Introduction for algebra statistics:

Algebraic statistics are the uses of algebra to advance statistics. Algebra has been useful for the experimental design, parameter estimation, and hypothesis testing. Traditionally, algebraic statistics have been associated with the design of experiments and multivariate analysis (especially time series). In recent years, the terms “algebraic statistics" has been sometimes restricted, sometimes being used to label the use of algebraic geometry and commutative algebra in statistics. (Source.Wikipedia)

  

Algebra statistics:

 

Algebraic statistics function: (Cumulative Distribution function)

The distribution functions of a random variable X is

F(x) = P(`<=` x) =   `sum_(x (i))^x` p(xi) : (− ∞ < x < ∞).

 

Examples for algerbra statistics:

 

Example 1:

Find the chance algebra mass function, and the cumulative distribution functions for getting ‘3’s when two dice are thrown.

Solution:

Sample space

Two dice are thrown. Let X be the random variable of getting same

‘3’s. Therefore X can take the value random number 0, 1, 2.

P(no ‘3’) = P(X = 0) =`25/36`

P(one ‘3’) = P(X = 1) =`10/36`

P(two ‘3’s) = P(X = 2) =`1/36`

Probability mass function is given by

X  value                

0

1

2

(X = x)   value  

`25/36` `10/36`

`1/36`

Cumulative distribution function:

We have F(x) = `sum_(x i)^-oo` xP(X = xi)

Function of F(0) = P(X = 0) =`25/36`

Function of F(1) = P(X = 0) + P(X = 1) =`25/36` +`10/36` =`35/36`

Function of F(2) = P(X = 0 ) + P(X = 1) + P(X =2) =`25/36` +`10/36` +`1/36` =`36/36` = 1

X  value                

0

1

2

P(X = x)   value    

`25/36 ` 

 `35/36 `     

1

 

 

Example 2:

A random variable X has the following chance mass function

 X  value

 0

 1

 2

 3

 4

 5

 6

 P(X = x)value

 t

 3t

 5t

 7t

 9t

 11t

 13t

(1) Find t value

(2) Evaluate P(X < 4), P(X ≥ 5) and P(3< X `<=` 6) value

Solution:

(1) Since P(X = x) is a chance mass function value`sum_(x=0)^6` P(X = x) = 1

    P(X=0) + P(X = 1) +P(X = 2) +P(X = 3) +P(X = 4) +P(X = 5)+P(X = 6) = 1.

⇒ t + 3t + 5t + 7t + 9t + 11t + 13t = 1 ⇒ 49 t = 1 ⇒ t =`1/49`

(2)Function P(X < 4) = P(X = 0) + P(X = 1 ) + P(X = 2) + P(X = 3)

=`1/49 ` +`3/49` +`5/49 ` +`7/49 ` =`16/49`

Function P(X `>=` 5) = P(X = 5) + P(X = 6) =`11/49` +`13/49` =`24 /49`

Function P(3 < X `<=` 6) = P(X = 4) + P(X = 5) + P(X = 6) =`9/49` +`11/49` +`13/49` =`33/49`