91Ó°ÊÓ

The intensity function \(\lambda(t)\) of a nonhomogeneous Poisson process is given by the derivative of the mean value function \(m(t)\): $$ \lambda(t) = \frac{d m(t)}{d t} $$ First, we need to find the derivative of the given mean value function: $$ m(t) = t^2 + 2t $$ $$ \frac{d m(t)}{d t} = 2t + 2 $$ So the intensity function \(\lambda(t)\) of the Poisson process is: $$ \lambda(t) = 2t + 2 $$

Now, we need to calculate the number of events in the interval \([4,5]\), which can be done by subtracting the mean number of events at \(t = 4\) from the mean number of events at \(t = 5\). So, we get: $$ m(5) = 5^2 + 2(5) = 25 + 10 = 35 $$ $$ m(4) = 4^2 + 2(4) = 16 + 8 = 24 $$ Then, subtract \(m(4)\) from \(m(5)\): $$ m(5) - m(4) = 35 - 24 = 11 $$ So, there are 11 expected events between time \(t = 4\) and \(t = 5\).

Now that we have the expected number of events in the interval, we can use the Poisson distribution to find the probability of \(n\) events occurring between the times \(t=4\) and \(t=5\). The probability mass function of a Poisson distribution is given by: $$ P(X = n) = \frac{e^{-\lambda} \lambda^n}{n!} $$ where \(X\) is the random variable representing the number of events, \(n\) is the specified number of events, \(\lambda\) is the expected number of events (in this case, it's equal to \(m(5) - m(4)\)), and \(e\) is the base of the natural logarithm. So, the probability of \(n\) events occurring between time \(t=4\) and \(t=5\) is: $$ P(X = n) = \frac{e^{-11} \cdot 11^n}{n!} $$ This is the final equation you would use to calculate the probability of \(n\) events occurring in the time interval \([4,5]\). To find the probability for a specific number of events, just substitute the desired value of \(n\) into the equation.