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

For 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 p_b (b)

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

= ^N∑_{(b =0)} b ^N C_b p_^b (1 – p)^{N – b}

= 0 +^N∑_{(b =1)}b ^N C_b p^b (1 – p)^{N – b}

= ^N∑_{(b =1)}b N! / (b! (N – b)! p^b (1 – p)^{N – b}

Substituting a = b – 1

= ^{N – 1}∑_{(a =0)}(a +1) N! / ((a + 1)! (N – a)! p^{a + 1}. (1 – p)^{N – a – 1}

= ^{N– 1}∑_{(a =0)}(a +1) N (N – 1)! / ((a + 1) a! ((N –1) – a)! p p^a (1 – p)^{N – a – 1}

=Np ^{N – 1}∑_{(a =0)}(N – 1)! /(a! ((N –1) – a)! p^a (1 – p)^{N – a – 1}

= Np ^{n – 1}∑_{^{(a =0)}}p_a(a)

And ^{n – 1}∑_{(a =0)}p_a (a) = 1

So, E[b] = Np

2. Proof of variance:

E [b²] = ^N∑_{(b =0)}b^{2 N} C_b p^b (1 – p)^{N – b}

= ^N∑_{(b =1)}b² N! / (b! (N – b)! p^b (1 – p)^{N – b}

Substituting a = b – 1,

= ^{N – 1}∑_{(a =0)}(a +1) (a +1) N! / ((a + 1)! (N – a)! p^a + 1. (1 – p)^{N – a – 1}

= ^{N– 1}∑_{(a =0)}(a +1) (a +1) N (N – 1)! / ((a + 1) a! ((N –1) – a)! p p^a (1 – p)^{N – a – 1}

=Np ^{N – 1}∑_{(a =0)}(a +1) (N – 1)! /(a! ((N –1) – a)! p^a (1 – p)^{N – a – 1}

= Np ^{N – 1}∑_{(a =0)}(a +1) p_a (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)