Minimize variance of $aX + bY + cZ$ for independent variables $X,Y,Z$

Question

There are three independent random variables $X, Y, Z$.

The goal is to minimize the variance of $aX + bY + cZ$ where $a+b+c = 1$ and $0\le a,b,c \le1$.

$$\operatorname{var}(X) = 2\operatorname{var}(Y) = 3\operatorname{var}(Z)$$

I did $$K = aX + bY + cZ\quad,\quad \operatorname{var}(K) = a^2\operatorname{var}(X) + b^2\operatorname{var}(Y) + c^2\operatorname{var}(Z)\,,$$ but it looks like one more equation might be needed to solve the problem.

Is there anything I missed?

Answer 1

Without loss of generality, we can suppose that $Var(X) = 6, Var(Y) = 3, Var(Z) = 2$.

The problem can be rewritten to this form without probability theory background, i.e. to minimize $f(a, b, c) = 6a^2 + 3b^2 + 2c^2$, subject to constraints $a,b,c \in [0, 1], a + b + c = 1$

Firstly, note that we can remove the constraint $a,b,c \in [0, 1]$ and still get the right answer. To see this, suppose one of $a, b, c$ gets a negative value, we will show that this won't lead to a minimal of $6a^2 + 3b^2 + 2c^2$. If $a < 0, b > 0$, we let $a' = a + \varepsilon, b' = b - \varepsilon (\varepsilon > 0 \text{ a small positive amount})$, then we will decrease $a^2$ while at the same time decreasing $b^2$. Thus $6a^2 + 3b^2 + 2c^2$ is not a minimum.

Then the remaining constraint is $a + b + c = 1$, which is the same as letting $g(a, b, c) = 0$, where $g(a, b, c) = a + b + c - 1$. This form of problem can be easily solved using Lagrangre Multiplier. $F(a, b, c, \lambda) = f(a, b, c) + \lambda g(a, b, c)$. Solve $\frac{\partial F}{\partial a} = \frac{\partial F}{\partial b} = \frac{\partial F}{\partial c} = \frac{\partial F}{\partial \lambda} = 0$, we arrived at: $a = \frac 1 6, b = \frac 1 3, c = \frac 1 2$, and the minimal value is actually $1$.

Please forgive my poor English:)

Answer 2

To minimize $6a^{2}+3b^{2}+2c^{2}$ subject to $a,b,c \geq 0$ and $a+b+c=1$ note that $1^{2}=(a+b+c)^{2}=(\frac 1 {\sqrt 6}\sqrt 6 a +\frac 1 {\sqrt 3}\sqrt 3 a +\frac 1 {\sqrt 2}\sqrt 2 a)^{2} \leq (\frac 1 6 +\frac 1 3+\frac 1 2) (6a^{2}+3b^{2}+2c^{2})$ by C-S inequality. Hence $6a^{2}+3b^{2}+2c^{2} \geq 1$ and equality holds when $(a,b,c) =(\frac 1 6, \frac 1 3,\frac 1 2)$. Hence the minimum value is $1$.