Poisson Process in 2D and 3D

107 Views Asked by At

I have a homework saying :

When Picard, Troi, Data and La Forge are returning to the Enterprise, they must avoid
bubbles of temporal disruptions, inside of which time runs at different speeds. Assume
the bubbles are stationary relative to the Enterprise, their centers are distributed with
a Poisson Pointe Process with intensity λ, and they are all spherical with radius r. Take
the runabout containing Picard et. al to be a square with side length l.
(a) Choose a point at random for the location of the center of the runabout.
What is the probability it is intersecting a time bubble? (2D and 3D)
(b) From this random point, the runabout must travel a distance d to reach the
Enterprise. What is the probability they intersect (must maneuver around) more
than 3 time bubbles while following a straight route to the Enterprise at speed c?
(2D and 3D)

And I know for Poisson Process N(t) the number of events happened in time [0,t] with rate $\lambda$ has the property : $p(N(t + \delta t) - N(t) = m) = e^{-(\lambda * \delta t)} * \frac{(\lambda * \delta t)^m}{m!}$ so for part (a), in 2D, N(A) = # of events in area A, then $P(N(A) >= 0) = 1 - e^{-\lambda A}= 1- e^{-\lambda (4lr + \pi r^2 + l^2)}$ and in 3D , N(A) = # of events in volume V, then $P(N(V) >= 0) = 1 - e^{-\lambda V}= 1- e^{-\lambda (6l^2 r + 2 \pi r^2 l+ l^3)}$. Is that right? And how can I do the part(b) of this problem, since I approach it with speed c, so I will use $t = \frac{d}{c}$ time to get to the destination and does that mean for example the 2D case, I can just use $P(N(d/c) >= 3) = 1 - e^{-(\lambda * d/c)} * \frac{(\lambda * d/c)^2}{2} - e^{-(\lambda * d/c)} - e^{-(\lambda * d/c)} * (\lambda * d/c)$ I feel I am confused here. Thank you for any help.