Urgent! Expected Value, total variance

Question

The number of winter storms in a good year is a Poisson random variable with mean 3, whereas the number in a bad year is a Poisson random variable with mean 5. If next year will be a good year with probability .4 or a bad year with probability .6, find the expected value and variance of the number of storms that will occur

Solution:

Let X be the number of storms and let G (resp. B) be the event that it is a good (resp. bad) year. Then conditioning on whether it’s a good or bad year and applying the law of total expectation give E[X] = E[X | G]P(G) + E[X | B]P(B) = 3(.4) + 5(.6) = 4.2 If Y is a Poisson random variable with mean λ, then E[Y^2] = λ + λ^2.

Therefore, E [X^2] = E [X^2 | G] P(G) + E [X^2 | B] P (B) = 12(.4) + 30(.6) = 22.8 Consequently, Var(X) = 22.8 − (4.2)*2 = 5.16

Why we need to care \lambda + \lambda^2 = E[Y^2] but not just simply use Var(X) = 3^2 * 0.4 + 5^2 * 0.6?

Answer 1

Let X be the number of storms. We expect X to be 3 with a probability of 0.4 and 5 with a probability of 6, so E(X) = 3(0.4)+5(0.6) = 4.2

Now, we must weigh the conditioning for E(X²) for both good and bad years. It's given to us that the next year will be good with a probability of 0.4 and bad with a probability of 0.6. Thus, the expected variance for a good year will be E(X²|G) = 3 + 3² = 12, and E(X²|B) = 5 + 5² = 30 for a bad year. So, the expected variance of the number of storms that will occur is E(X²) = 12(0.4) + 30(0.6) = 22.8

Therefore, the variance of the number of storms that will occur is Var(X) = E(X²) - E(X)² = 22.8 - 4.2² = 5.16

You cannot simply use Var(X) = 3²(0.4) + 5²(0.6) because that assumes the mean and variance in a Poisson distribution are equal, but are not in this case, so you must find E(X²) using λ + λ².

Hope this helped!