Poisson Thinning and Pareto Distribution

Question

was hoping to get some advice on this one. It is a two part question.

The number of days(D) in any given month has a Poisson Distrubtion with $\lambda$ =10 days that rain per month. Let $X_1,X_2$... be independent amounts of rain(mm) that are independent of D and have a Pareto distribution with parameters $\alpha = 5$ $\delta = 30$. Also, let Z= $\sum_{i=1}^D$ $X_i$ be the total rainfall for any given month.

a, Calculate the mean and standard deviation of Z.

b, Find the probability that at least one day in a month has 80mm or more rain. (Question suggests looking at Poisson Thinning)

To calculate the mean of Z, I multiplied the mean of the Poisson $\lambda$=10 by the mean of the Pareto, $\delta / \alpha - 1$ = 7.5, to give 16*7.5=120.

For the variance, the same applies, Var Poisson = 10 & Var Pareto = $\frac {\alpha \delta^2}{(\alpha-1)^2(\alpha-2)}$ = 93.75, Var Z= 937.5 and $\sigma$= $\sqrt{937.5}$ =30.6186

My question is with part b. To find the probability that any given day has 80mm or more rain, would I use the cdf of the Pareto? My thinking is that P(X$\gt$80mm) = 1- P(X$\lt$80mm) using this I got the answer, P(X$\gt$80)= 1-(1-($\frac {\delta}{\delta+y})^\alpha$ = 0.0015(seems too low but i'm not sure) Regardless of whether or not this is the answer to the probability, i'm sure there is a step I am yet to complete. This probability is calculated given that it's a rainy day, whereas the question asks, what is the probability that at least one day will have that much rain. Any advice would be greatly appreciated.

Answer 1

Your calculation of the variance is incorrect, because $$\operatorname{Var}[Z] \ne \operatorname{Var}[D]\operatorname{Var}[X_i].$$ Instead, you need to use the law of total variance: $$\operatorname{Var}[Z] = \operatorname{E}[\operatorname{Var}[Z \mid D]] + \operatorname{Var}[\operatorname{E}[Z \mid D]].$$ In your particular case, $D$ is a counting distribution, with $Z = X_1 + \cdots + X_D$ and the $X_i$ are IID, so the conditional variance of $Z$ given $D$ is simply $$\operatorname{Var}[Z \mid D] \overset{\text{iid}}{=} D \operatorname{Var}[X_i].$$ This is because the variance of a sum of independent random variables is equal to the sum of their variances; and because the variables are identically distributed, each has the same variance; and there are $D$ such $X_i$. Consequently, $$\operatorname{E}[\operatorname{Var}[Z \mid D]] = \operatorname{E}[D \operatorname{Var}[X_i]] = \operatorname{Var}[X_i]\operatorname{E}[D].$$ Note this expectation is taken with respect to the distribution of $D$, as the variance of $X_i$ is a fixed constant. Next, we have $${E}[Z \mid D] \overset{\text{id}}{=} D \operatorname{E}[X_i],$$ by linearity of expectation and the fact that the $X_i$ are identically distributed. Then $$\operatorname{Var}[\operatorname{E}[Z \mid D]] = \operatorname{Var}[\operatorname{E}[X_i] D] = \operatorname{E}[X_i]^2 \operatorname{Var}[D].$$ So putting all this together, $$\boxed{\operatorname{Var}[Z] = \operatorname{Var}[X_i] \operatorname{E}[D] + \operatorname{E}[X_i]^2 \operatorname{Var}[D].}$$ This is the proper formula to use.

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As for the second part of your question, the idea is to think of the Poisson events (a rainy day) as belonging to one of two classes: either on that rainy day, the total rainfall was less than $80$ mm, or the total rainfall was at least $80$ mm. Then because of Poisson thinning, each of these two classes of events is itself a Poisson process with a Poisson distribution, in which the intensity parameter is some function of the proportion of the respective class of event relative to the overall intensity of occurrence.

For example, if I conduct an experiment in a neutrino detector, in which I model the number of neutrinos passing through the detector per second as following some Poisson distribution with intensity $\lambda$, but I know that on average, only $1\%$ of such neutrinos actually interact in the detector and produce an observable reaction, then the rate of observed events is modeled by a Poisson distribution with intensity $0.01\lambda$.

Therefore, in your case, if you compute the probability that the amount of rain is at least $80$ mm on a given rainy day, you would use this as a way to model the random number of rainy days with rainfall at least $80$ mm in a month. Then use the Poisson distribution to compute the probability that no such events occur, and then take the complementary probability--which is the probability that at least one such day occurs.