Bi-linear Property of Covariance.

Question

Covariance between $X$ and $Y$. Let $U, V, W$ be independant random variables with equal variances. $X = 2U + 3W$ and $Y = 3V - 4W$.

Answer 1

Hints.

• $\text{Covar}(,)$ is bilinear.
• If $R,S$ are independent random variables then $\text{Covar}(R,S)=0$
• $\text{Covar}(R,R)=\text{Var}(R)$

Answer 2

Maybe you have just lost interest in this. But it is a fundamental exercise on convariances, and the solution may be of general interest.

The first hint of @drhab (on bi-linearity) accounts for all but the last line below:

\begin{eqnarray} Cov(X,Y) &=& Cov(2U+3V, 3V-4W)\\ &=& Cov(2U,3V) + Cov(2U,-4W) +Cov(3V,3V)+Cov(3V,-4W)\\ &=& 6Cov(U,V) - 8Cov(U,W) + 9Cov(V,V) - 12Cov(V,W)\\ &=& 0 - 0+ 9Var(V) - 0 = 9Var(V). \end{eqnarray}

The second and third hints are used in the last line.

---

The brief sampling experiment (in R statistical software) below illustrates the result for the particular case $U,V,W \stackrel{iid}{\sim} \mathsf{Norm}(\mu=0, \sigma=2),$ giving $Cov(X,Y) = 9\sigma^2 = 36$ and $3SD(V) = 3\sigma = 6,$ correct to about three significant digits.

set.seed(1215);  m = 10^6
u = rnorm(m, 0, 2);  v = rnorm(m, 0, 2);  w = rnorm(m, 0, 2)
x = 2*u + 3*v;  y = 3*v - 4*w;  cov(x, y);  3*sd(v)
## 36.0797    # aprx Cov(X,Y) = 36
## 6.002979   # aprx 3SD(V) = 6