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Foundation of Statistics Sec. 3 under Shirin Cheat Sheet (DRAFT) by

Continuation from Foundation of Statistics Sec. 1 & 2 under Shirin

This is a draft cheat sheet. It is a work in progress and is not finished yet.

Binomial & Bernoulli Distri­butions

Introduction:
Given an experiment with 2 possible outcomes (Success & Failure ie. 1 and 0 ie. Binary) ran 5 times.
Sample space Ω becomes a combin­ation of the 5 results ie. Ω = {0,1} x {0,1} x {0,1} x {0,1} x {0,1}.
With the inform­ation that event A is when any one experiment is a success we are able to deduce that set A = {(0,0,0,0,1),(0,0,0,1,0),(0,0,1,0,0),(0,1,0,0,0),(1,0,0,0,0)}
Equate the probab­ility of success happening (not necess­arily in A, only in general as a result of the experiment being ran any 1 time) is given by p, and conseq­uently (as there are only 2 possib­ili­ties) failure is given by 1-p. Remember p is the probab­ility of a single result occuring that is considered a success. In the case we are flipping a coin and heads is a success p = P(H) = 1/2, in the case of a dice where 5 is considered a success p = P(5) = 1/6
We can obviously see that P(A) = 5(1-p)4p or in plain english - "­There are 5 possible combin­ations of A occuring, contained within each 4 failures (1-p)4 and 1 success p"
Bernoulli Distribution...
- Used to model experi­ements with binary outcomes (success or failure)
- Defini­tion: Discrete rv X has a Bernoulli distri­bution with parameter p (p is the probab­ility of success occuring in any given experi­ment) where 0<=­p<=1 if its probab­ility mass function (pmf) is given by...
px(1) = P(X=1) = p
and
px(0) = P(x=0) = 1-p
ie. Probab­ility of success occuring is p and probab­ility of failure is 1-p
Binomial Distribution:
A discrete rv X has a binomial distri­bution with parameters n and p where n = 1,2... and 0<=­p<=1, if the pmf is given by... Px(k) = P(X = k) = (
k
n)pk(1-p)n-k

Denoted by Bin(n,p)
Expectation of binomial distri­bution Bin(n,p) is E(X)=np
Variance is Var(X)­=np­(1-p)
 

Hyperg­eom­etric Random Variable

Introduction:
When an experi­ement consists of...
1. Drawing n random elements WITHOUT REPLAC­EMENT from a set of N elements.
2. s elements of N are special.
3. N - s are regular
Result: Our Hyperg­eom­etric rv X is the number of special elements in our random draw of n ie s∩n (sortof)

The Geometric Distri­bution

Introductin:
1. Observ­ations have 2 possible values Success and Failures{{nl]}2. Probab­ility of success (p) is the same for each observation.
Observations are all independent.
We are looking for the number of trials required to find the first success.

This is an example of an infinite experi­ment...
A discrete rv X has a geometric distri­bution with parameter p, where 0<=­p<=1, if its pmf is given by
px(k) = P(X=k) = (1-p)k-1p
for k=1,2,....
This distri­bution is denoted by Geo(p){{nl}E(X) = Σ
k=1
∞kp(1-p)k-1=1/p
Var(X) = (1-p)/p2
Memoryless Property:
P(X>n+k|X>k) = P(X>n)

Poisson Distri­bution

Introduction:
Used when we are interested in the number of X per the number of Y. eg. The number of cars on a road per hour. This is phrase as counts per unit interval.
When count is relatively low it can be modelled as a Poisson Distri­bution.
Assumptions:{{n}}1. Homoge­neity: The rate λ (the rate at which the event occurs) is constant over time/s­pace. This implies that at any unit interval E(X)=λ
2. Indepe­ndence: Number of events in disjoint intervals are indepe­ndent of one another.
Please refer to Defini­tions and Other Formulae sheet for more inform­ation.
Property of Poisson:
If we sum 2 Poisson rv X and Y, our result is also Poisson.{{nl}x~Poisson(λ) and Y~Pois­son(µ) then X+Y~Po­iss­on(λ+µ)