Question about the Poisson distributed of random variable

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  • #1
user366312
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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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  • #2
StoneTemplePython
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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.
 
  • #3
user366312
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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?
 
  • #4
StoneTemplePython
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'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.

- - - - -
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.

- - - - -

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
 
  • #5
Ray Vickson
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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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  • #6
user366312
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