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Let us model the number of winter storms in a given year as a Poisson random variable. Suppose that in a good year the average number of storms is 3, and in a bad year the average is 5. If the next year will be good with probability 40% and bad with probability 60%, What's the variance?

Well I thought for Poisson random variables, the expectation and the variance are the same. In this case the expectation is 4.2 so I thought presumably the variance would 4.2

What am I missing?

Jilly
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Let $W$ be the number of winter storms. We condition on whether it is a good year. The expectation of $W$ is $3$ with probability $0.4$ and $5$ with probability $0.6$. It follows that $E(W)=4.2$.

Now we will be cautious in our evaluation of the variance of $W$. Given that it is a good year, the expected value of $W^2$ is $3+3^2$. Given that it is a bad year, the expected value of $W^2$ is $5+5^2$. Thus $E(W^2)=(0.4)(12)+(0.6)(30)=22.8$.

Thus the variance of $W$, which is $E(W^2)-(E(W))^2$, is $5.16$.

Remark: The problem with the argument of the OP is that it tacitly assumes that the random variable $W$ has Poisson distribution. It doesn't. That follows from the above calculation, since if $W$ had Poisson distribution it would indeed have variance $4.2$.

Alternately, we can show $W$ does not have Poisson distribution by computing the probability that $W=k$ for $k=0$ and $k=1$.

André Nicolas
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  • so in the case of finding E(W^2) we should weigh both the E(W^2) for a good year and the E(W^2) for a bad year – Jilly Apr 29 '14 at 20:21
  • Yes, that is what the calculation of the answer does. – André Nicolas Apr 29 '14 at 20:24
  • awesome thanks! i was assuming the distribution since it was a poisson random variable which should not have been assumed – Jilly Apr 29 '14 at 20:25
  • You are welcome. There is a somewhat related situation in which we do get a Poisson. Let us suppose that customers come in, with the number coming in a randomly selected hour having Poisson distribution parameter $\lambda$. Suppose that with probability $p$ a customer is bad. Let $Y$ be the number of bad customers coming in an hour. Then $Y$ is Poisson, parameter $p\lambda$. For a moment I thought the problem of your post would use that fact. But your situation is quite different. – André Nicolas Apr 29 '14 at 20:30
  • @andrenicolas Could you explain why $E(W^2)=3+3^2$ given that it is a good year? Thanks! – user137481 Apr 29 '14 at 20:52
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    Given that it is a good year, the variance is $3$, and the mean is $3$. Note that $\text{Var}(z)=E(Z^2-(E(Z))^2$, So if the mean is 3 and the variance 3, then $3=E(Z^2)-3^2$, so $E(Z^2)=3+3^2$. – André Nicolas Apr 30 '14 at 00:22