minimize variance $X_1$ and $X_2$ are independently distributed random variables with
$$P(X_1=\Theta+1) = P(X_1=\Theta-1) = 1/2 \\
P(X_2=\Theta-2) = P(X_2=\Theta+2) = 1/2$$


*

*Find the values of a and b which minimize the variance of $Y=aX_1 + bX_2$ subject to the condition that $E[Y]=\Theta$.

*What is the minimum value of this variance?


The answers at the back of the book says that $a$ and $b$ is $4/5$ and $1/5$ respectively and the variance is $4/5$.
I don't know how they got that answer and I'm not even sure where to start...
 A: *

*Compute the variance based of $Y = aX_1+bX_2$  on the formula in your textbook - you should get a function which is quadratic in both $a$ and $b$.

*Compute the expectation of $Y$ using the formula from the same book. You will obtain the function which is linear in $a$ and $b$.

*Solve the problem of minimization of quadratic function with linear constraints: either by expressing $a$ through $b$ using the equation provided by linear constraints, or with the use of Lagrange multipliers - whatever technique you are familiar with.

A: Assume that $\Theta \neq 0$.
Then, since $X_1$ and $X_2$ both have expected value $\Theta$, we
have from the linearity of expectation that 
$$E[Y] = E[aX_1 + bX_2] = aE[X_1] + bE[X_2] = (a+b)\Theta 
~\text{equals}~  \Theta ~\text{if and only if}
~ a+b = 1.$$
Since $X_1$ and $X_2$ are independent random variables
and therefore have zero covariance,
the standard variance formula
$$\text{var}(Y) = \text{var}(aX_1 + bX_2)
= a^2\text{var}(X_1) + b^2\text{var}(X_2) + 2ab\text{cov}(X_1,X_2)$$
reduces to
$$\text{var}(Y) = a^2\text{var}(X_1) + b^2\text{var}(X_2) = a^2 + 4b^2.$$
Since $a$ and $b$ are constrained to satisfy $a+b=1$, we have that
$$\text{var}(Y) = a^2 + 4(1-a)^2 = 5a^2 -8a + 4.$$
You can obtain the solution given at the back of your book
by finding the value of $a$ that minimizes $5a^2 - 8a +4$,
and the minimum value attained by $5a^2 - 8a +4$.
You may also want to consider where the above method breaks down
when $\Theta = 0$, for which case the answer in the back of your book
is not correct.
