Assume that random variables π andπ are normally distributed.
π~$π ( π_ π , π_π^2 )$
π ~$π ( π_π , π_π^2 )$
The correlation between πand π is π. How can you choose constants $π$ and $ π $such that you minimize the variance of the random variable sum $π=ππ+ππ $ under the constraints that $a+π=1$, $0β€πβ€1$ πππ $0β€πβ€1$?
My idea:
$$Var(aX+bY)=a^2Var(X)+b^2Var(Y)+2abCov(X,Y)$$
$$=a^2\sigma_X^2+b^2\sigma_Y^2+2ab\rho\sigma_X\sigma_Y$$
$$=a^2\sigma_X^2+(1-a)^2\sigma_Y^2+2a(1-a)\rho\sigma_X\sigma_Y$$
$$=a^2(\sigma_X^2+\sigma_Y^2-2\rho\sigma_X\sigma_Y)-a(2\sigma_Y^2-2\rho\sigma_X\sigma_Y)+\sigma_Y^2$$
Then, I just follow the root of quadratic equation to get the minimum. How should I use $0β€πβ€1$? And I think the variance has nothing to do with $\mu$ here. Thank you!
Best Answer
First notice that there is a typo in your algebra,
$$ \operatorname{Var}(aX+bY) = a\operatorname{Var}(X)+b\operatorname{Var}(Y)\boldsymbol{\color{red}{+}}2ab\operatorname{Cov}(X,Y).$$
It might be helpful to then set this up as a constrained maximization problem, i.e.
$$ \min_a (a^2\sigma_{X}^2+(1-a)^2\sigma_{Y}^2+2a(1-a)\rho\sigma_{X}\sigma_{Y}) $$
subject to
$$ a \ge 0 \quad \text{and} \quad a \le 1. $$
Notice that the second derivative of the objective function is positive,
$$ 2(\sigma_{X}^2+\sigma_{Y}^2-2\rho\sigma_{X}\sigma_{Y})=2\operatorname{Var}(X-Y)>0 ,$$
and since it is quadratic in $a$ it is a convex parabola and its minimum is going to be either the critical point of the parabola or the boundary. Accordingly, setting the first derivative of the objective function with respect to $a$ to zero leads to
$$(2\sigma_{X}^2-4\rho\sigma_{X}\sigma_{Y}+2\sigma_{Y}^2)a+(2\rho\sigma_{X}\sigma_{Y}-2\sigma_{Y}^2) = 0 $$ and therefore the critical point of the parabola is given by
$$ a^{c} = \frac{\sigma_{Y}^2-\rho\sigma_{X}\sigma_{Y}}{\sigma_{X}^2-2\rho\sigma_{X}\sigma_{Y}+\sigma_{Y}^2}. $$
and the solution is
$$ a^* = \begin{cases} 0\;\quad\text{, if}\quad a^c \le 0\\ 1\;\quad\text{, if}\quad a^c \ge 1\\ a^{c}\quad\text{, otherwise.} \end{cases} $$
Finally, notice that
$$ a^c \le 0 \iff \sigma_{Y}\ge\rho\sigma_{X},$$ and $$ a^c \ge 1 \iff \rho\sigma_{Y}\ge \sigma_{X}.$$