# minimize variance

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

1. Find the values of a and b which minimize the variance of $Y=aX_1 + bX_2$ subject to the condition that $E[Y]=\Theta$.
2. 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...

-
1. Compute the variance based of $Y = a X_1+b X_2$ on the formula in your textbook - you should get a function which is quadratic in both $a$ and $b$.  2. Compute the expectation of $Y$ using the formula from the same book. You will obtain the function which is linear in $a$ and $b$.  3. Solve the problem of minimization of quadratic function with linear constraints: either by expressing $a$ through $b$ in linear constraints, or with the use of Lagrange multipliers. – Ilya Aug 21 '12 at 14:02
@Ilya Why don't you copy and paste this as an answer? – Sasha Aug 21 '12 at 14:47
@Sasha: thanks for advice, done. I expected somebody to put more... elaborated answer. – Ilya Aug 21 '12 at 14:50

1. 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 $b$.
2. Compute the expectation of $Y$ using the formula from the same book. You will obtain the function which is linear in $a$ and $b$.
3. 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.
-

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.

-