Structure of the CAPM Covariance Matrix

Transcription

1 Structure of the CAPM Covariance Matrix James Bridgeman University of Connecticut UConn Actuarial Science Seminar September 10, 2014 (UConn Actuarial Science Seminar) CAPM September 10, / 24

2 INTENTIONS SEE HOW CAPM COVARIANCE MATRIX: ACTUALLY HOLDS THE EXPECTED RETURNS ACTIVELY REFLECTS THE MARKET WEIGHTS MOSTLY PEDAGOGY A TINY BIT OF LIGHT ON LITERATURE (UConn Actuarial Science Seminar) CAPM September 10, / 24

3 CAPM SET-UP MARKET COVARIANCE MATRIX - σ ij = covariance[r i,r j ] 8 9 σ 11 σ σ 1n >< >= σ Σ = 21 σ σ 2n invertible, σ ij = σ ji, σ ii > 0, and σ ii σ jj > σ 2 ij >: >; σ n1 σ n2... σ nn MARKET WEIGHTS & EXPECTED RETURNS w 1 1 w = Bw 2 A market weights, e = B 1 A, et w = 1, w n 1 w i > 0 (you 1 can t short what no one owns) r 1 r = Br 2 A expected returns, r f risk-free rate, (r r f e) market risk premia r n (UConn Actuarial Science Seminar) CAPM September 10, / 24

4 CAPM CONCLUSIONS ASSUMING MARKET EFFICIENT (ON FRONTIER) (FOR EXAMPLE, ASSUME THAT 9 RATIONAL REP AGENT or THAT MARKET IS BIVAR NORMAL or ALL HAVE QUADR UTIL) COVARIANCE-RETURNS-WEIGHTS RELATIONSHIP (r r f e) = Σ w w = Σ 1 (r r f e) e T Σ 1 (r r f e) (denominator ensures et w = 1) RETURNS-BETAS RELATIONSHIP (r r f e) = β w T r r f = β (r M r f ) where β = Σ w h i w T Σ w = Cov [r,r M ] = Cov r,w T r and r M = w T r is the expected market return (UConn Actuarial Science Seminar) CAPM September 10, / 24

5 QUICK PROOF FOR COVARIANCE-RETURNS-WEIGHTS RELATIONSHIP - use Lagrange Multipliers FOR BETA (r r f e) = Σ w Σ w = w T w T Σ w Σ w = β w T (r r f e) = β w T r r f w T e = β w T r r f = β (r M r f ) (UConn Actuarial Science Seminar) CAPM September 10, / 24

6 THE LAZY PEDAGOGUE 8 < Σ = : EXAM QUESTION Be Sure To Emphasize E ect Of Negative Covariance = <.02= ;, r =.10 : ;, r f =.03, what is w? WAIT UNTIL GRADING TO WORK OUT THE ANSWER ANSWER < = <.27= Σ 1 = : ;, w = Σ 1 (r r f e) e T Σ 1 (r r f e) =.80 : ;, Rats!.47 WHAT WENT WRONG? AHA! Forgot Requirement σ ii σ jj > σ 2 ij. Easy To Fix Next Semester. (UConn Actuarial Science Seminar) CAPM September 10, / 24

7 ENLIGHTENED PERHAPS, BUT STILL LAZY 8 < Σ = : EXAM TIME THE FOLLOWING SEMESTER Still Be Sure To Emphasize E ect Of Negative Covariance = <.02= ;, r =.10 : ;, r f =.03, what is w? STILL WAIT UNTIL GRADING TO WORK OUT THE ANSWER ANSWER < = < 1.92= Σ 1 = : ;, w = Σ 1 (r r f e) e T Σ 1 (r r f e) = 2.03 : ;, Worse!.89 WHAT WENT WRONG THIS TIME? God Knows, Just Avoid The Negative Covariances Next Semester (UConn Actuarial Science Seminar) CAPM September 10, / 24

8 JUST PLAIN STUBBORNLY LAZY THIRD SEMESTER S A CHARM? Let Everything Be Positive, Including All Of The Risk Premia < = <.04= Σ = : ;, r =.10 : ;, r f =.03, what is w? CONFIDENTLY WAIT TO WORK OUT THE ANSWER ANSWER < = < 1.47 = Σ 1 = : ;, w = Σ 1 (r r f e) e T Σ 1 (r r f e) =.15 : ;, No!.62 THERE S MORE TO THIS THAN MEETS THE EYE! WHAT DOES Σ HAVE TO LOOK LIKE TO BE A MARKET? (UConn Actuarial Science Seminar) CAPM September 10, / 24

9 A FAMILIAR PHENOMENON IN THE LITERATURE BEST & GRAUER (1985, 1992) E cient Markets with all w i > 0 form a segment of the Frontier with length! 0 as number of assets n! Brennan & Lo (2010) For any given (r r f e), Impossible Σ are those with some w i < 0. Then P [Σ Impossible] % geometrically with n for reasonable distribution assumptions on Σ. LEDOIT & WOLF (2004, 2013, 2014) Introduce statistical Shrinkage techniques to transform Empirical Σ into a Frontier Portfolio LEVY & ROLL (2010) Empirical Σ has "high" P of being "close" to a Frontier Portfolio, with "close" attained by ɛ on σ ii and r i. Note ρ ij can remain xed. BOYLE (2012, 2014) On large class of Σ, Frontier Portfolio equivalent to Σ Almost Positive (UConn Actuarial Science Seminar) CAPM September 10, / 24

10 IS THERE A SIMPLE-MINDED WAY TO SEE ALL THIS? GIVEN w and r r f e WITH ALL w i > 0 WHAT DO Σ THAT SIT ON FRONTIER LOOK LIKE? w = Σ w = (kσ) w = = So it is enough to nd all Σ such that and then multiply by any constant. Σ 1 (r r f e) e T Σ 1 (r r f e) so (r r f e) e T Σ 1 and for any k (r r f e) k (r r f e) e T Σ 1 (r r f e) (r r f e) e T (kσ) 1 (r r f e) Σ w = (r r f e) (UConn Actuarial Science Seminar) CAPM September 10, / 24

11 IS THERE A SIMPLE-MINDED WAY TO SEE ALL THIS? EASY TO FIND A MATRIX R WITH Rw=r r f e Just let R = n(r o r f e) e T with n identical column vectors each equal to (r r f e). We can also express R as n row vectors 8 (r 1 r f ) e >< 9 >= (r R = 2 r f ) e T.... >: >; (r n r f ) e T UNFORTUNATELY THE CHOICE Σ=R IS NOT INVERTIBLE Rows and columns clearly fail to be independent COULD EASILY FAIL TO HAVE σ ii > 0 IS UNLIKELY TO HAVE σ ij = σ ji OR σ ii σ jj > σ 2 ij FOR ALL i, j (UConn Actuarial Science Seminar) CAPM September 10, / 24

12 IS THERE A SIMPLE-MINDED WAY TO SEE ALL THIS? BUT MAYBE THERE IS A MATRIX A WITH Aw=0 So (R + A) w = R w = (r r f e) AND WITH Σ= (R+A) INVERTIBLE AND SATISFYING σ ij =σ ji, σ ii >0, and σ ii σ jj >σ 2 ij VIEW A AS ROW VECTORS 8 a >< T 9 1 >= a A = T 2, with each a T... i = a i1 a i2... a in >: >; a T n THE CONDITION FOR Aw=0 IS a T i w = 0 for all i in other words, all n of the row vectors a T i must be in the (n 1)-dimensional hyperplane through the origin perpendicular to the market weight vector w. (UConn Actuarial Science Seminar) CAPM September 10, / 24

13 THERE IS A SIMPLE-MINDED WAY TO UNDERSTAND! IF ALL Σ ARISE THIS WAY AND IF ALL OF THE OTHER CONDITIONS CAN BE MET (PERHAPS TWO BIG IFs) THEN The odds of an (n 1)-dimensional hyperplane through the origin n 1 being perpendicular to a vector w with all w i > 0 is 12 Half the lines through the origin in 2-space are perpendicular to something in rst quadrant. A quarter of the planes through the origin in 3-space are perpendicular to someting in the rst octant. And so on... The lazy pedagogue had at best 1 chance in 4 to write down a valid Σ even after he respected all of the σ constraints It s immediately clear that the odds to randomly encounter a valid Σ disappear at least exponentially in n, independent of (reasonable) distribution assumptions (UConn Actuarial AsScience n increases, Seminar) empirical A,RCAPM will tend to be closer September to a valid 10, 2014 Σ 13 / 24

14 SIGMA CONDITIONS ON A EASY TO UNDERSTAND σ ij = σ ji means that (r i r f ) + a ij = (r j r f ) + a ji so a ij = a ji (r i r j ) and A can be expressed also as consisting of n column vectors A = c 1 c 2... c n where c j = a j (r r j e). σ ii > 0 means that σ ii σ jj > σ 2 ij means that (r i r f ) + a ii > 0 so a ii > (r i r f ). (a ii + (r i r f )) (a jj + (r j r f )) > (a ij + (r i r f )) 2 so a jj > (r j r f ) + (a ij + (r i r f )) 2 a ii + (r i r f ). Looks like an induction might be possible. (UConn Actuarial Science Seminar) CAPM September 10, / 24

15 INVERTIBILITY CONDITIONS 8 (r 1 r f ) e >< T + a T 9 1 >= (r Σ = R+A = 2 r f ) e T + a T 2... >: >; (r n r f ) e T + a T n IS INVERTIBLE IFF ITS n ROW VECTORS ARE INDEPENDENT INDEPENDENCE HOLDS IF AND ONLY IF 1 The n row vectors a T 1, at 2,...,aT n span the (n 1)-dimensional hyperplane perpendicular to the market weight vector w, and 2 w T (r r f e) 6= 0, i.e. the market risk premium vector (r r f e) is not perpendicular to the market weight vector w. PROOF 1 (r 1 r f ) e T,...,(r n r f ) e T are collinear so the row vectors of R+A cannot span an n-space unless a T 1,...,aT n span an (n 1)-space, and the (n 1)-dimensional hyperplane perpendicular to the market weight vector w is the (n 1)-space they are in. (UConn Actuarial Science Seminar) CAPM September 10, / 24

16 INVERTIBILITY CONDITIONS PROOF (continued) 1 (prior slide) 2 Requires a slightly fussy proof that essentially follows from the fact that e T and w both have all components positive and a T 1,...,aT n are all perpendicular to w. CONDITION 2 HOLDS IN ANY REASONABLE MODEL w T (r r f e) = w T r r f = (r M r f ) > 0 where r M = w T r is the expected return on the risky market. SO Σ = R+A IS INVERTIBLE IN A REASONABLE MODEL IF AND ONLY IF 1 The n row vectors a T 1, at 2,...,aT n span the (n 1)-dimensional hyperplane perpendicular to the market weight vector w (UConn Actuarial Science Seminar) CAPM September 10, / 24

17 HOW HARD IS IT TO CHOOSE A? WORK BACKWARDS TO CHOOSE a T 1, at 2,...,aT n Suppose you already have chosen a T 1, at 2,...,aT n 1. Then there is no choice about what a T n must be. By symmetry: a n 1 = a 1 n (r n r 1 ) a n 2 = a 2 n (r n r 2 )... a n n 1 = a n 1 n (r n r n 1 ). Since a T n w = 0 the choice for a n n also is xed. The requirement is w 1 a n w n 1 a n n 1 + w n a n n = 0 so a n n = 1 w n (w 1 a n w n 1 a n n 1 ), which nishes the complete determination of a T n. (UConn Actuarial Science Seminar) CAPM September 10, / 24

18 HOW HARD IS IT TO CHOOSE A? WORK BACKWARDS TO CHOOSE a T 1, at 2,...,aT n But a n n has to satisfy some σ conditions: a nn > (r n r f ) and for all i < n a nn > (r n r f ) + (a in + (r i r f )) 2 a ii + (r i r f ) That means the choice of a T n 1 couldn t have been completely free 1 (w 1 a n w n 1 a n n 1 ) > (r n r w f ) and for all i < n n 1 (w 1 a n w n 1 a n n 1 ) > (r n r w f ) + (a in + (r i r f )) 2 n a ii + (r i r f ) where a n 1 = a 1 n (r n r 1 )... a n n 1 = a n 1 n (r n r n 1 ). Since all w i > 0, just pick a small enough a n 1 n (negative if need be) (UConn Actuarial Science Seminar) CAPM September 10, / 24

19 HOW HARD IS IT TO CHOOSE A? WORK BACKWARDS TO CHOOSE a T 1, at 2,...,aT n One possible problem at i = n 1: 1 w n (w 1 a n w n 1 (a n 1 n (r n r n 1 ))) > > (r n r f ) + (a n 1 n + (r n 1 r f )) 2 a n 1 n 1 + (r n 1 r f ) so a n 1 n on both sides, and squared (so made positive) on the right. Does picking a n 1 n small enough (negative if need be) still work? YES! a T n 1 is perpendicular to w with all w j > 0 so picking a small enough a n 1 n (negative if need be) forces a n 1 n 1 to increase and it turns out (some delicate analysis) to be enough to make the inequality work. 8 9 < =... " # : ;... # " (UConn Actuarial Science Seminar) CAPM September 10, / 24

20 HOW HARD IS IT TO CHOOSE A? THE HARD PART IS DONE Given a T 1, at 2,...,aT n 2 we saw that we can choose at n 1, at n that satisfy the σ conditions. GO BY INDUCTION STARTING AT a T 1 For 1 i n 2, given a T 1, at 2,...,aT i 1 always free to choose a ii big enough to satisfy the σ conditions, then a i i +1,...,a i n any values that keep a T i w = 0 SOME SPAN RESTRICTIONS ON CHOICES OF a T i To ensure that a T 1, at 2,...,aT i spans at least an (i 1)-dimensional subspace of the (n 1)-dimensional hyperplane perpendicular to w we have to disallow some choices in the induction. The whole set of disallowed matrices A altogether has dimension n(n 1) 2 3. The whole set of allowed matrices A has dimension xed choice of w and (r r f e) and still requiring Σ w = (r r f e) n(n 1) 2, still for a (UConn Actuarial Science Seminar) CAPM September 10, / 24

21 BACK TO THE LAZY PEDAGOGUE A FEW MORE DEGREES OF FREEDOM We can multiply any Σ = R+A developed above by any constant k. We can choose w 1,...,w n subject to e T w = 1 and w i > 0 So altogether the space of possible solutions Σ has dimension n(n 1) n 1 = (n+1)n 2, now with a possible disallowed set of dimension (n+1)n 2 3. A w = 0 IS THE MAIN NON-OBVIOUS RESTRICTION The σ restrictions are all visible in Σ and true for any empirical Σ Invertibility of Σ is apparent or easy to check, at least for small n. The space of possible n n symmetric matrices has dimension n + (n 1) = (n+1)n 2, same as the space of solutions The disallowed set of matrices has measure (probability) 0. But the odds of an (n 1)-dimensional hyperplane through the origin n 1 being perpendicular to a vector w with all w i > 0 is 12 n 1 So solutions seemingly are rare only because of that factor 12 (UConn Actuarial Science Seminar) CAPM September 10, / 24

22 ONE MORE SOURCE OF MEANINGFUL CONSTRAINT REMEMBER THAT RETURNS ARE CONSTRAINED In meaningful models w T (r r f e) = w T r r f = (r M r f ) > 0 If there are any risky assets with expected return r i < r f then this inequality cuts o a fraction (call it f ) of the otherwise possible w. Now the only possible w are in the intersection of the set having all w i > 0 with the half-space having w T (r r f e) > 0. This in turn eliminates the same fraction f of the set of otherwise possible matrices A, whose row vectors have to live on the plane n perpendicular to w, so possible Σ are now rare by a factor f This further militates against the lazy pedagogue s chance of success, since he likes to illustrate negative covariances and negative risk premia. It impairs by the same factor the probability of a random empirical Σ to satisfy CAPM, even if it satis es all the σ constraints and invertibility. If I have understood Phelim Boyle s work, the "almost positive" condition does not yet contemplate this possibility. How should it be modi ed/generalized to this case? (UConn Actuarial Science Seminar) CAPM September 10, / 24

23 ALMOST FORGOT - What Is That Matrix A Anyway? h Claim: A = Cov r, r T r M e T i where bar means random, r is the random vector of actual asset returns in the market and r M is the random actual return on the market as a whole, i.e. Proof: r M = r T w h i Cov r, r T = Σ by de nition. h i Cov r, r M e T = = h i Cov r, r T r M e T = Σ R = A = = n o Cov [ r, r M ] e T, with n identical columns n h i o Cov r, r T w e T n h i o Cov r, r T w e T n o n o (Σ w) e T = (r r f e) e T = R so (UConn Actuarial Science Seminar) CAPM September 10, / 24

24 THANKS PHELIM BOYLE MY STUDENTS THIS SUMMER (UConn Actuarial Science Seminar) CAPM September 10, / 24