If x Bin use the binomial expansion to find the mean and var

If x Bin(), use the binomial expansion to find the mean and variance of X. To find the variance, use the second factorial moment E[X(X - 1 )] and note that x/x! = 1/(x - 1)! When x GE 1.

Solution

The probability function for a binomial random variable is

b(x; n, ) = n C x * ^x * (1 ) ^ nx

This is the probability of having x successes in a series of n independent trials when the probability of success in any one of the trials is .

E (X) = x * n C x * ^x* (1 ) ^ nx ( x is from 0 to n)

= x * ( n! / x!*(n-x)! ) * ^x *(1 ) ^ nx ( x is from 0 to n)

= n! / [(x-1)!*(n-x)!] * ^x* (1 ) ^ nx (x is from 1 to n)

Since x=0 term vanishes.

Let y=x-1 and m=n-1 it becomes x=y+1 and n=m+1 into the last sum.

E(X) = (m+1)! / [y!*(m-y)!] * ^(y+1) * (1-)^(m-y) ( y is from 0 to m)

= (m+1)* p * m!/(y!*(m-y)!) * ^(y) * (1-)^(m-y) (y is from 0 to m)

= np * m!/y!*(m-y)! * ^(y) * (1-)^(m-y) (y is from 0 to m)

We know that the result of binomial theorem,

(a+b)^m = m!/y!*(m-y)! * a^y * b^(m-y) ( y is from 0 to m)

By setting a = and b = 1 -

m!/y!*(m-y)! * ^y * (1-)^(m-y) =    m!/y!*(m-y)! * a^y * b^(m-y) ( y is from 0 to m)

(a+b)^m = (+1-)^m = 1^m = 1

E(X) = n

To find a variance,

Consider E(X(X-1)) = x*(x-1) * (nCx)* ^x* (1-)^n-x ( x is from 0 to n)

= x*(x-1) * (n!/x!*(n-x)!)* ^x *(1-)^n-x (x is from 0 to n)

= n!/( (x-2)!*(n-x)! ) *^x* (1-)^n-x (x is from 2 to n)

= n*(n-1)*^2 * (n-2)! / (x-2)!*(n-x)! ( x is from 2 to n)

Similarly here we using y=x-2 and m=n-2

= n*(n-1)*^2 * m!/(y!*(m-y)!) *^y* (1-)^m-y (y is from 0 to m)

= n*(n-1)*^2* (+(1-))^m

= n*(n-1)*^2

Variance(X) = E(X²) - [E(X)]²

= E(X(X-1)) + E(X) - E(X)²

= n*(n-1)*^2 + n - (n)²

= n*(1-)