A Introduction to Statistics Question on Normally distributed RVs

[CaptainBlack]

Question from Jim on Yahoo Answers:

Let X and Y be independent random variables with X ~ N(3, 1)
and Y ~ N(6, 9). Put Z = Y − 4X + 7.

(a) State the distribution of Z.
(b) Calculate Cov(X, Z).
(c) Calculate corr(X, Z)

The things you should know are:

1. The sum of normal random variables is a normal random variable with mean equal to the sum of the means and variance equal to the sum of the variances.

2. Multiplying a normal RV by a constant gives a normal RV with mean equal to the constant times the mean of the original RV and variance equal to the square of the constant times the variance of the original RV

3. Adding a constant to a normal RV gives you a RV with mean equal to the original plus the constant and variance unchanged.

(a) So applying these to \(Z\) you find \(Z \sim N( (6-12+7), (1+9) ) = N(1,10) \)

For (b) you need to write:

\[Cov(X,Z)= E( (X-\mu_X)(Z-\mu_Z) )= E(X*Z - \mu_X Z-\mu_Z X +\mu_X \mu_Z)=E(X*Z) - \mu_X \mu_Z\]

Now you expand the remaining expectation above replacing \(Z\) by \(Y- 4X+7\) using the linearity properties of the expectation operator and the independence of \(X\) and \(Y\)

(c) is a trival application of the definition once you have done (a) and (b)
.

 

[jvicens]

(c) Using the definition of correlation, we have:

\[corr(X,Z) = \frac{Cov(X,Z)}{\sqrt{Var(X) Var(Z)}} = \frac{E(X*Z) - \mu_X \mu_Z}{\sqrt{Var(X) Var(Z)}}\]

Plugging in the values calculated in (a) and (b), we get:

\[corr(X,Z) = \frac{E(X(Y-4X+7)) - 3*1}{\sqrt{1 * 10}} = \frac{E(XY) - 12}{\sqrt{10}}\]

To calculate \(E(XY)\), we can use the independence of \(X\) and \(Y\) and the linearity property of the expectation operator to get:

\[E(XY) = E(X) * E(Y) = 3 * 6 = 18\]

Thus, we have:

\[corr(X,Z) = \frac{18 - 12}{\sqrt{10}} = \frac{6}{\sqrt{10}} \approx 1.89\]