Two Stocks at 0.7 With the Same Index, and Still −0.02 With Each Other

The question usually arrives in one sentence: two stocks each move with the same index at a correlation of 0.7. How correlated are the two stocks with each other? You get about ten seconds, and the person asking already knows the answer.

Three replies come back most often. That it is 0.7 again, because both stocks are close to the index so they must be close to each other. That it is 0.49, from multiplying. Or the shrug: no information, could be anything. The last one costs candidates the most, because the two numbers you were handed rule out 49% of the dial. They just do not pick a single point on what is left.

This turns up on first-round screens for quant trading and quant research seats, and again much later on risk and portfolio desks, where the same fact walks in disguised as a covariance matrix somebody wants signed off. It survives as a screen because it costs ten seconds to ask and splits the room. Nobody is checking whether you can quote a formula. They want to know whether you treat three correlations as three separate facts, or notice that they are three angles in one room and cannot be chosen independently.

Which tables of correlations are real

Write the three numbers into a table, each variable against each other variable, ones down the diagonal:

Not every such table comes from real data. Suppose I tell you the first pair is perfectly in step and the second pair perfectly opposed. I have already decided the third pair, and if I write anything else in that cell I have described a world that does not exist. The condition separating the possible tables from the impossible ones is short: $R$ has to be positive semi-definite. That is not an extra technical requirement bolted on, it is forced. Standardise the three variables to unit variance and any weights $w$ give

A portfolio cannot have negative variance, so any correlation matrix satisfies this, and Higham takes it as the definition: a correlation matrix is a symmetric positive semi-definite matrix with unit diagonal. For semi-definiteness you need every principal minor to be non-negative, not just the leading ones. Here the $1 \times 1$ minors are all 1, the $2 \times 2$ minors are $1 - a^2$, $1 - b^2$ and $1 - c^2$, and everything else is carried by the determinant:

Now read that as a statement about $c$, which is the only unknown. It is a quadratic, and the sign in front of $c^2$ is negative, so the determinant is non-negative exactly between the two roots:

The discriminant tidies itself up. What ends up under the square root is $a^{2}b^{2} + 1 - a^{2} - b^{2}$, which factors as $(1-a^{2})(1-b^{2})$, and both factors are non-negative because a correlation lives in $[-1,1]$. So the roots are real, always, and they are

The remaining condition, $|c| \le 1$, comes for free. Read $c_{+}$ as the dot product of the unit vectors $(a, \sqrt{1-a^{2}})$ and $(b, \sqrt{1-b^{2}})$; Cauchy–Schwarz caps it at 1, and the same argument floors $c_{-}$ at $-1$. Put $a = b = 7/10$ into (2). The product is $49/100$, and the square root is $\sqrt{(51/100)^{2}} = 51/100$, a rational number because the two factors are equal. That is where the exactness comes from:

Correlation is an angle wearing a disguise

The algebra gives the answer and explains nothing. Here is the same result as a picture, and the picture is what you want in the room, because you can draw it while you talk.

Take a return series and subtract its average, so it wobbles around zero. Ask what covariance does to two of those. It is symmetric, so the order does not matter. It is linear in each slot. And the covariance of something with itself is its variance, which is never negative and is zero only for something that never moves. Those three properties are the axioms of an inner product, so centred square-integrable random variables form an inner-product space with $\langle X, Y\rangle = \operatorname{Cov}(X,Y)$. Lengths and angles come with the structure, whether or not the objects started life as arrows. The length is the standard deviation, and

The centring is not decoration. Drop it and the cosine of the raw series measures the angle between two clouds that both sit far from the origin, so two perfectly opposed return streams with the same average level come back at $+0.9999$. Correlation is the cosine after the means are removed.

So 0.7 is not a proportion of anything. It is $\arccos(0.7)$, which is 45.573°. Point the index arrow wherever you like; nothing depends on that. The first stock leans 45.573° away from it, and that is the whole content of the number, so it can sit anywhere on a cone of half-angle 45.573° around the index. The second stock gets the same instruction and lives on the same cone. The question has quietly become a question about two arrows on a cone: how far apart can they get?

Slide one arrow on top of the other and the angle is zero: correlation 1, the top of the interval. Slide it to the opposite side and the two stocks lean away from the index in exactly opposite directions, so the angle between them is 45.573° twice over, which is 91.146°. That is past the right angle, and the cosine of anything past 90° is negative. Its value is $\cos(2\arccos(0.7)) = 2(0.7)^2 - 1 = -\tfrac{1}{50}$.

This is not a rhyme with the algebra, it is the algebra. Write $a = \cos\mu$ and $b = \cos\nu$. Then $\sqrt{1-a^{2}} = \sin\mu$, likewise for $\nu$, and equation (2) reads

The two endpoints are the two ways of stacking the angles, and the feasible set is every angle in between. Castano, Paksoy and Zhang prove exactly this identity on the way to a broader result about correlation matrices and metric-preserving functions; the equivalence between angles and feasibility is old enough to have a name, Kreĭn’s inequality, the triangle inequality for angles in a Hilbert space.

One restriction has to travel with the picture. The plain angular triangle inequality lines up with positive semi-definiteness only while the two known angles add to no more than 180°. Past that, a second spherical condition starts to bite, and a literal reading of the triangle inequality lets through triples that no random variables realise: $a = b = -0.7$ with $c = -0.5$ satisfies all three inequalities on the angles and still has a negative determinant. Here both known angles are 45.573°, so their sum is 91.146° and we are nowhere near the trouble. The cone argument is exact for this problem, and I am claiming it for this problem.

The candidate who multiplies is right about the wrong thing

Average the two endpoints in (2) and the square roots cancel:

So 0.49 is not a wrong guess that happens to be close. It is the exact centre of the set of admissible answers, and it would be the exact centre for any other pair of correlations too. The candidate who multiplies has found the midpoint of the answer without noticing there was an answer to have a midpoint of, which is a stranger place to land than being wrong.

There is a reason, and it is worth carrying around. Split each stock into the part explained by the index and the part that is not. In the geometry that is just resolving an arrow into a component along the index and a component perpendicular to it:

Take the inner product. The index parts are shared, so they always agree and contribute $ab$ no matter what else happens. The leftover parts contribute whatever they feel like:

The free quantity $\rho_{AB \cdot I}$ is the partial correlation of the two stocks given the index, and it ranges over the whole of $[-1,1]$ with no constraint whatsoever from $a$ and $b$. That is the honest content of the puzzle: the two numbers you were handed say precisely nothing about how the residuals relate. For 0.7 and 0.7 the shared part is worth 0.49 and the private parts are worth 0.51 in either direction, which is the whole interval in one line. The width is $2\sqrt{(1-a^2)(1-b^2)} = 1.02$ out of a possible 2, so the two given numbers eliminate 49% of the dial and leave the rest genuinely open.

The lens shape says something the algebra hides. The band pinches to a single point at $b = \pm 1$: if the second stock is the index, there is nothing left to choose and the third correlation is forced to be $a$. It is widest at $b = 0$, where knowing that $B$ is uncorrelated with the index tells you almost nothing about how it sits against $A$. Information about the third number comes from the two given ones being extreme, not from them being large. And since correlation is a cosine, it is blind to shifts and rescalings of either series, which is why a change of units moves nothing here, and why volatilities combine with the law of cosines rather than by adding.

Where the floor stops being negative

The useful version of this question is not the arithmetic, it is knowing in advance whether the answer can dip below zero at all. Multiply the two roots of the quadratic. The product of the roots of $c^{2} - 2abc - (1-a^{2}-b^{2})$ is the constant term, so

One identity, and the whole sign question falls out of it. The two endpoints share a sign exactly when $a^{2} + b^{2} \ge 1$, which is to say exactly when the pair $(a,b)$, plotted as a single point on a page, sits outside the unit circle. Which sign it is comes from the sum of the roots, $2ab$. So the criterion has two halves and needs both:

Getting there directly is a good habit to have: $c_{-} \ge 0$ means $ab \ge \sqrt{(1-a^{2})(1-b^{2})}$, and since the right side is never negative you need $ab \ge 0$ before you are allowed to square. Squaring then gives $a^{2}b^{2} \ge 1 - a^{2} - b^{2} + a^{2}b^{2}$, which is the circle condition. The sign proviso is not fine print, it is the step that makes the squaring legal.

Now the shortcut people carry out of this problem, which is that both correlations have to clear $1/\sqrt{2} \approx 0.707$ for the third to be forced positive. That is the $a = b$ case and nothing more. Set $a = b$ in the circle condition and you get $2a^{2} \ge 1$, hence $a \ge 1/\sqrt{2}$. Off the diagonal it fails in both directions.

Take $a = 0.9$ and $b = 0.5$. Only one of the two clears 0.707, so the shortcut says the third correlation could go negative. It cannot. Here $a^{2}+b^{2} = 1.06 \ge 1$ and $ab = 0.45 > 0$, so (9) applies and the feasible interval is

Strictly positive at the bottom, and strictly below 1 at the top. That second half is easy to miss: $c_{+} = \cos(\mu - \nu)$ equals 1 only when the two angles coincide, so the ceiling reaches 1 if and only if $a = b$. With 0.9 and 0.5 the two stocks can be at most 83% correlated. There is no world in which they are the same thing.

The failure in the other direction is sharper. Take $a = 0.9$ and $b = -0.9$. Then $a^{2}+b^{2} = 1.62$, comfortably outside the circle, and yet

Perfect opposition, the worst the dial has. Clearing the circle test on its own forces nothing; it only says the two endpoints agree in sign, and with $ab < 0$the sign they agree on is negative. Two strong positives force a positive, one of each forces a negative, and that is what the two shaded corners in Fig. 3 are for.

One more piece of small print on the shortcut. Squaring the two numbers, adding, and taking away one returns the floor itself only when $a = b$. Equation (8) says $c_{-} = (a^{2}+b^{2}-1)/c_{+}$, and the denominator is 1 exactly on the diagonal. For 0.7 and 0.7 the recipe gives $0.49 + 0.49 - 1 = -0.02$, which is right. For 0.9 and 0.5 it gives 0.06, while the floor is 0.0725. The sign is still correct, because the numerator carries the sign, so the trick survives as a sign test and dies as a formula.

Building a world where it happens, and what it costs

An inequality tells you a value is permitted. It does not hand you the world in which it happens, and an interviewer who is enjoying himself will ask for one. Take the index $I$ and one more thing $N$ with nothing to do with it, both centred with unit variance, independent. Then set

Each has variance $0.49 + 0.51 = 1$, so each correlates with the index at exactly 0.7. The leftovers point in opposite directions, so $\rho(A,B) = 0.49 - 0.51 = -0.02$. That is the picture from Fig. 1 written out as a recipe. Simulated over four million days and measured back out of the numbers as if nobody had seen how they were made, the three correlations come back at 0.7, 0.7 and −0.02 with a maximum error of $1.8 \times 10^{-5}$, and the cosine of the angle between the centred sample vectors agrees with the Pearson coefficient to $3.3 \times 10^{-16}$. Somebody could hand you that risk report and every number on it would be honest.

What the simulation does not do is certify the fraction. At four million draws the standard error on a measured correlation is around $4 \times 10^{-4}$, so a construction aiming at −0.021 would have sailed through the same check. The exactness of $-\tfrac{1}{50}$ rests on the rational arithmetic in (3) and nowhere else. Simulation is how you confirm a value is reachable, not how you learn what it is.

Standing on the floor costs something, and this is the part worth saying out loud in the room. Add the two stocks in (10) and the noise cancels: $A + B = 1.4\,I$, so $B = 1.4\,I - A$. The second stock is not a third thing at all. It is a deterministic combination of the other two, the matrix is singular, and its rank has dropped from 3 to 2. The clean way to see this is to price the portfolio that is long both stocks and short 1.4 units of the index, which is $w = (-1.4, 1, 1)$ in standardised units:

That portfolio has zero variance at $c = -0.02$, and negative variance below it. One single portfolio is enough to prove the floor, and it is the same portfolio that vanishes when you get there. Ask the recipe for −0.0201 and it has nowhere to go: the leftovers are already perfectly opposed and there is no more opposition left to spend. The determinant is $-10201/10^{8}$ and the smallest eigenvalue is $-5.05 \times 10^{-5}$, small enough that a risk system might round it away and carry on.

Everything cannot hedge everything

Three arrows was the small case. The version that matters on a desk is what happens with a lot of them. Suppose you want a book where every position pushes against every other one by the same amount, so all $\binom{n}{2}$ pairwise correlations equal a single $\rho$. How negative is that allowed to be?

No determinant needed. Add all $n$ standardised positions together and demand that the total has non-negative variance:

The spectrum confirms it and adds detail: an equicorrelation matrix has eigenvalue $1 + (n-1)\rho$ once, along the all-ones direction, and $1-\rho$ with multiplicity $n-1$, so the determinant is $(1-\rho)^{n-1}\left(1+(n-1)\rho\right)$ and the first eigenvalue is the one that runs out. With three positions the floor is $-\tfrac12$, which is three arrows in a plane at 120° to each other. With ten it is $-\tfrac19$. With a hundred it is $-\tfrac{1}{99}$, about one percent.

Mutual hedging does not scale, and that is a fact about geometry rather than about markets. You can arrange a handful of things to genuinely oppose one another, but there is not enough room in any number of dimensions for a hundred arrows all pointing away from each other. The bigger the book, the closer the average pair is forced towards zero from below, which is one of the reasons diversification stops paying at some point instead of improving forever.

The bound is attained when the distributions are unconstrained. Put one red marble and $n-1$ green ones in an urn and draw the lot without replacement; the indicator that position $i$ holds the red marble has $\mathbb{E}[X_iX_j] = 0$ and mean $1/n$, which gives a pairwise correlation of exactly $-1/(n-1)$. Fix the marginals instead and the floor lifts. Take three symmetric coin flips written as $\pm 1$, all pairs at $\rho$. Their sum is an odd number, so $S^2 \ge 1$ on every outcome, and therefore $\mathbb{E}[S^2] = 3 + 6\rho \ge 1$, which gives $\rho \ge -\tfrac13$. Not $-\tfrac12$. Parity got there before positive semi-definiteness did.

The interval assumes the distributions are free

Everything above answers one precise question: which correlation matrices exist, over all possible joint distributions. That is the right question when nobody has told you what the variables are. Fix the marginals and the answer changes, sometimes a lot, and this is the part of the problem that separates a candidate who has the formula from one who knows what it is a formula about.

Take the same setup with indicators instead of returns: three events, each happening half the time, the first two correlated 0.7 with the third. For two symmetric binary variables the correlation has a plain reading. Writing $p_{11} = \mathbb{P}(X = Y = 1)$, the covariance is $p_{11} - \tfrac14$ and each standard deviation is $\tfrac12$, so

A correlation of 0.7 means the two indicators disagree 15% of the time, no more and no less. And disagreement obeys a triangle inequality for the cheapest of reasons: if $A \ne B$ then $A \ne I$ or $I \ne B$, so the union bound gives $\mathbb{P}(A \ne B) \le 0.15 + 0.15 = 0.30$, and (13) turns that into $\rho(A,B) \ge 1 - 0.60 = 0.40$. The bound is reached: let $A = I \oplus D_A$ and $B = I \oplus D_B$ with two disjoint flip events of probability 0.15 each, independent of $I$. Both stay symmetric, both sit at 0.7 with the index, and they disagree with each other exactly 30% of the time.

The ceiling can go missing too, which is the version that shows up in a real stress test. Take two events, one happening half the time and the other three times in ten. The joint law of two binary variables has exactly one free parameter, again $p_{11}$, and the correlation is an increasing affine function of it, so it is maximised at the largest feasible $p_{11}$. You cannot have both events more often than the rarer one occurs, so $p_{11} \le \min(p,q)$— the upper Fréchet–Hoeffding bound, realised by the comonotone coupling in which the rarer event always arrives inside the commoner one. For a rarer event $q$ and a commoner one $p$:

Line those two events up as tightly as anything can be lined up and they still are not in lockstep, because the commoner one turns up alone two days in ten. So a model that stresses a portfolio by turning every correlation up to 1 may be asking for a scenario that does not exist, and quietly returning a number for it anyway. The dial runs from −1 to +1. What actually lives on the dial is decided by everything else in the table.

Four things the interval does not say

Everything here is about population correlations, the true ones. A 0.7 that arrives in an interview is a number somebody typed; a 0.7 that arrives on a desk is an estimate with a standard error attached, and the bound says nothing about how far the estimate might be from the truth.

That gap has consequences you can see in production. A matrix assembled cell by cell, with each pair estimated on whatever history both series happen to share, or over windows of different lengths, routinely comes out with a negative eigenvalue. It is not a rounding artefact, it is three numbers that no world produces, sitting in the same table. Repairing that is a genuine computational problem with a literature of its own, which is what Higham was writing about.

The interval is Pearson’s. Spearman’s rank correlation inherits it unchanged, being a Pearson correlation computed on ranks, so equation (2) applies verbatim. Kendall’s tau does not. Its natural constraint is linear rather than quadratic: $1 - \tau$ is proportional to a metric on orderings, so $\tau_{AB} \ge \tau_{AI} + \tau_{IB} - 1$, and two taus of 0.7 put the floor at $+0.4$ or above. Positive, where the cosine bound was negative. Asking which coefficient the interviewer means is not a stalling tactic; it changes the sign of the answer.

And the interval describes what is possible, not what is likely. Nothing above says a third correlation near −0.02 is common, or that a pair of equity names behaves that way. It says that if somebody claims one, you cannot refute them with the two numbers you were given, and that if somebody claims −0.03 you can.

What the question was actually for

The interviewer is not collecting the number. He wants to know whether you treat three correlations as three independent facts, or notice that they are three angles in one room and cannot be chosen separately. That instinct outlives the question, because every risk table you will ever be handed is a claim about geometry, and the bad ones are usually bad because somebody typed a value into a cell that no world could produce.

Which is why the good answer out loud is not “minus two percent”. It is: the answer is a range, here is the range, here is the one line that produces it, and here is a world sitting on each end of it.