Question

1. Let X.. xs ^ N(-1,4) and Y.. y, " N(0.1) be independent. Using properties of the normal distribution, derive the distribution of the following random variables (b) Wa = 2.1(X + 1)

Answer 1

(a)

We know that the linear combination of Normal random variable is a Normal random variable.

Thus, W1 is a Normal random variable.

E(W1) = E(2X1 + 4Y2 - X2) = 2E(X1) + 4E(Y2) - E(X2) = 2 * - 1 + 4 * 0 - (-1) = -1

Var(X1) = Var(2X1 + 4Y2 - X2) = 2²Var(X1) + 4²Var(Y2) + (-1)² Var(X2) = 4 * 4 + 16 * 1 + 1 * 4 = 36

W1 ~ N(-1, 36)

(b)

We know that the linear combination of Normal random variable is a Normal random variable.

Thus, W2 is a Normal random variable.

$E(W_2) = E\left [ \frac{1}{2}\sum_{i=1}^{3}(X_i + 1) \right ] = \left [ \frac{1}{2}\sum_{i=1}^{3}(E[X_i] + 1) \right ] = \left [ \frac{1}{2}\sum_{i=1}^{3}(-1+ 1) \right ]$

=> E[W2] = 0

$Var(W_2) = Var\left [ \frac{1}{2}\sum_{i=1}^{3}(X_i + 1) \right ] = \left [ \frac{1}{2^2}\sum_{i=1}^{3}(Var[X_i]) \right ] = \left [ \frac{1}{4}\sum_{i=1}^{3}4 \right ] = 3$

W2 ~ N(0, 3)

(c)

Let us consider the variable Z1 = (X1 + 1)/2

E[(X1 + 1)/2] = (E(X1) + 1 )/2 = (-1 + 1) / 2 = 0

Var[(X1 + 1)/2] = (Var(X1) )/2² = 4 / 4 = 1

Thus, Z1 = (X1 + 1)/2 ~ N(0, 1)

and Y3 ~ N(0,1)

We know that sum of squares of k standard normal variables follow Chi square distribution with k degree of freedom.

Thus, W3 = Z1² + Y3² follows Chi square distribution with 2 degree of freedom.

(d)

We know that sum of squares of k standard normal variables follow Chi square distribution with k degree of freedom.

$\sum_{i=1}^{4} Y_i^2$ follow Chi square distribution with 4 degree of freedom.

Student's t-distribution with ν degrees of freedom can be defined as the distribution of the random variable T with

$T = Z/ \sqrt{V/v}$

where

• Z is a standard normal with expected value 0 and variance 1;
• V has a chi-squared distribution with ν degrees of freedom;
• Z and V are independent.

Now,

$W_4 = \frac{2Y_5}{\sqrt{\sum_{i=1}^{4} Y_i^2}} = \frac{Y_5}{\sqrt{\sum_{i=1}^{4} Y_i^2}/4}$

As, Y5 ~ N(0,1) and $\sum_{i=1}^{4} Y_i^2$ follow Chi square distribution with 4 degree of freedom.

W4 follows Student's t-distribution with 4 degrees of freedom