# Question about the Poisson distributed of random variable

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Gold Member
The following is related to Poisson process:

$$P(N_1=2, N_4=6) = P(N_1=2, N_4-N_1=4) = P(N_1=2) \cdot P(N_3=4)$$

Why is $$(N_3=4)=(N_4-N_1=4)$$?

Can anyone explain?

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Gold Member
I can guess that the answer is memorylessness -- in fact ##(N_3=4)=(N_4-N_1=4)## screams exactly that even though I don't know what these symbols stand for.

To say much more requires you to define some of the symbols you are throwing around and provide some background.

Gold Member
There is not much background. A lambda value is given and I have to find the value of $$P(N_1=2, N_4=6)$$.

What I understand is:

1. $$P(N_1=2, N_4=6)$$ means "the probability of count of 2 items arriving at step-1, AND 6 items arriving at step-4.

2. $$P(N_1=2, N_4-N_1=4)$$ means that "the probability of count of 2 items arriving at step-1 AND the 4 items as the difference of steps 4 and 1.

3. which is same as $$P(N_1=2, N_3=4)$$ i.e. "the probability of count of 2 items at step-1 AND 4 items at step-3".

I understand (2) as (1) is represented as a "difference" term in (2).

But, how is (2) and (3) equivalent?

Gold Member
'step' isn't exactly standard terminology for continuous time processes like a Poisson and it is also ambiguous as to how they 'steps' overlap or don't. So I am once again guessing.

My guess is step one is ##(0, t]##, step two is ##(0, 2t]##, step 3 is ##(0, 3t]## and step 4 is ##(0,4t]##

So first things first, simplify and show

##P(N_4 - N_1) = P(N_3)##
i.e. Pr ##k## arivals in ##(t, 4t]## is same as in ## (0, 3t]## -- why is this true? memorylessness (see stationary and independent increments). A slick way to see this is to look at the renewal function ##m(t)## which is linear for Poisson processes and in general uniquely characterizes a renewal process (at all points of continuity). There are more direct routes -- one way or another you need to get your arms around memorylessness.

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now consider ##P(N_1, N_4 - N_1) = P(N_1, N_3)##
This doesn't hold given my guess of your term 'step'.

consider

##k = \lceil 1000000 \lambda \rceil##
(or insert some other natural number that is much larger than ##\lambda##)

and ##r \geq k##

##P(N_1 =k , N_4 - N_1 \geq r) = P(N_1 =k) P( N_4 - N_1\geq r \big \vert N_1=k) ##
##\neq P(N_1=k) P( N_3\geq r\big \vert N_1=k) = P(N_1=k, N_3\geq r)##

why? because if we divide out ##P(N_1 = k)##
we have
##\epsilon \lt 1##
where ## \epsilon## may be made arbitrarily small (and in any case it is certainly ##\lt 1##).

(note if you don't like e.g.,
##P(N_1 =k , N_4 - N_1 \geq r) ##
then consider

##P(N_1 =k , N_4 - N_1 \geq r) = \sum_{i=r}^\infty P(N_1 =k , N_4 - N_1 = i) ##
so we can look at point wise bounds and sum over the bound to recover the above inequality.

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note: I saw you were having trouble LaTeX -- as you've discovered double dollars works here but single dollars don't -- you can enclose with double hashtags instead of single dollars though

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Dearly Missed
The following is related to Poisson process:

$$P(N_1=2, N_4=6) = P(N_1=2, N_4-N_1=4) = P(N_1=2) \cdot P(N_3=4)$$

Why is $$(N_3=4)=(N_4-N_1=4)$$?

Can anyone explain?
There are different types of "1,2,3,..." here, but you are confusing them because of inadequate notation. Presumably, your ##N_a## means N[0,a] = number of events in the interval ##0 \leq t \leq a.## More generally, let us write ##P(N[a,b] = k)## for the probability of ##k## events in the interval ##a \leq t \leq b.##

So, what you meant to write at the start (in full notation) is
$$P(N[0,1]=2, N[0,4]=6) = P(N[0,1]=2, N[0,4]-N[0,1] = 4) \\= P(N[0,1]=2) P([N[1,4] = 4).$$ If you use a shorthand notation (but without explanation) you could write this last factor as ##P(N[1,4]=4) = P(N_3=4),## where here ##P(N_b = k)## means the probability of ##k## events in an interval of length ##b##, with un-specified starting and ending points. So your ##N_3## is not the same as ##N[0,3]##, but it is the same as ##N[1,4].##

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Gold Member