Question

Let X and Y be two real-valued random variables with
\(E(X)=1\), \(E(Y)=2\), \(E(X^2)=4\), \(E(Y^2)=9\), and \(E(XY)=0.9\), where E denotes the expectation operator.
The value of \(\alpha\) that minimizes \(E((X-\alpha Y)^2)\) is (round off to one decimal place).

Hint:

Differentiate the mean square error expression with respect to alpha and set it to zero.

Answer 1

Correct Answer: 0.1

Step 1: Use the orthogonality principle.
When $\alpha Y$ is the best linear estimate of $X$ in the least mean square sense, the estimation error $X-\alpha Y$ must be uncorrelated with $Y$. This gives
$E\big[(X-\alpha Y)Y\big]=0$.

Step 2: Expand this condition.
$E(XY)-\alpha E(Y^2)=0$

Step 3: Solve for $\alpha$.
$\alpha=\dfrac{E(XY)}{E(Y^2)}$

Step 4: Plug in the known values.
$E(XY)=0.9$ and $E(Y^2)=9$, so
$\alpha=\dfrac{0.9}{9}=0.1$

Step 5: Note which quantities were not needed.
The means $E(X)=1$ and $E(Y)=2$ do not enter this formula at all, since minimizing $E((X-\alpha Y)^2)$ only involves the second moments $E(X^2)$, $E(Y^2)$ and $E(XY)$, and here $E(X^2)$ also drops out because it does not contain $\alpha$.
\[\boxed{0.1}\]