Find the first four moments of binomial distribution
Answers
Recall that if X∼Bin(n,p), then E[X]=np and Var(X)=np(1−p). Given E[X]=4 and Var(X)=3, we have np=4 and np(1−p)=3. Hence n=16, p=
1
4
. So the distribution of X is given by
P(X=k)=(
16
k
)(
1
4
)k(
3
4
)16−k,k=0,1,…,16.
The second moment of X is
E[X2]=Var(X)+E[X]2=3+42=19.
The generating function of X is
E[sX] =
∞
∑
k=0 P(X=k)sk.
Since X is the sum of 16 i.i.d. Ber(p) random variables, each with generating function 1−p+ps, we have
E[sX]:=P(s)=(1−p+ps)n=(
3
4
+
1
4
s)16.
The third central moment of X is
E[(X−E[X])3] =E[(X−4)3] =E[X3−12X2+48X−64] =E[X(X−1)(X−2)−9X2+46X−64] =E[X(X−1)(X−2)]−9E[X2]+46E[X]−64.
Now,
E[X(X−1)(X−2)] = lims↑1 P(3)(s) = lims↑1 (
1
4
)316⋅15⋅14(
3
4
+
1
4
s)13 =
105
2
.
Hence
E[(X−E[X])3] =E[X(X−1)(X−2)]−9E[X2]+46E[X]−64 =
105
2
−9⋅19+46⋅4−64 =
3
2.
I hope it will help you