Expected Value and Variance

Let X is a numerically values discrete random variable by the resources of sample space and distribution function m(x). The expected value E(X) is defined by

E(x) =  "sum_(x=0)^(n) (x*m(x))

Provided this calculation covers absolutely. Pass on the expected value as the mean value for the given discrete random variable, and to represent E(X) as  μ. If the above sum does not cover it completely, then we say that X does not have an expected value.

Variance:

Let X be a numerically values discrete random variable with expected value μ = E(X). Then the variance of X, denoted by V(X), is

V (X) = E((X - mu)^2)  .

V (X) can also given by,

V (X) = sum_(x=0)^(n)(x - mu)^2m(x)

here m is the  distribution function of X

Expected value of probability density function:

Let x be the probability density function, integrating x with the given function with using the limits, The expected value for the probability density function is given as,

E(x) = int_a^b x f(x) dx

Variance of probability density function:

Let x be the probability density function, integrating squared difference between x and expected value and with the function using the limits. The variance is given by,

V(x) = int_a^b ((x-E(x))^2) f(x) dx

Expected value and Variance - Example Problems:

Expected value and Variance - Problem 1:

Find the expected value and the variance for the discrete random variable. (1/5). Where x is from 0 to3.

Solution:

Expected value of the discrete random variable,

E(x) =  sum_(x=0)^(n) (x* m_x)

E(x) = 1(1/5) + 2(1/5) + 3(1/5)

E(x) = 1.2

Variance for the discrete random variable,

V (X) = sum_(x=0)^(n)(x - mu)^2m(x)

V (X) = (0-1.2)^2(1/5)+(1-1.2)^2(1/5)+(2-1.2)^2(1/5)+(3-1.2)^2(1/5)

V(x) = 1.07200

Expected value and Variance - Problem 2:

Find the expected value and variance for the probability density function.  2x  0<= x <= 1

Solution:

Expected value,

"E(x) = int_a^b xf(x)dx

E(x) = int_0^1 x (2x) dx

E(x) = int_0^1(2x^2) dx

E(x) = [(2x^3) /3 ]_0^1

E(x) = [2 / 3]

Variance:
"V(x) = int_a^b ((x-E(x))^2)f(x) d(x)

"V(x) = int_0^1 (x-(2/3))^2 (2x)dx

V(x) = int_0^1(x-(2/3))^2 (2x) dx

"V(x) = int_0^1[x^2+(4/9)-(4/3(x))] (2x) dx

V(x) =int_ 0^1[2x^3 +(8/9x)-(8/3)(x^2)]dx

"V(x) = [2(x^4/4)+(8/9)(x^2 /2)-(8/3)(x^3 /3)]_0^1

V(x) = [(2/4) +(4/9)-(8/9)]

V(x) = [0.5+0.44444-0.888888]

V(x) = 0.05555