Quiz 10 Solution

Let $(X,Y)$ be a vector random variable satisfying:

• $E(X)=E(Y)=0$
• $\text{VAR}(X)=\text{VAR}(Y)=1$, and
• $\text{COV}(X,Y)=\frac{1}{2}$.

Let $W=X+Y$ and let $Z=X-Y$ denote the sum and difference of $X$ and $Y$, respectively.

Note that all random variables in this quiz have a mean value of zero, thus variances are equal to second moments, and covariances are equal to correlations.

Question 1:  True or false?  $X$ and $Y$ are uncorrelated.

False;   $X$ and $Y$ are uncorrelated if and only if $\text{COV}(X,Y)=0$.

Question 2:  Find $\text{VAR}(W)$

$\text{VAR}(W)=E(W^2)=E(X^2+2XY+Y^2)=E(X^2)+2E(XY)+E(Y^2)=1+2\cdot \frac{1}{2}+1=3$.

Question 3:  Find $\text{VAR}(Z)$

$\text{VAR}(Z)=E(Z^2)=E(X^2-2XY+Y^2)=E(X^2)-2E(XY)+E(Y^2)=1-2\cdot \frac{1}{2}+1=1$.

Question 4:  Determine the correlation coefficient ${\rho }_{W,Z}$ of $W$ and $Z$.

$\text{COV}(W,Z)=E(WZ)=E((X+Y)(X-Y))=E(X^2-Y^2)=E(X^2)-E(Y^2)=0$, thus ${\rho }_{W,Z}=0$.

Question 5:  True or false?  $W$ and $Z$ are uncorrelated.

True.

Question 6:  Let $\hat{Y}=\alpha X+\beta$ be an affine predictor of $Y$ from $X$, where the constants $\alpha$ and $\beta$ are chosen to minimize the mean-squared error $E((\hat{Y}-Y{)}^2)$.  Find the value of $\hat{Y}$ when $X=1$.

The affine minimum mean-squared error predictor of $Y$ from $X$ is

$\hat{Y}={\rho }_{X,Y}{\sigma }_Y\left(\frac{X-m_X}{{\sigma }_X}\right)+m_Y$

$=\frac{1}{2}\cdot 1\cdot X=\frac{1}{2}X$.  When $X=1$, $\hat{Y}=\frac{1}{2}$.

Question 7:  Let $\hat{X}=\gamma W+\delta$ be an affine predictor of $X$ from $W$, where the constants $\gamma$ and $\delta$ are chosen to minimize the mean-squared error $E((\hat{X}-X{)}^2)$.  Find the value of $\hat{X}$ when $W=2$.

We have $\text{COV}(X,W)=E(XW)=E(X(X+Y))=E(X^2+XY)=E(X^2)+E(XY)=1+\frac{1}{2}=\frac{3}{2}$.  We then have ${\rho }_{X,W}=\frac{\text{COV}(X,W)}{{\sigma }_X{\sigma }_W}=\frac{3}{2}\cdot \frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{2}$. The affine minimum mean-squared error predictor of $X$ from $W$ is $\hat{X}={\rho }_{X,W}{\sigma }_X\left(\frac{W-m_W}{{\sigma }_W}\right)+m_X=\frac{\sqrt{3}}{2}\cdot 1\cdot \frac{W}{\sqrt{3}}=\frac{W}{2}$. When $W=2$, we get $\hat{X}=1$.