Question about the Poisson distributed of random variable

In summary, the given equation shows the equivalence of probabilities for the number of events occurring in different intervals within a Poisson process, using memorylessness property and proper notation.
  • #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
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
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
'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
user366312 said:
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

1. What is the Poisson distribution?

The Poisson distribution is a probability distribution that describes the number of events that occur in a fixed interval of time or space. It is often used to model rare events, such as the number of customers arriving at a store in a given hour or the number of accidents on a particular stretch of road in a day.

2. What is a random variable?

A random variable is a numerical quantity whose value is determined by the outcome of a random event. In the context of the Poisson distribution, it represents the number of events that occur in a given interval.

3. What are the key characteristics of the Poisson distribution?

The key characteristics of the Poisson distribution are that the events must occur independently of each other, the average number of events must be constant, and the probability of an event occurring must be the same for any given interval.

4. How is the Poisson distribution different from other probability distributions?

The Poisson distribution is different from other probability distributions, such as the normal distribution, because it is used to model rare events rather than continuous data. It also assumes that the events occur independently of each other, whereas other distributions may have different assumptions.

5. How is the Poisson distribution used in real-world scenarios?

The Poisson distribution is commonly used in various fields, such as finance, biology, and engineering, to model events that occur randomly and infrequently. It can be used to make predictions about the likelihood of rare events and to analyze data from experiments or observations.

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