Differentiating Portfolio Variance with Respect to an Asset Weight

Summary

The answer explains why differentiating portfolio variance with respect to a single asset weight removes the full summation from view: only terms involving that weight contribute to its derivative. It expands the result using covariance bilinearity, then combines the terms as the covariance between the asset return and the weighted portfolio return.

The derivation also adds a deterministic risk-free component inside the covariance without changing its value, connecting the result to the tangent portfolio return. This is a focused mathematical clarification for mean-variance portfolio calculations. It does not provide empirical evidence or discuss investment performance, and its explanation assumes the risk-free return is deterministic and the portfolio setup in the referenced derivation.

Key ideas

• A partial derivative with respect to one portfolio weight acts only on terms that depend on that weight.
• Covariance bilinearity lets the derivative terms be combined into one covariance expression.
• Adding a deterministic quantity inside a covariance does not change the covariance.
• The resulting expression links an asset’s return to the tangent portfolio return.

Tags

# MPT Tangent Portfolio: Buck for the Bang Ratio


# MPT Tangent Portfolio: Buck for the Bang Ratio












The $R_{TP}$ is the tangent portfolio return, but I don't understand the step regarding $\frac{dV(R)}{dw_n}$, you apply this, and how come it get rids of the summation?

## Answer by JejeBelfort (score 2, accepted)

https://quant.stackexchange.com/a/34352

You get rid of the sum because you are computing the derivative of the variance with respect to one weight only (i.e. $\omega_n$)!

This implied that you take the derivative relatively to one single term of the sum, not all. You can basically compute $$\dfrac{dV(R)}{d\omega_n}$$ for all $n = [1, \cdots, N]$.

Regarding the calculation, and starting from the 3rd row of the derivation:

$$2 \omega_n Cov(R_n,R_n) + 2 Cov (R_n, \sum_{m \neq n} \omega_m R_m)$$

$$ = 2 Cov(R_n,\omega_n R_n) + 2 Cov (R_n, \sum_{m \neq n} \omega_m R_m)$$ (bilinearity of the covariance) $$ = 2 Cov (R_n, \sum_{m} \omega_m R_m)$$ $$ = 2 Cov (R_n, \sum_{m} \omega_m R_m - (1 - \sum_{m} \omega_m) R_f)$$ (as the variance of a constant is zero. The constant I am referring to is $- (1 - \sum_{m} \omega_m) R_f$ since $R_f$ is deterministic. I have just added it in the covariance) $$ = 2 Cov (R_n, R_{TP})$$

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.