``` 最小化方差的加权平均 ```

``` 最小化方差的加权平均 ```
Weighting an average to minimize variance

原始链接: https://www.johndcook.com/blog/2025/11/12/minimum-variance/

本文讨论了优化投资配置以最小化投资组合波动率的方法。在选择两个独立的资产（A和B）时，如果A比B更具波动性，最佳策略不是将所有资金投入其中任何一个。相反，混合配置是最佳的，并且应偏向波动性较低的资产（B）。最佳配置通过数学确定，取决于每个资产的方差。资产Y的方差越低，对其的配置就越大。如果Y没有方差（是常数），则完全投资于Y。方差相等意味着配置相等。这个原则可以扩展到*n*个资产。每个资产（Xi）的最佳配置与其方差成反比。具体来说，分配给每个资产的权重是其反方差除以所有反方差之和。这确保了给予低风险资产更高的权重，从而最小化整体投资组合的波动率。

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原文

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.

Multiple variables

Now suppose we have n independent random variables X_i 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 t_i 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, j ≤ n. These (n − 1) equations along with the constraint that all the t_i sum to 1 give us a system of equations whose solution is

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

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