Sales Toll Free No: 1-800-481-2338

# How to Derive the Expectation and Variance of a Binomial Distribution?

TopFor any Random Variable “b” let us see how to derive the expectation and variance of a Binomial Distribution:

1. Proof of expectation:

Where, 'N' is the number of mutually exclusive Bernoulli Trials and 'p' is the Probability of success.

E [b] = ∑(b ∈R)b pb (b)

Where, pb (b) is the probability mass function.

= N(b =0) b N Cb pb (1 – p)N – b

= 0 +N(b =1)b N Cb pb (1 – p)N – b

= N(b =1)b N! / (b! (N – b)! pb (1 – p)N – b

Substituting a = b – 1

= N – 1(a =0)(a +1) N! / ((a + 1)! (N – a)! pa + 1. (1 – p)N – a – 1

= N– 1(a =0)(a +1) N (N – 1)! / ((a + 1) a! ((N –1) – a)! p pa (1 – p)N – a – 1

=Np N – 1(a =0)(N – 1)! /(a! ((N –1) – a)! pa (1 – p)N – a – 1

= Np n – 1(a =0)pa(a)

And n – 1(a =0)pa (a) = 1

So, E[b] = Np

2. Proof of variance:

E [b2] = N(b =0)b2 N Cb pb (1 – p)N – b

= N(b =1)b2 N! / (b! (N – b)! pb (1 – p)N – b

Substituting a = b – 1,

= N – 1(a =0)(a +1) (a +1) N! / ((a + 1)! (N – a)! pa + 1. (1 – p)N – a – 1

= N– 1(a =0)(a +1) (a +1) N (N – 1)! / ((a + 1) a! ((N –1) – a)! p pa (1 – p)N – a – 1

=Np N – 1(a =0)(a +1) (N – 1)! /(a! ((N –1) – a)! pa (1 – p)N – a – 1

= Np N – 1(a =0)(a +1) pa (a)

So, E[b2] = Np (E[a] + 1) = Np ((N-1)p + 1) = N2p2 + Np (p-1)

So, Variance = E[b2] - E[b]2 = N2p2 + Np (p-1) - N2p2 =Np (p-1)