Answer:
(a) No.
(b) Yes.
Step-by-step explanation:
(a)
Let N(t) denote the number of points on the interval [0,t].
Consider the random variables N(1/4) and N(1/2)-N(1/4).
P(N(1/4) = 1) = P(0<X≤1/4) = 1/4,
P(N(1/2)-N(1/4) = 1) = 1/4.
However,
P(N(1/4) = 1, N(1/2)-N(1/4) = 1) = 0,
as the process cannot have two points on the interval [0,1/2].
Since 0 = P(N(1/4) = 1),
P(N(1/2)-N(1/4) = 1) = 6 ≠ P(N(1/4) = 1) * P(N(1/2)-N(1/4) = 1) = 1/16,
the counting process does not have independent increments
(b)
N(s+t)-N(s) and N(t) have the same distribution.
First consider N(t). Since the distance between points in the counting process is exactly 1,
N(t) must have either [t] or [t]+1 points, where [x] is the greatest integer less than or equal to x.
Since N(t)=[t]+1 only if there is a point on the interval ([t],t)
by this the counting process have stationary increments
Hope this helps!
