Suppose that x has the binomial distribution with probability mass function

Question

f(x)=(5 x)(1/4)^x(3/4)^5-x, x=0,1,2.....5

1.Deduce the moment generating function of x

2.Use your moment generating function above to Calculate the mean and variance of the distribution of X.

Answer 1

I interpret the binomial distribution as

∑⁵C ͓(¼)ˣ(¾)⁵⁻ˣ) where 0≤x≤5 as the expansion of (¼+¾)⁵. I have used ⁵C ͓ In place of (5 x).

The Binomial Theorem states that (a+b)ⁿ=∑ⁿCᵣaʳbⁿ⁻ʳ for r in {0,n}.

The mean is 5×¼=5/4 and the variance is 5×¼×¾=15/16 where mean=np and n=5 and p=¼, and variance=np(1-p).

Calculating the mean and variance through M ͓(t) should give these results.

1. M ͓(t)=E(eᵗˣ)=∑eᵗˣ(⁵C ͓)(¼)ˣ(¾)⁵⁻ˣ= ∑(⁵C ͓)(¼eᵗ)ˣ(¾)⁵⁻ˣ = (¼eᵗ+¾)⁵ by the Binomial Theorem. So the MGF is M ͓(t)=(¼eᵗ+¾)⁵.
2. To find the mean and variance we need to find dM/dt=5(¼eᵗ+¾)⁴・¼eᵗ. Now we need to evaluate the first moment, M'(0)=5・1⁴・¼=5/4. This is the mean. Now we need d²M/dt²=M''(t), the second moment. M''(t)=vu'+uv' where u(t)=5(¼eᵗ+¾)⁴ and v(t)=¼eᵗ. So u'=5(¼eᵗ+¾)³eᵗ and v'=¼eᵗ. So M''(t)=5e²ᵗ(¼eᵗ+¾)³/4+5eᵗ(¼eᵗ+¾)⁴/4, M''(0)=5/4+5/4=5/2. Variance = M''(0)-(M'(0))² = 5/2-(5/4)² = 5/2-25/16 = 40/16-25/16 = 15/16.