Let X and Y be two real-valued random variables with E(X) = 1,E(Y) = 2, E(X²) = 4,E(Y²) = 9, and E(XY) = 0.9. The value of α that minimizes E((X−αY)²) is _________ (Round off to one decimal place)

The correct option is 0.1.

To find the value of $\alpha$ that minimizes the expectation, we define the objective function as:

$$f\left(\alpha \right)=E\left[{(X-\alpha Y)}^2\right]$$

First, we expand the squared term inside the expectation:

$${(X-\alpha Y)}^2=X^2-2\alpha XY+{\alpha }^2{Y}^2$$

Using the linearity property of expectation, we can distribute the expectation operator $E$ over each individual term:

$$f\left(\alpha \right)=E\left[X^2\right]-2\alpha E\left[XY\right]+{\alpha }^{2}E\left[Y^2\right]$$

We are given the following values in the problem description:
• $E\left[X^2\right]=4$
• $E\left[Y^2\right]=9$
• $E\left[XY\right]=0.9$

Substituting these values into our expression for $f\left(\alpha \right)$ gives:

$$f\left(\alpha \right)=4-2\alpha \left(0.9\right)+9{\alpha }^2$$

Simplifying the terms, we obtain a quadratic function of $\alpha$:

$$f\left(\alpha \right)=9{\alpha }^2-1.8\alpha +4$$

To find the value of $\alpha$ that minimizes this quadratic function, we take the first derivative with respect to $\alpha$ and set it to zero:

$$\frac{df}{d\alpha }=18\alpha -1.8=0$$

Solving for $\alpha$ yields:

$$18\alpha =1.8$$

$$\alpha =\frac{1.8}{18}=0.1$$

To verify that this value corresponds to a minimum, we check the second derivative of the function:

$$\frac{d^{2}f}{d{\alpha }^2}=18$$

Since the second derivative is positive ($18>0$), the function is convex, and $\alpha =0.1$ indeed minimizes $E\left[{(X-\alpha Y)}^2\right]$.