Each night different meteorologists give us the probability that it will rain the next day. To judge how well these people predict, we will score each of them as follows: If a meteorologist says that it will rain with probability p, then he or she will receive a score of 1−(1−p)2 if it does rain 1−p2 if it does not rain We will then keep track of scores over a certain time span and conclude that the meteorologist with the highest average score is the best predictor of weather. Suppose now that a given meteorologist is aware of our scoring mechanism and wants to maximize his or her expected score. If this person truly believes that it will rain tomorrow with probability p⋅, what value of p should he or she assert so as to maximize the expected score?

Question

Each night different meteorologists give us the probability that it will rain the next day. To judge how well these people predict, we will score each of them as follows: If a meteorologist says that it will rain with probability p, then he or she will receive a score of  1−(1−p)2 if it does rain  1−p2 if it does not rain  We will then keep track of scores over a certain time span and conclude that the meteorologist with the highest average score is the best predictor of weather. Suppose now that a given meteorologist is aware of our scoring mechanism and wants to maximize his or her expected score. If this person truly believes that it will rain tomorrow with probability p⋅, what value of p should he or she assert so as to maximize the expected score?

Nancy Johnson · Accepted Answer

Define random variable X that marks the score of a certain meteorologist at some day. If the says that it will rain (he says it with probability p⋅) and it really rains (with probability p), he gets score 1−(1−p)2. Also, if he says that it will not rain and it really does not rain (with probability 1−p⋅) and it really does not rain (with probability 1-p), he get score 1−p2. Hence  E(X)=p⋅⋅(1−(1−p)2)+(1−p⋅)⋅(1−p2)  =2pp⋅−p⋅−p2+1  We now have to analyse the function p↦E(X) and find its maximum on (0,1). We have that  dE(X)dp=2p⋅−2p=0  ⇔p=p⋅  So we have that the meteorologist maximezes it chances when p=p⋅, which is according to our intuition.

Andre BalkonE · Accepted Answer

To find the value of p that maximizes the expected score, let&#039;s denote the expected score as E(p). We can calculate the expected score by considering the two possible outcomes: rain and no rain.If it rains, the meteorologist will receive a score of 1−(1−p)2. The probability of rain is p, so the expected score for this case is p×(1−(1−p)2).If it does not rain, the meteorologist will receive a score of 1−p2. The probability of no rain is 1−p, so the expected score for this case is (1−p)×(1−p2).Now we can calculate the overall expected score E(p) by summing up the expected scores for the rain and no rain cases:E(p)=p×(1−(1−p)2)+(1−p)×(1−p2)To maximize the expected score, we need to find the value of p that maximizes E(p). We can do this by taking the derivative of E(p) with respect to p, setting it to zero, and solving for p:dE(p)dp=0Let&#039;s differentiate E(p) with respect to p:dE(p)dp=(1−(1−p)2)+p×2(1−p)(−1)+(1−p2)×(−1)Simplifying the expression:dE(p)dp=2p−2+2p2−1+p2=3p2+2p−3Now we set the derivative equal to zero and solve for p:3p2+2p−3=0Using the quadratic formula, we find the solutions for p:p=−2±22−4×3×(−3)2×3Simplifying further:p=−2±4+366=−2±406=−2±2106=−1±103The two possible values for p are:p1=−1+103andp2=−1−103Since the probability p cannot be negative, we conclude that the meteorologist should assert the value of p=−1+103 to maximize the expected score.Thus, the value of p that maximizes the expected score is −1+103.

fudzisako · Accepted Answer

Step 1: Let&#039;s denote the average score as S(p). To find the maximum of S(p), we can take the derivative of S(p) with respect to p and set it equal to zero.S(p)=(1−(1−p)2)·p+(1−p2)·(1−p)To simplify the equation, let&#039;s expand and rearrange terms:S(p)=p−p3+p2−2p+1Taking the derivative of S(p) with respect to p:dS(p)dp=1−3p2+2pSetting the derivative equal to zero:1−3p2+2p=0Simplifying the equation further:3p2−2p−1=0Step 2:We can solve this quadratic equation using the quadratic formula:p=−b±b2−4ac2aFor our equation, a = 3, b = -2, and c = -1. Substituting these values into the quadratic formula:p=−(−2)±(−2)2−4·3·(−1)2·3Simplifying further:p=2±4+126=2±166=2±46Step 3:This gives us two possible solutions:p1=2+46=66=1p2=2−46=−26=−13However, since we&#039;re dealing with probabilities, p must be between 0 and 1. Therefore, the valid solution is:p=66=1Hence, the meteorologist should assert a probability of 1 (p = 1) to maximize the expected score.

Each night different meteorologists give us the probability that it wi

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