Weighting an average to minimize variance

Suppose you have $100 to invest in two independent assets, A and B, and you want to minimize volatility. Suppose A is more volatile than B. Then putting all your money on A would be the worst thing to do, but putting all your money on B would not be the best thing to do.

The optimal allocation would be some mix of A and B, with more (but not all) going to B. We will formalize this problem and determine the optimal allocation, then generalize the problem to more assets.

Two variables

Let X and Y be two independent random variables with finite variance and assume at least one of X and Y is not constant. We want to find t that minimizes

\text{Var}[tX + (1-t)Y]

subject to the constraint 0 ≤ t ≤ 1. Because X and Y are independent,

\text{Var}[tX + (1-t)Y] = t^2 \text{Var}[X] + (1-t)^2 \text{Var}[Y]

Taking the derivative with respect to t and setting it to zero shows that

t = \frac{\text{Var}[Y]}{\text{Var}[X] + \text{Var}[Y]}

So the smaller the variance on Y, the less we allocate to X. If Y is constant, we allocate nothing to X and go all in on Y.  If X and Y have equal variance, we allocate an equal amount to each. If X has twice the variance of Y, we allocate 1/3 to X and 2/3 to Y.

Muliple variables

Now suppose we have n independent random variables Xi for i running from 1 to n, and at least one of the variables is not constant. Then we want to minimize

\text{Var}\left[ \sum_{i=1}^n t_i X_i \right] = \sum_{i=1}^n t_i^2 \text{Var}[X_i]

subject to the constraint

\sum_{i=1}^n t_i = 1

and all ti non-negative. We can solve this optimization problem with Lagrange multipliers and find that

t_i \text{Var}[X_i] = t_j \text{Var}[X_j]

for all 1 ≤ i, jn. These (n − 1) equations along with the constraint that all the ti sum to 1 give us a system of equations whose solution is

t_i = \frac{\prod_{j \ne i} \text{Var}[X_i]}{\sum_{i = 1}^n \prod_{j \ne i} \text{Var}[X_i]}

Incidentally, the denominator has a name: the (n − 1)st elementary symmetric polynomial in n variables. More on this in the next post.

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