Covariance of three dice rolls

Question

I understand this question has been asked but I have a different comment to make on the matter and wondering if someone could help me.

Let Z1,Z2,Z3 be values resulting from three tosses. X=Z1^2+Z2^2+Z3^3 and Y=4X-7. What is the correlation coefficient. Now I've reduced the Covariance equation to be 4Var(X) making the correlation coefficient 4Var(X)/(S.D(X)*S.D(Y).

My question is there a simpler way to calculate the variance of X and Y rather than compute all the different possibilities by hand and compare them to the expected values of X and Y which I have computed properly.

So could the variance of x = V(Z1)^2+V(Z2)^2+V(Z3)^3

Sorry if this is dumb I feel I'm missing a much simpler way of calculating these rather than computing every combination of Z1,Z2,and Z3 and comparing it to the expected value.

Thanks!

Answer 1

You're almost there. The useful rule is the following: $$ \text{Var}(aX+bY+c)=a^2\text{ Var}(X)+b^2\text{ Var}(Y)+2ab\text{ Cov}(X,Y) $$ Apply it to your $Y$ variable: $$ \text{Var}(Y) = \text{Var}(4X-7)=4^2\text{ Var}(X) = 16\text{ Var}(X)\\ \sigma(Y) = 4\sigma(X) $$ Now write the correlation coefficient: $$ \rho(X,Y) = \dfrac{\text{Cov}(X,Y)}{\sigma(X)\sigma(Y)}=\dfrac{\text{Cov}(X,4X-7)}{\sigma(X)\sigma(4X-7)}=\dfrac{4\text{Cov}(X,X)}{4\sigma(X)\sigma(X)}=1. $$

Answer 2

There is almost nothing to do. The variance of $4X-7$ is $16$ times the variance of $X$. And the covariance of $X$ and $4X-7$ is $4$ times the variance of $X$. It follows that the correlation coefficient of $X$ and $Y$ is $1$. No need to know the variance of $X$!

More simply, $Y=aX+b$ with positive $a$, so the correlation coefficient is $1$.