David,

Please help with the following question.

You are given that X and Y are random variables, each of which follows a standard normal distribution with covariance (X, Y) = 0.4 What is the variance of (5X + 2Y)?

How would I go about in answering the above question?

JB.

Hi jb,

Often it is helpful to skip the facts and work backwards from the question. Here,
What is the variance of (5X + 2Y)?

we may think of this as variance(w+z) where w=5x and z = 2y. Now the question is:
what is variance(w+z)? which is familiar from Gujarati.

variance (w+z) = var(w)+var(z): if w,z are independent,
variance (w+z) = var(w)+var(z)+2covariance(w,z): if w,z are not independent (i.e., covariance <>0)

so can substitute:
= var(5x)+var(2y)+2covar(5x,2y)
= 25var(x)+4var(y)+(2)(5)(2)cov(x,y)
=(25)(1)+(4)(1)+(2)(5)(2)(0.4)

Note:
var(5x) = 5^2var(x)
covar(5x,2y)=(5)(2)cov(X,Y), and
when we needed the “variance of x” and variance of y,” we are temporarily stuck until we read carefully that they are *standard* normal variables. If question said “normal variables,” cannot be solved. But standard normal implies N(0,1); i.e., variance of 1.

Key formulas from Gujarati:
variance (x+y) = var(x)+var(y): if x,y are independent,
variance (x+y) = var(x)+var(y)+(2)cov(x,y): if x,y are not independent (i.e., correlated)
var(aX) = a^2var(x) where a = constant
covar(a+bX,c+dY) = (b)(d)cov(x,y). Gujarati 3.25
A *standard* normal has mean of 0 and variance of 1.0 (and therefore, std deviation = 1)

David