The Correlation Equals the Ratio of the Swings, and Diversification Stops Paying

Two stocks with equal expected returns, variances 0.10 and 0.40, and correlation 0.5: the reflex differentiates the portfolio variance and reports an interior weight. The minimum sits at 100% in the calmer stock, and the usual explanation for that, which blames the no-shorting rule, is wrong. The vertex of the variance parabola lands exactly on w = 1, so the constraint does no work at all and the answer survives dropping it.

Lambdia
August 23, 2026 · 11 min read

Two stocks have the same expected return. One has variance $0.10$ , the other $0.40$ , and their correlation is $0.5$ . You may hold any mix with no short selling. Which mix has the least risk?

Almost everyone reaches for the standard result. A correlation strictly below one leaves room to diversify, so there ought to be an interior weight that beats both assets, and the way to find it is to differentiate the portfolio variance and set the derivative to zero. The instinct is sound in general and wrong here. The least risky portfolio holds the calmer stock and nothing else, at a standard deviation of $31.62\%$ , and a fifty-fifty split is strictly worse at $41.83\%$ .

The numbers are hiding a coincidence

Convert the variances before doing anything else, because the correlation lives on the scale of standard deviations and the problem statement does not.

\sigma_1 = \sqrt{0.10} = 0.31623\ldots, \qquad \sigma_2 = \sqrt{0.40} = 0.63246\ldots = 2\sigma_1

(1)

The wild stock swings exactly twice as wide as the calm one. So the ratio $\sigma_1/\sigma_2$ is exactly $1/2$ , which is exactly the stated correlation. That equality is the entire question, and it is easy to walk past because nothing in the wording points at it.

Feed it into the covariance. Since $\rho = \sigma_1/\sigma_2$ ,

\operatorname{Cov}(R_1, R_2) = \rho\,\sigma_1\sigma_2 = \frac{\sigma_1}{\sigma_2}\,\sigma_1\sigma_2 = \sigma_1^2 = 0.10 .

(2)

The covariance of the two assets equals the variance of the calm one. That single line is the whole result in compressed form, and everything below is unpacking it.

The wild stock is the calm stock plus something you cannot remove

Write $D = R_2 - R_1$ for the difference in returns and compute its covariance with the calm asset:

\operatorname{Cov}(R_1, D) = \operatorname{Cov}(R_1,R_2) - \operatorname{Var}(R_1) = 0.10 - 0.10 = 0 .

So $R_2 = R_1 + D$ with $D$ uncorrelated with $R_1$ , and the variance of that leftover piece is

\operatorname{Var}(D) = 0.40 - 2(0.10) + 0.10 = 0.30 .

(3)

Be careful about what has been shown. Zero covariance is not independence, and the article will not claim it is. Uncorrelatedness is enough, because variance minimisation only ever sees second moments, so nothing stronger is needed and nothing stronger is available.

The geometric picture is a right triangle rather than a stacked bar. Standard deviations of uncorrelated pieces add in quadrature, so $0.3162$ and $0.5477$ do not sum to $0.6325$ (they sum to $0.864$ ); their squares do. The right angle in the figure is the zero covariance, drawn.

Fig. 1 — The wild asset decomposed. The leg lengths are standard deviations, so they combine in quadrature and the right angle is exactly the vanishing covariance.

Read the problem through that decomposition and the answer is immediate. Holding the wild asset means holding the calm asset with an extra, unhedgeable rider attached. Buying any of it buys the rider too.

An identity that removes the calculus

Let $w$ be the weight on the calm asset. With $\operatorname{Cov} = \sigma_1^2$ , the two-asset variance collapses:

w^2\sigma_1^2 + (1-w)^2\sigma_2^2 + 2w(1-w)\sigma_1^2 \;=\; \sigma_1^2 + (1-w)^2\!\left(\sigma_2^2 - \sigma_1^2\right)

(4)

This is an algebraic identity, true for every $w$ , and it is worth expanding by hand once to see that nothing has been assumed. The right-hand side is $\sigma_1^2$ plus a square times a positive constant. Every dollar moved out of the calm asset increases the second term and changes nothing else. There is no trade-off to optimise, so there is nothing for a derivative to find.

Corner solution

A constrained optimum sitting on the boundary of the feasible set rather than in its interior. The first-order condition $f'(w) = 0$ is the wrong test at such a point: what matters is the sign of the derivative as the boundary is approached. A corner may or may not coincide with the unconstrained optimum, and the distinction is the subject of the next section.

Where the constraint story goes wrong

Substituting the numbers gives the variance as a function of the weight,

V(w) = 0.10w^2 + 0.40(1-w)^2 + 0.20\,w(1-w) = 0.3w^2 - 0.6w + 0.4, \qquad V'(w) = 0.6w - 0.6 .

(5)

The derivative is strictly negative for every $w < 1$ and vanishes only at $w = 1$ . Now notice what that means. The vertex of the parabola sits exactly on $w = 1$ , and $w = 1$ is a perfectly ordinary feasible portfolio. Nothing has escaped anywhere.

A popular way to describe this case says the interior optimum has left the feasible set and the no-shorting rule is what forces you to the corner. That account is false here, and pleasantly so: it would predict that relaxing the constraint improves the portfolio. Relaxing it does nothing at all. Allow arbitrary short positions and the minimiser of $0.3w^2 - 0.6w + 0.4$ over the whole real line is still $w = 1$ , because $w = 1$ is where the unconstrained vertex already is. The correct diagnosis is the knife edge $\rho = \sigma_1/\sigma_2$ , at which the unconstrained optimum and the corner coincide. This is a stronger statement than the constraint story, since it survives dropping the constraint.

Fig. 2 — The variance parabola on the feasible interval. Its vertex lands on w = 1, a feasible point, so the corner and the unconstrained minimum are the same portfolio.

The fifty-fifty trap is worth pricing precisely, because a candidate who reaches for it is not making an arithmetic slip. Evaluating (5) at $w = 1/2$ gives $V = 7/40 = 0.175$ , so the standard deviation is $\sqrt{0.175} = 41.83\%$ against $31.62\%$ for the single-asset holding. The computation is exact; the belief underneath it is what fails.

The general condition, and the case that behaves normally

Solving $V'(w) = 0$ in symbols gives the familiar minimum-variance weight,

w^\star = \frac{\sigma_2^2 - \rho\sigma_1\sigma_2}{\sigma_1^2 + \sigma_2^2 - 2\rho\sigma_1\sigma_2}, \qquad w^\star \ge 1 \iff \rho \ge \frac{\sigma_1}{\sigma_2} .

(6)

The trichotomy is clean. Below the knife edge there is a genuine interior optimum that beats both assets. At the knife edge the optimum is the calm asset alone. Above it the unconstrained optimum asks for a short position in the wild asset, and only then does the no-shorting rule actually bind and push you back to $w = 1$ .

To see that this problem is about the knife edge and not about diversification in general, change one number. Take standard deviations of $20\%$ and $30\%$ at the same correlation $0.5$ . Now $\rho = 0.5 < 20/30 = 2/3$ , so (6) gives an interior weight of $w^\star = 6/7$ and a portfolio standard deviation of $19.64\%$ , strictly below the calmer asset's own $20\%$ . Diversification pays there, exactly as the textbook picture promises. The original numbers were rigged so that it does not.

One more check on the original problem, since the reflex weight is the thing to refute. Plugging $w = 6/7$ into the first problem's variance gives $26/245 = 0.1061$ , above $0.10$ . The interior candidate is available, feasible, and worse.

Where this stops being the right question

The whole analysis assumes equal expected returns, which is what licenses pure variance minimisation. Give the wild asset any excess return and the calm asset no longer wins automatically, because the problem becomes a trade-off rather than a minimisation and the answer depends on how much variance the investor will accept per unit of return.

Two more limits deserve naming. The knife edge is exact and therefore fragile: at $\rho = 0.49$ the optimum moves back inside the interval, so the result is a statement about one point in parameter space rather than a rule of thumb. And variances and correlations are estimated, not given. An estimated correlation of $0.5$ with a standard error of a few hundredths cannot tell you which side of the knife edge you are on, which is a good reason to prefer the identity in (4) to a numerical optimiser. The identity says something a candidate can defend under questioning: with the covariance equal to the smaller variance, the wild asset carries a rider of variance $0.30$ and no allocation removes it.

Sources and further reading

Harry Markowitz, “Portfolio Selection”, The Journal of Finance7 (1952), 77–91, where variance minimisation over a set of weights was first posed in this form.
Wikipedia: Modern portfolio theory for the two-asset variance formula and the minimum-variance weight.
Wikipedia: Covariance and Uncorrelatedness, on why zero covariance is weaker than independence.
Wikipedia: Karush–Kuhn–Tucker conditions, for the general reason a boundary optimum needs a sign condition rather than a vanishing derivative.

Every figure above was checked in exact rational arithmetic and against an exhaustive scan of the weight interval at a resolution of one hundred-thousandth, which confirms that the variance is strictly decreasing on the whole interval and that no interior optimum exists to be missed.

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