1 Description of variables

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1 1 Description of variables We have three possible instruments/state variables: dividend yield d t+1, default spread y t+1, and realized market volatility v t+1 d t is the continuously compounded 12 month dividend yield on the value-weighted NYSE. Data is from CRSP s index file. ( ) total dividends paid over months t 11 to t d t = ln price at the end of month t y t is the log of the difference between the monthly yields on corporate bonds rated Aaa and Baa by Moody s. Data is from the St Louis Fed s FRED website. v t is calculated as: v t = ln ( Nt j=1 y t = ln(aaa yield t Baa yield t ) ( ln(1 + Rj,t ) ) N t j=1 ln(1 + R j,t ) ln(1 + R j,t+1 ) where N t is number of days in month t, and R j,t is daily market return. The daily market return is taken from Ken French s website. We inspect the histograms of the raw and the logged variables, and settle on the log forms because each appears more normal. After construction, the three instruments are standardized so that each has a distribution with mean 0 and variance 1. The assets used are three portfolios formed on the basis of book-to-market ratio. They are value-weighted portfolios of the bottom three, the middle four, and the top three BE-ME decile portfolios available from Ken French s website. The decile portfolios themselves are value-weighted. We take the log of [1 plus] the returns. The data period used is from 31 Dec 1926 to 31 Dec 2015, yielding 1069 observations. Since our regressions are on instruments lagged one period, the data series in our estimations are generally 1068 observations long. 2 Framework Start with an instrument series, Z t+1. This consists of one or more of dividend yield d t+1, default spread y t+1, and realized market volatility v t+1. To specify how the instruments evolve, we run a VAR: Z t+1 = a Z + b Z Z t + e Z t+1 (1) The log return series is r t+1. We consider two different specifications for its conditional expectation. Returns are either predictable by dividend yield: r t+1 = a r + b r d t + e t+1 1 )

2 or they are not predictable: r t+1 = a r + e r t+1 We call e r t+1 the return residual and e Z t+1 the residual from the instrument VAR. Finally, we regress e r t+1 = b e e Z t+1 + u r t+1 (2) 3 Outline of solution method We will solve the portfolio problem by building a numerical model in the framework laid out in the previous subsection. We discretize the AR(1) process which generates the Z. In this discretization, the e Z can take on a finite number of values, and the evolution Equation 1 then implies that the state variables Z too take on only a finite number of possible values, and constitute a Markov chain with discrete states and a transition probability matrix. The technique we use for discretization is Gaussian quadrature, as described by Tauchen and Hussey. Previous work (e.g., Lynch (2000)) has used this technique with great success under the assumption that all variables in the model are normal. Working under this assumption, Lynch (2000) discretizes both the e Z (and therefore the Z) as well as the u r. The e r are automatically rendered discrete, since they are constructed from the e Z and the u r using Equation 2. In this model, all variables are assumed to be homoscedastic. In our work, we maintain the assumption that the e Z are normal and homoscedastic, but we allow the u r to be nonnormal and heteroscedastic. While we could adapt the Tauchen and Hussey procedure to work with nonnormal variables, such an adaptation would be fraught with problems and is likely to work poorly. Instead, we simulate a large number of points for the u r, and, in combination with the discretized, normal, homoscedastic e Z, this gives us the possible values of the e r. 4 Normalizing the e Z Our goal, then, is to be able to simulate the u r so that when these simulated values are used together with the discretized e Z, we obtain e r which have the same properties as the e r in the data. An initial step would be to specify distributions for the u r and to estimate the parameters of these distributions. However, before we can run estimations involving the u r, we need to ensure that the e Z have the properties in the data that we will assume they have in the quadrature: specifically, that they are homoscedastic and normal. In the data, it turns out that the e Z are heteroscedastic and nonnormal. If we were to use these data e Z, and estimate Equation 2, some, perhaps most, of the nonnormality in the e r will be accounted for by the e Z. Estimating the parameters of the resulting u r, and attempting to simulate from those u r parameters with normal e Z, will produce simulated e r which have different nonnormal behavior than shown by the e r in the data. Such differences may show up in the level of nonnormality (where we would expect the simulated e r to have less nonnormality than in the data), but they may also show up in how that nonnormality varies over time. 2

3 Because a principal aim of this paper is to estimate hedging demands driven by time-varying nonnormality, it is essential that the nonnormality of the model e r closely match that in the data. As a first step towards resolving this issue, we normalize the e Z. We consider two combinations of the instruments Z: first, dividend yield and default spread, and second, default spread and realized market variance. For concreteness, in the description below I focus on the first of these combinations. We start with the data Z, and we estimate Equation 1. This gives us the data e Z, a T 2 matrix. We then fit a conditional variance model to each of these e Z. The model we use is: (e Z t+1) 2 = exp(k) + exp(a + b Z t ) + ν t+1 where the model is estimated under the constraint that the unconditional variance equals the average conditional variance: E(e Z t+1) 2 = exp(k) + E exp(a + b Z t ) Please see Section 5.1 for a description of this model, and the reasoning behind this specification. With this model, we can estimate the conditional variance of each of the e Z as: V t (e Z ) = exp(k) + exp(a + b Z t ) We divide each e Z by its conditional standard deviation at each point in time. Our expectation is that this removes any time-varying variance from the e Z. If our conditional variance model were perfect, the standardized series would have mean 0 and variance 1. In our estimations these conditions are almost met. We force them to be met exactly. We then orthogonalize the two e Z with respect to each other, by postmultiplying by the inverse of the Cholesky decomposition of their covariance matrix. The Cholesky decomposition is lower triangular, and so the first column of the e Z will not be affected by the orthogonalization. The orthogonalization results in series which are uncorrelated, but possibly dependent. We attempt to make these orthogonalized series bivariate normal. Bivariate normality of two variables implies first, that each variable is univariate normal, and second, that the conditional distribution of the second variable given the first is normal with a known mean and variance. For mean-zero variables which are uncorrelated, the mean of the conditional distribution is zero and the variance is the unconditional variance. We render the first e Z unconditionally normal in the following way. We obtain the percentiles each observation occupies in the empirical CDF. We then apply the inverse normal CDF (with parameters mean zero and standard deviation 1) to this series and get the values that this series would take if it were normal. We standardize the result to have mean 0 and standard deviation 1. When we run the usual tests for normality on this transformed series, we are unable to reject. We then sort the first series into 20 bins. Because there are 1068 observations, this will result in an uneven number of observations in each bin. We assign the leftover observations, 3

4 one to a bin, starting from the most extreme bins and moving inward. This is done because if forced to choose, we would like the extreme values to be better behaved. Assigning more values to those bins may achieve this. For each bin, we take the subset of values that the second series takes when the first series takes values in that bin, and we make that subset of values unconditionally normal. Again we standardize the result to have mean 0 and standard deviation 1. Once we have processed all bins, we unconditionally normalize the entire second e Z series. We check for time-varying variance by estimating a conditional variance model on these series. We find little evidence of time-varying variance. We then postmultiply by the Cholesky decomposition of the covariance matrix of the original e Z. This gives us an e Z series which we hope has a distribution more close to normal than previously, but which has a similar covariance structure as the previous series. Using this e Z series, we reconstruct the Z. Given that the e Z are distributed normally with mean zero and known variance, say Σ ez, Equation 1 implies the population distribution of Z: it is normal with a mean of (I b Z ) 1 a Z and a variance given by the solution to the following matrix equation: V ar(z) = b ZV ar(z)b Z + Σ ez To solve this equation, we pick an initial guess for V ar(z), and then iterate on the equation above until convergence. We us Σ ez as the initial guess. We then reconstruct the Z series. We start from the initial values in the data Z series, but we modify them to be more consistent with normality. We know what percentiles each initial value occupies in the empirical CDF of the Z. We replace it with the value the inverse normal CDF assigns to that probability value. In this calculation, the inverse normal CDF has parameters equal to the population parameters of the Z. Starting from these values, and given the e Z series, we can construct the Z series using Equation 1. Once we have these Z, we repeat the entire normalization procedure using them as the data Z. We do this to make the Z as normal as we can. In theory we could iterate on the Z, repeatedly normalizing until we find that the values don t change much. In practice this leads to poor properties: in some instances the autoregressive coefficient on the Z becomes larger than 1. With the Z output from the second normalization, we reestimate Equation 1. The coefficients we estimate are close to but not exactly the same as the coefficients when using the original Z. Using the coefficients we obtain the e Z. We test these e Z for univariate normality, and find that we are unable to reject the null of normality, with p-values in excess of 0.8. A necessary condition for two variables to be bivariate normal is that all linear combinations are normal. We test the following linear combinations: [ 1, 1], [ 1, 2], [ 2, 1 ], and once again we are unable to reject the null of normality. The smallest p-value we obtain is

5 5 Model specification 5.1 Conditional variance specification The first task is to model the time-varying conditional variance of the u r t+1 and the e r t+1. (Recall that the conditional mean has already been modelled, either by assuming that it is a constant or that it is linear in the lagged dividend yield.) We would like to specify the conditional variance as some function of the lagged state variables, because this is how it will be modelled in the quadrature. Once we have obtained estimates of the conditional variance of the u r t+1 and the e r t+1 at each point in time, we can remove the effects of the time-varying variance by dividing by the appropriate conditional standard deviation. We can then further model the behavior of these standardized processes. The conditional variance specifications must satisfy a variety of conditions: 1. Once estimated, it must be easy to extract the conditional variance at each point in time. 2. The conditional variance must not be negative. Further, it is desirable that the conditional variance not go too close to zero, otherwise the standardized series (creating by dividing by the conditional standard deviation) may blow up. 3. The average conditional variance must be equal to the unconditional variance. This is necessary because of the relation, for example for the u below, that the unconditional variance equals the expected conditional variance plus the variance of the conditional expectation: V (u r t+1) = E(V t (u r t+1)) + V (E t (u r t+1)) In our application, we assume that the u r and the e r have a constant conditional mean of zero, which implies that V (u r t+1) = E(V t (u r t+1)) and V (e r t+1) = E(V t (e r t+1)). 4. The conditional variance processes specified for the u r and the e r must be consistent with each other, in that they obey the relations implied by (a) our assumptions about e Z : that it is homoscedastic and normal and (b) Equation 2. Together, these restrictions imply that: V t (e r t+1) = b ev (e Z t+1)b e + V t (u r t+1) (3) A simple way to ensure that this condition is met is to model the conditional variance of the u r, and to use Equation 3 to back out the conditional variance of the e r. As an example, we could model the conditional variance of the u r t+1 by estimating the following equation: ln(u r t+1) 2 = a + b Z t + ν t+1 This imposes positivity, but it hard to force the conditional variance to be above a bound. Further, the conditional variance at each point in time, E t ((u r t+1) 2 ), is difficult to obtain: the equation above gives us an estimate only of E t (ln(u r t+1) 2 ). While it is possible to adjust 5

6 this by multiplying by a Jensen s inequality term, in practice we find that such adjustments produce conditional variance estimates with poor properties. Instead, we consider the following specification for the conditional variance of the u r (u r t+1) 2 = exp(k) + exp(a + b Z t ) + ν t+1 It is easy to estimate the conditional variance at any point, as E t (u r t+1) 2 = exp(k) + exp(a + b Z t ) The parameters k, a, b can be estimated through nonlinear least squares (NLS). NLS minimizes Eν 2 t+1 = E ( (u r t+1) 2 exp(k) exp(a + b Z t ) ) 2 with first order conditions: w.r.t. k : 2 exp(k)eν t+1 = 0 w.r.t. a : 2E exp(a + b Z t )ν t+1 = 0 w.r.t. b : 2E exp(a + b Z t )Z t ν t+1 = 0 In the absence of a corner solution (i.e., k = ), the first of these FOCs guarantees that Eν t+1 = 0, or that E(u r t+1) 2 = exp(k) + E exp(a + b Z t ) meaning that we may get the desired constraint, that the unconditional variance of the u will be equal to the expected estimated conditional variance, for free. To ensure that this constraint is always met, we estimate the constrained NLS: min E(ν t+1) 2 subject to E(ν t+1 ) = 0 k,a,b We can run this estimation using the sample average as the expectation in the constraint. This will mean that the sample average conditional variance equals the unconditional variance. What we care about, however, is that the quadrature expected conditional variance equal the unconditional variance. One way to proceed is to observe that we know what the population distribution of Z is: it is normal with a known mean and variance. This implies that the conditional variance, given by exp(k) + exp(a + b Z t ), is a constant plus a lognormal variable with known parameters, and its expected value is just exp(k) + exp(a + b EZ t + 0.5V ar(b Z t )). To the extent that the quadrature faithfully reproduces this population expectation, this is the appropriate constraint to impose. In the quadrature, however, the use of discrete points for the Z implies that the distribution of Z is truncated at both ends. In running the estimation of the conditional variance, we 6

7 would like the conditions we impose to resemble the quadrature as closely as possible. Accordingly, we estimate the NLS with the constraint that the truncated population expected conditional variance equals the unconditional variance of the u r. It is fairly simple to obtain the symmetrically truncated expectation of the lognormal variable. 1 Therefore, our final specification for the conditional variance of the u r involves solving the following NLS problem: min k,a,b E(ν t+1) 2 subject to E(u r t+1) 2 = exp(k) + E [exp(a + b Z t ) U > Z t > L] To get the truncation points, we look at the extreme values that our normalized Zs take, and find the corresponding percentiles in their known population distributions. These may be different for the minimum and the maximum of each Z. We take the laxer of the percentiles as the cutoffs in the truncation. Using the truncated population expectation in the constraint produces coefficient estimates that are close to when we use the sample expectation. Using the untruncated population expectation produces values which are different. This is likely because the expectation of the exponential function of a variable is severely affected by the very large values in the extreme right tail. 1 We make use of the following relations: we know how to decompose an integral and observing that for example, we can decompose an expectation xdf (x) = L E(x x L) = U xdf (x) + xdf (x) + L U 1 L xdf (x) P (x L) xdf (x) E(x) = P (x L)E(x x L) + P (L < x U)E(x L < x U) + P (U < x)e(x U < x) Further, we know: If y N(µ, σ), and x = exp(y), then where U 0 = (ln(u) µ)/σ, and E(x U < x) = E(x) Φ(σ U 0) Φ( U 0 ) E(x x L) = E(x) Φ( σ + L 0) Φ(L 0 ) where L 0 = (ln(l) µ)/σ. Now we want the expectation of a symmetrically truncated lognormal. From the above, E(x L < x U) = We can work out each part of this separately. E(x) P (x L)E(x x L) P (U < x)e(x U < x) P (L < x U) 7

8 Given the specification for the conditional variance of the u r, we obtain the conditional variance of the e r from Equation 2: V t (e r t+1) = b ev (e Z )b e + V t (u r t+1) We then standardize the u r and the e r by their conditional standard deviations, and standardize the resulting series to have unconditional mean 0 and unconditional standard deviation of Distribution specification Given the homoscedastic, standardized u r and e r, we can proceed to model their distributions. In each case, we might use a single multivariate distribution, but copula theory allows us the more flexible alternative of modelling the marginal and joint distributions separately. We consider conditional and unconditional models. A conditional model is one whose parameters vary over time. Time-varying parameters are modelled by allowing them to be functions of the lagged state variables. In particular, we model them as [possibly nonlinear] functions of linear combinations of the lagged state variables. For the marginals we consider the following specifications: 1. Normal: with parameters mean µ and variance σ 2. Since the series have already been standardized to have constant conditional mean and variance of 0 and 1, the parameters of the normal marginals are determined. 2. Skew-t: with parameters degrees of freedom ν and skewness λ. For the copulas we consider the following specifications: 1. Normal: with three correlation parameters ρ i,j. 2. t: with three correlation parameters ρ i,j, and with a degrees of freedom parameter ν 3. Mixture: where the copula can be either t with a certain probability, or Clayton with one minus that probability. The parameters are the four parameters of the t-copula, one parameter κ of the Clayton copula, and one mixture probability. In principle, any or all of these six parameters can be time-varying. Each of the parameters of the marginals and the copula have certain bounds on them. For example, the correlation parameters have to lie between 1 and 1. In Table 1 we list the parameters, the bounds on them, and the functional forms that we use to ensure that they stay within the bounds. When estimating the copulas, we would like to ensure that the unconditional correlations of implied by the copula parameters are the same as in the data. Unfortunately, there is no analytical formula linking the copula parameters to the unconditional correlations. We impose the constraint via simulation. 8

9 Consider a particular marginal-copula pair. For example, take the skew-t marginals and the normal copula. Marginal parameters are unaffected by correlations. The normal copula has three correlation parameters. 2 We model them as ρ i,j,t = g(β 0 + β Z t ) where g() is a nonlinear function with range ( 1, 1). We first estimate the copula with no constraints, and obtain estimates of β 0 and β. We then simulate 100,000 observations from the e Z and the Z (with a burn-in period of 1000 observations), and for the three assets from the copula-marginal pair using the estimated parameters. We measure the sample unconditional correlations of the simulated variables. If they do not equal their values in the data, we adjust each of the three β 0, the level parameter of the correlations, until they do. When looking at the other copulas which have more parameters than just correlations, we maintain this approach: we only ever adjust the level parameters of the three time-varying correlations. 6 Specifications considered We consider three principal specifications: 2 Even in this simple case the correlation parameters don t have an analytical link to the sample correlations. 9

10 A Description of marginal distributions used to model return residuals A.1 Normal marginals The univariate standard normal distribution with parameters µ and σ has a PDF given by A.2 Skew-t marginals f(u) = 1 σ 2π e (u µ) 2 2σ 2 The univariate skew-t distribution, for a variable u with mean 0 and standard deviation 1, has parameters ν and λ and has a PDF given by { bc ( ν 2 f(u) = ( bu+a)2) (ν+1)/2 1 λ u < a/b, bc ( ( bu+a)2) (ν+1)/2 ν 2 1+λ u a/b. The constants a, b and c are defined as a = 4λc ( ) ν 2, b 2 = 1 + 3λ 2 a 2, c = ν 1 Γ ( ) ν+1 2 π(ν 2)Γ( ν 2 ) B Description of copula distributions used to model return residuals B.1 Normal copula Let u be a vector-valued random variable with dimension N 1. Let Φ(u; R) be the N- dimensional multivariate normal CDF with correlation matrix R. Let Φ(u i ) be the univariate standard normal CDF. Then the normal copula is given by and its PDF is given by: B.2 t copula B.3 Clayton copula B.4 Mixture copula C(u, R) = Φ(Φ 1 (u 1 ),..., Φ 1 (u n ); R) (4) 10

11 Distribution Parameters and bounds Implementation as functions of lagged Z Panel A: Marginals Normal mean µ Fixed at 0 in standardized series standard deviation σ Fixed at 1 in standardized series Skew-t degrees of freedom ν, always above 2 νi,t+1 = 2 + e Z t β i skewness λ, in ( 1, 1) λi,t+1 = /(1 + e Z t β i) Panel B: Copulas Normal Correlations ρ, in ( 1, 1) ρi,j,t+1 = /(1 + e Z t β i,j) t Correlations ρ, in ( 1, 1) ρi,j,t+1x = /(1 + e Z t β i,j) Degrees of freedom ν, always above 2 νt+1 = 2 + e Z t β Clayton copula kappa κ, always above 0 κt+1 = (1 1./(1 + e Z t β ) We bound it below by 0.01 and above by to help convergence Mixture Probability p pt+1 = 1 1/(1 + e Z t β ) All parameters of Clayton and t Table 1: Distribution parameters, bounds on them, and functional forms: The Zt are assumed to include a constant. 11