Sum of Uniform Distributions

Uniform distribution is one of  the part of probability distribution function. The simplest distribution of uniform distribution is also known as uniform random distribution. Two types of uniform distribution:

1.Discrete Uniform distribution,

2.Continues Uniform distribution.

Two types of Continuous variables distributions

Uniform distribution
Exponential distribution.

Uniform distribution:

In probability theory and statistics, the continuous uniform distribution is a family of probability distributions such that for each member of the family, all intervals of the same length on the distribution's support are equally probable.

Exponential distribution:

In probability theory and statistics, the exponential distributions are a class of continuous probability distributions. They describe the times between events in a Poisson process, Study process in which events occur continuously and independently at a constant average rate.

Uniform distribution probability density function (PDF) is:

phi(x) = {(1/ (b - a ) if a
Uniform distribution  cumulative distribution function (PDF) is:

phi(x) =(x-a)/(x-b)

Expectation or mean of uniform distribution:

Let X be a random variable. We represent the probable value (expectation or mean) of X as either or E(X). Mean describes the (probability-weighted) average value for the random variable.

Probability density functions (PDFs) for mean uniform distribution is denoted by two random variables. The mean of each are indicated on the x-axes

If X is discrete random variable means, we define its expectation as

E(x) = sum x phi(x)

Where phi is the probability function.

If X is continuous random variable means, we define its expectation as

E(x) = int_-oo^ooxphi(x)dx

Where phi is the probability density function.

Standard deviation  uniform distribution:

The variance of a random variable X, denoted by sigma^2 or var(X).

Variance of uniform distribution is

Var ( X ) = E [( X- mu )2]

Where mu  = E ( X )

Standard deviation of uniform distribution is denoted by sigma .

mu = ( a + b) / 2

sigma= (b -a) / 2 sqrt(3)

Finally we get

Mean = ( a +b ) / 2

Variance = (b-a)2 / 12

Moment generating function = (etb - eta) / t(b - a)

Types of Uniform Distribution:

Uniform Distributions are classified in to two types,

Continues uniform distributions, and
Discrete  uniform distributions

1. Continues uniform distributions:

The continues uniform distributions is the density function of the random variable. Continues interval between a and b. The density function of  continues uniform distribution is ,

 f(x) ={( 1/(b-a) "when a<= x<=b") , (0 "when xb"):}

2. Discrete Uniform Distributions :

Discrete distributions are also known as  statistical distributions , It is the simplest way of Probability distribution functions.

Probability distributions functions  of p(Xm)  is defined  over m= 1,2,….,N ,

The discrete distribution function as

D(Xn)= sum_(m=1)^n p(Xm)
Uniform Random Distribution General Formula:

The general formula of probability density function of the uniform random distribution function is defined as follows:

f(x) = 1 /b-a               for  a<= x<= b

Where a is the position parameter and (b-a) is the scale parameter. In case where a = 0 and b = 1 is called the standard Uniform random distribution:

Uniform random distribution  is called as simple distribution or rectangular distribution function. In this distribution  their may be chance tooccurring constant probability distribution function. Uniform distribution analysis only numerical data value. That all of the data are  arranged in systematic order is called uniform random distribution.

The equation of the standard uniform random distribution is

f(x) = 1          for   0 <=x <= 1 .

These all are important in the Uniform random distributions.

Examples Problems for Uniform distributions

Example 1: f(x) = 4x for 0 x  1. Find the expected value continuous of given f(x).

Solution:

E (X) = X’ =  x f(x) dx

=   x (4x) dx

=    4x2 dx

= [(4X^3) /3]

=(4/3) - 0

E(X) = 1.333

Example 2: f(x) = x^2 for 0 x  1. Find the expected value continuous of given f(x).

Solution:

E (X) = X’ =  x f(x) dx

=   x (x^2 ) dx

=    x^3 dx

=[(x^3)/3]

=(1/3) - 0

E(X) = 0.333

Example 3: Find the continuous variables of given f(y) = 4e-3y.

Sol:

f(Y) =   ? y (4e -3y) dy

= (-4ye-3y – e -3 / 3)

= ½ (-4ye-3– e -3) – ½ (-4 (0) e0-e0)

f(Y) = ½

Example 4: Find the continuous variable of given f(x) = 4x for 0 x  1.

Sol:

f (X)       =  x f(x) dx

=   x (4x) dx

=    4x2 dx

= [4/ 4  x3]

= 1 - 0

f(X) = 1

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Example 5: Find mean and variance value. From the given value a = 2, b = 4

Solution:

The probability density function F(x) = 1/(4 -2)

F(x) = 1/2 = 0.5

Mean = (a +b ) / 2

Here a = 2 and b = 4

Mean = (2 + 4) / 2

After simplify, we get

= 6 / 2 = 3

Mean = 3

Variance = ( b - a)2 / 12

= (4 - 2)2 / 12

=  (2)2 / 12

After simplify, we get

= 4 / 12 = (1 / 3) = 0.333

Variance = 0.333

Exercise Problems:

Problem -1: f(x) = x^3 for 0 x  1. Find the expected value continuous of given f(x).

Answer:  0.25

Problem -2: f(x) = 2x^3 for 0 x  1. Find the expected value continuous of given f(x).

Answer: 0.5

Problem -3: f(x) = 5x for 0 x  1. Find the expected value continuous of given f(x).

Answer:  2.5