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: Expand the expression to be minimized.
We need to minimize
\[ g(\alpha)=E\big((X-\alpha Y)^2\big) \]
Expanding the square,
\[ g(\alpha)=E(X^2)-2\alpha E(XY)+\alpha^2E(Y^2) \]
Note that \(E(X)\) and \(E(Y)\) do not appear in this expansion, so their values are not needed for this problem.

Step 2: Substitute the given values.
We are given
\[ E(X^2)=4,\qquad E(XY)=0.9,\qquad E(Y^2)=9 \]
So
\[ g(\alpha)=4-2\alpha(0.9)+9\alpha^2=4-1.8\alpha+9\alpha^2 \]

Step 3: Differentiate with respect to \(\alpha\).
\[ \frac{dg}{d\alpha}=-1.8+18\alpha \]

Step 4: Set the derivative to zero.
For a minimum,
\[ -1.8+18\alpha=0 \]
\[ 18\alpha=1.8 \]
\[ \alpha=\frac{1.8}{18}=0.1 \]

Step 5: Confirm it is a minimum.
The second derivative is
\[ \frac{d^2g}{d\alpha^2}=18>0 \]
which is positive, so \(\alpha=0.1\) does give the minimum value of \(g(\alpha)\), not a maximum.

Final Answer:
\[ \boxed{0.1} \]