Poisson distribution from first principles

What is the probability that a dice will roll 6?

{6,6} {x,6} {6,x} where x ∈ [1,5] Total there are 11 outcomes where a 6 rolled.

p = 11/36.

What is the probability that there are 2 rolls of 6 in 3 rolls?

1      2      3
yes -- yes -- yes
           -- no

    -- no  -- yes
           -- no
           
no  -- yes -- yes
    -- no

    -- no  -- yes
           -- no

Let X be number of events where dice is rolled to 6.

We wish to find P(X=2).

P(X=2) = Number of events where dice rolled to 6 * Probability of event occurring. P(X=2) = 3 * Probability of event occurring. P(X=2) = 3 * p^2 * (1-p)^(3-2)

P(X=2) = 3 * p^2 * (1-p)^(3-2)

We note from the diagram above that there are 3 occasions where dice could roll six. We can come to this conclusion by just choosing 2 occasions where dice rolls six out of 3. 3C2 = 3.

Then just substitute general variables to obtain binomial distribution. P(X=2) = (3 c) * p^2 * (1-p)^(3-2) P(X=k) = (N k) * p ^ k * (1-p)^(N-k)

Binomial to poisson.

P(X=k) = (N k) * p^k * (1-p)^(N-k)

Given:

  1. p, probability of a specific event occurring.
  2. k, the number of times it occurs.
  3. N, the upper bound on number of times it can occur.

We can obtain binomial distribution.

Now suppose we want to deal with continuous domain. How??? there can be infinitely many events? N=∞

We can use limits.

P(X=k) = (N k) * p^k * (1-p)^(N-k)

Define “choose”

First define (N k). We start of able to choose N objects. After we choose 1 from the N, We want to choose N-1, all the way until N-(k-1).

For example for k = 2, N = 3, We have, k-1 = 1 => N-(k-1) = 3-1 = 2

So we choose 1 from 3, then choose 2 from 3.

How many unique combinations are there? 3 * 2?

Nope, that’s wrong.

We can have duplicates, so we need to deduplicate them.

To gain some intution let’s use an example:

3 choose 2 for ABC: AB,AC BA,BC CA,CB

We have 3 duplicates.

Poisson

INCOMPLETE

See derivation here instead: https://www.le.ac.uk/users/dsgp1/COURSES/LEISTATS/poisson.pdf

let p = x/μ, then as N->∞,

P(X=k) = N! / (k!(N-k)!) * p^k * (1-p)^(N-k)
P(X=k) = N! / (k!(N-k)!μ^k) * x^k * (1-p)^(N-k)
P(X=k) = N! / ((N-k)!μ^k) * k! * x^k * (1-p)^(N-k)
P(X=k) = 1 * k! * x^k * (1-p)^(N-k)
P(X=k) = k! * x^k * (1-p)^(N-k)
P(X=k) = k! * x^k * (1-p)^N * (1-p)^-k
P(X=k) = k! * x^k * (1-p)^N * (1-p)^-k

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