Suppose we have a bus station. Each bus stops in this station for a pre-determined amount of time. due to traffic and other reasons, each bus can be late or early to arrive at the station, and we have the PDF (Probability Density Function) that describes each deviation in time that each bus could have. We would like to find what are the probabilities that $k$ buses will be in the station at the same time, if we have $n$ buses in total ($k<=n$).
To put the problem in mathematic variables: Note $n$ is the number of buses that are supposed to stop in the station. Note each bus with a letter ("A", "B", "C" and so on). Define the following for each $I\in$ {A,B,C...} :
- The duration a bus stays in the station is $L_I$.
- The start time that the bus arrives will be $s_I$ and the end time $e_I$.
- The deviation of the bus in time will be noted as $\Delta I$. Note that if the bus is early then $\Delta I <0$ and if late then $\Delta I >0$
- The PDF of the bus deviation in his times is $f_I$ and the CDF is $F_I$.
Also note the time difference between 2 buses $I$ and $J$ is: $$ T_{IJ} = |e_I - s_J| $$ (note that $T_{JI}$ does not equal to $T_{IJ}$)
My Idea on how to solve this: for the 2 buses case - say we have 2 buses: A and B, and assume that A arrives before B. Now we can have 2 situations: A and B are not set to be at the station together at all, and the situation in which they are. We can use convolution to define: $$ g_{A-B} = f_A*f_{-B} $$ $$ f_{-B}(b)=f_B(-b) $$
So for the first situation A and B can only arrive at the station at the same time if $ T_{AB}<\Delta A-\Delta B < T_{BA}$ and so: $$ P(sitaution 1 -> Arrive At The Same Time) = \int_{T_{AB}}^{T_{BA}}{g(t)dt} $$ and for situation 2 the condition for the buses to still arrive at the station at the same time will be: $-T_{AB} < \Delta A - \Delta B < T_{BA}$ , so: $$ P(sitaution 2 -> Arrive At The Same Time) = \int_{-T_{AB}}^{T_{BA}}{g(t)dt} $$
Now, I tried to solve it for the 3 buses case, but just couldn't figure out how to do it in a way that will help me figure the $n$ buses case.
Is there any way to solve for the $n$ buses case? if so, how to do it? Thanks.
$$P(\text{two there at the same time}) = P(\text{second bus arriving}|\text{first bus is there})\times P(\text{first bus being there})$$
$$P(\text{three there at the same time}) = P(\text{third bus arriving}|\text{two there at the same time})\times P(\text{two there at the same time})$$