lambda=0.9;p=0.2;mu=1/p;sigma2=(1-p)/p^2
m1=mu*lambda*50;m2=(sigma2+mu^2)*lambda*50
1-pnorm(250,225,sqrt(m2))[1] 0.2892574PoissonProcess
Law of Total Variance:
\[ Var(X) =E[Var(X|Y)]+Var[E(X|Y)] \]
Proof: The variance can be decomposed into expected values as follows: \[ \operatorname{Var}(X)=\mathrm{E}\left(X^2\right)-\mathrm{E}(X)^2 . \]
This can be rearranged into: \[ \mathrm{E}\left(Y^2\right)=\operatorname{Var}(Y)+\mathrm{E}(Y)^2 . \]
Applying the law of total expectation, we have: \[ \mathrm{E}\left(Y^2\right)=\mathrm{E}\left[\operatorname{Var}(Y \mid X)+\mathrm{E}(Y \mid X)^2\right] . \]
Now subtract the second term from (2): \[ \mathrm{E}\left(X^2\right)-\mathrm{E}(X)^2=\mathrm{E}\left[\operatorname{Var}(X \mid Y)+\mathrm{E}(X \mid Y)^2\right]-\mathrm{E}(X)^2 . \]
Again applying the law of total expectation, we have: \[ \mathrm{E}\left[\operatorname{Var}(X \mid Y)+\mathrm{E}(X \mid Y)^2\right]-\mathrm{E}[\mathrm{E}(X \mid Y)]^2 . \]
With the linearity of the expected value, the terms can be regrouped to give: \[ \mathrm{E}[\operatorname{Var}(X \mid Y)]+\left(\mathrm{E}\left[\mathrm{E}(X \mid Y)^2\right]-\mathrm{E}[\mathrm{E}(X \mid Y)]^2\right) . \]
Using the decomposition of variance into expected values, we finally have: \[ \operatorname{Var}(X)=\mathrm{E}[\operatorname{Var}(X \mid Y)]+\operatorname{Var}[\mathrm{E}(X \mid Y)] \]
Compound Poisson Process
\[ Z(t) = \sum_{k=1}^{X(t)}Y_k \]
Take its exptcation:
\[ E[Z(t)]= E[E[Z(t)|X(t)]]=E[\mu X(t)]= \mu\lambda t \]
And its variance:
\[ Var[Z(t)]=E[Var[Z(t)|X(t)]] + Var[E(Z(t)|X(t))] \]
Used the Law of total Variance, and sub in previous result \(E(Z(t)|X(t))=\mu X(t)\)
hence the result goes to
\[ E[\sigma^2X(t)] + Var[\mu X(t)]= \sigma ^2 \lambda + \mu^2 \sigma^2 \]
Note: use MGF + Law of toal expectation can reach the same result but it’s easier beyond 3rd moment
Example 13.1
Aftershocks constitute the greatest proportion of shocks in
an earthquake catalog。
Solution
\(Z(t)=\sum_{k=1}^{X(t)} Y_k\), for \(t \geq 0\), where \(X(t)\) is a homogeneous Poisson process with \(\lambda=0.9\) (year) and \(Y\) has a geometric distribution with parameter \(p=0.2\), mean \(\mu=1 / p\), and variance \(\sigma^2=(1-p) / p^2\). Note that \(E[Z(t)]=\mu \lambda t\) and \(\operatorname{Var}[Z(t)]=\left(\sigma^2+\mu^2\right) \lambda t\). Find the approximate probability that at least 100 aftershocks within the next 50 years.
Estimate \(\lambda\)
Taking mean on the shocks
Estimate \(p\)
Method of moment
Example 13.2
Suppose that families migrate to an area at a Poisson rate \(\lambda=2\) per week. Find the approximate probability that at least 240 people migrate to the area within the next 50 weeks.
\[ E[X(50)]= \lambda * 50 *E(Y_i) = 2*50 *5/2 \]
Example 13.3
The number of hours between successive train arrivals (T ) at the station is uniformly distributed on (0, 1).Passengers arrived according to a Poisson process with rate 7 per hour. Suppose a train has just left the station. Let X denote the number of people who get on the next train. Find \(E(X) ,Var(X)\)
Solution
\[E(X) = E(E(X|T)) = E(7T(t)) = 0.5*7\]
\[ Var(X)= E(Var[X|T])+Var[E(X|T)] \]
Simplifies:
\[ RHS= E(Var[X|T])+ 49Var[T] = 7*\frac{1}{2}+ \frac{49}{12} \]