On a particular highway cars pass as a possionprocess with intensity 3 cars per minute and motorcycles with intensity 1 motorcycles per minute. Other types of vehicles are not counted.
a) What is the probability that next vehicle that passes is a motorcycle?
b) What is the probability that there will pass exactly $5$ vehicles during the next minute?
c) What is the probability that there will pass exactly 3 cars before the next motorcycle?
I think I do understand what a possionprocess is. It's just impulses that happen (or points if you will) within an alotted period of time $t$ and the time inbetween these impulses $T_1,T_2...,T_n$ are exponentially distributed with parameter $\lambda t$.
a): Let $T_1$ be the time between cars and $T_2$ the time between motrocycles. So, mathematically we are looking for the probability that the time between cars is greater than time between motorcycles, that is $P(T_1>T_2)$. Computing this, in the book they say that
$$P(T_1>T_2)=\int_0^{\infty}(1-F_{T_1}(x))f_{T_2}(x) \ dx = \int_{0}^{\infty}e^{-3x}e^{-x}=\frac{1}{4}.\tag1$$
Questions for a):
- I don't understand $(1)$. Why do I get that integral?
- Is there another way to do this? Would it not, for example, be possiblt to compute $P(T_2<T_1)?$ How would this pan out?
For b) and c) I would just like some verification that everything is correct.
b): Let $X\sim \text{Poi}(3)$ be the number of cars that passes each minute, $Y\sim\text{Poi}(1)$ the number of motorcycles and $X+Y=Z\sim\text{Poi}(4)$ the number of vehicles. It can be proven by using PGF's that the sum of two random variables with Poisson distributions with parameters $\lambda$ and $\mu$ have the convoluted distribution that is a Poisson distribution with parameter $\lambda + \mu.$ So here we are looking for $$P(Z=5)=\frac{e^{-4}\cdot 4^5}{5!}\approx 0.156.$$
c) Here we can use the memoryless property of the exponential distribution. In a we have that the probability that the next vehicle that passes is a car is $1-1/4=3/4$. So this has to happen 3 times first, and the 4'th vehicle has to be a motorcycle, that is
$$P=\left(\frac{3}{4}\right)^3\left(\frac{1}{4}\right)=0.106.$$
The easiest way to see that the integral in (a) is correct is to look at a discrete example. Let's assume that $T_2$ takes on discrete values $1,2,\dots$ and so on. Then, by conditioning on $T_2$ and using the law of total probability, $$ P(T_1 > T_2) = \sum_{x=1}^{\infty} P(T_1 > T_2 | T_2 = x) P(T_2 = x) \\ = \sum_{x=1}^{\infty} P(T_1 > x | T_2 = x) P(T_2 = x) = \sum_{x=1}^{\infty} P(T_1 > x) P(T_2 = x). $$ The continuous analog would be an integral instead of a sum and $P(T_2=x)$ should be replaced by its density. I hope this makes the argument clearer to you!
Yes, you could do it differently by calculating $P(T_1<T_2)$ in a similar fashion. However, this doesn't make the argument easier in any sense.
b) and c) look fine to me.
Edit: I cannot find a proof of the exact verification of just replacing the summation by the integral as I suggested. I am not sure but it might be an application of the Law of total expectation. Some would call this actually the continuous analog of the total law of probability.