Anyone help state distribution of these?

Question

Suppose X1…Xn is a random sample from the N(μ, σ^2) distribution. State the distribution of each of the following statistics:

i) 2X1 - 3 μ

ii) (X1 – μ)/ 2σ

iii) X1 - 4X2 + 2X3

iv) Xn – 2 μ = 1/n ∑ (i=1 to n) Xi - 2 μ

v) (Xn – μ)/ σ

vi) 2Xn – 4

Answer 1

First, they're all still normal as any linear transformation of a normal distribution is still normal. So, just calculate the expectation and variance of each one.

i) E[2X1 - 3mu] = 2E[X1] - 3mu = 2mu - 3mu = -mu

Var[2X1 - 3mu] = 4Var[X1] = 3sigma^2

N(-mu, 3sigma^2)

ii) E[(x1 - mu)/(2sigma)] = 1/(2sigma) x E[x1 - mu] = 0

Var[(x1 - mu)/(2sigma)] = 1/(4sigma^2)Var[X1] = 1/(4sigma^2) x sigma^2 = 1/4

N(0,1/4)

iii) E(X1 - 4X2 + 2X3) = E[X1] - 4E[X2] + 2E[X3] = mu - 4mu + 2mu = -mu

Var(X1 - 4X2 + 2X3) = (independence) Var[X1] + Var[4X2] + Var[2X3] = Var[X1] + 8Var[X2] + 4Var[X3] = sigma^2 + 8sigma^2 + 4sigma^2

N(-mu, 13sigma^2)

iv) E[Xn - 2mu] = E[Xn] - 2mu = mu - 2mu = -mu

Var[Xn - 2mu] = var[Xn] = sigma^2

N(-mu, sigma^2)

and

E[1/n sum Xi - 2mu] = 1/nE[X1 + ... + Xn] - 1/n * n*2mu

= 1/n * n*mu - 1/n*n*2mu = -mu

Var[1/n sum Xi - 2mu] = Var[1/n sum Xi] = 1/n^2 Var[X1 + ... + Xn] =

n/n^2 sigma^2 = sigma^2/n

N(-mu, sigma^2/n) so that equality is wrong.

v) E[(Xn - mu)/sigma] = 1/sigma x E[Xn - mu] = 0

Var[(Xn - mu)/sigma] = 1/sigma^2 x Var[Xn] = sigma^2/sigma^2 = 1

N(0,1)

vi) E[2Xn - 4] = 2E[Xn] - 4 = 2mu - 4

Var[2Xn - 4] = Var[2Xn] = 4Var[Xn] = 4sigma^2

N(2mu - 4, 4sigma^2)