What is the distribution of x.

Question

suppose the random varible x has a binomial distribution with p>.5 then the distribution of x is

a) skewed to the left

b) skewed to the right

c) normal distribution

d) symmetric

Answer 1

It is skewed to the left, with less skew as n increases. You would think this immediately because of the longer tail to the left. However, you can work this out as well.

x^3 = x(x-1)(x-2) + 3(x)(x-1) + x

Thus, as f(x) = C(n,x)p^x(1-p)^x

E(x^3) = ∑x^3 C(n,x)p^x(1-p)^x =

∑(x(x-1)(x-2) + 3(x)(x-1) + x) n!/x!(n-x)!p^x(1-p)^x =

n(n-1)(n-2)p^3∑C(n-3,x-3)p^(x-3)(1-p)^(n-x) +

3n(n-1)p^2∑C(n-2,x-2)p^(x-2)(1-p)^(n-x) +

np∑C(n-1,x-1)p^(x-1)(1-p)^(n-x) =

n(n-1)(n-2)p^3 + 3n(n-1)p^2 + np

As x^2 = x(x-1) + x, we can calculate similarly that

E(x^2) = n(n-1)p^2+np and

E(x) = np

Thus, E(x - x-bar)^3 =

E(x^3) - 3 E(x^2)E(x-bar) + 3 E(x)(x-bar)^2 - E(x-bar)^3 =

n(n-1)(n-2)p^3 + 3n(n-1)p^2 + np - 3(n(n-1)p^2+np)np + 3 n^3p^3 -n^3p^3 =

n^3p^3 - 3n^2p^3 + 2np^3 + 3n^2p^2 - 3np^2 + np - 3n^3p^3 + 3n^2p^3 - 3n^2p^2 + 3 n^3p^3 -n^3p^3 =

2np^3 - 3np^2 + np =

np(2p^2 - 3p + 1) =

np(2p - 1)(p - 1)

This formula makes sense, as it equals 0 at 0, 1/2, and 1.

For p in (1/2, 1), p > 0, 2p - 1 > 0 and p - 1 < 0, so the skew is negative.

For p in (0, 1/2), p > 0, 2p - 1 < 0 and p - 1 < 0, so the skew is positive

(Skew = E(x - x-bar)^3 /(E(x-x-bar)^2)^(3/2), so this is also negative.)

As the denominator contains n^(3/2), this shows that the skewness decreases as n increase.