calls for sun, what is the probability it is sunny? (d) Compute H(W | A) and I(W; A) to a few significant digits. (e) Compute H(W | B) and I(W; B) to a few significant digits. Optional: Which forecaster is better? Let C be a cyclic code. We call e(x) an idempotent generator for C if e(x) elementof C and for all c(x) elementof C_l. e(x)c(x) = c.(x) in V^n[x].

Explanation / Answer

P(W =r ) = 0.25 and P(W = s) = 0.75

P( A=r | W =r ) = 0.8 and therefore P( A = s | W = r) = 0.2

P( A=s | W=s ) = 2/3 and therefore P(A = r | W =s ) = 1/3

P(B =s ) = 1

a) Probability that Agatha gets the forecast correct:

= P(A = r | W=r) P(W=r) + P(A = s | W=s) P(W=s)

= 0.8*0.25 + (2/3)*0.75 = 0.2 + 0.5 = 0.7

Therefore the probability that Agatha gets the forecast correct is 0.7

b) Probability that Bertrand gets it correct:

= P(B =s | W=s) P(W= s) = 0.75

Therefore the probability that Bertrand gets the forecast correct is 0.75

Therefore Bertrand's forecast is better

c) On days when agatha calls for rain, probability that it rains is represented as:

P( W=r | A =r ) = ?

First computing :

P(A= r ) = P(A=r | W=r ) P(W=r) + P(A = r | W=s )P(W=s) = 0.8*0.25 + (1/3)*0.75 = 0.2 + 0.25 = 0.45

Now using Bayes formula for conditional probability we get:

P( W=r | A =r ) P(A =r ) = P( A=r | W= r)P(W=r)

P( W=r | A =r ) * 0.45 = 0.8*0.25

P( W=r | A =r ) = 0.2 / 0.45 = 0.4444

Therefore 0.4444 is the required probability here.

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