33

The minimum variance solution loads up on securities that have low variances and co-variances. Theoretically you are correct that this should have a low expected return profile. However, it turns out - in contradiction to modern portfolio theory - that securities that have low-volatility or low-beta experience higher returns than high-volatility or high-...


15

The following papers may help. A New Look at Minimum Variance Investing by Bernd Scherer Minimum Variance Portfolio Composition by Clarke, De Silva & Thorley Under a multifactor risk-based model, if the global minimum variance portfolio dominates the market portfolio, the implication is that the market portfolio is not multifactor efficient and that ...


13

The PCA analysis does not really tell you what the bonds do but it tells you how the rates move together. The variations of $n$ rates (i.e. 1 y, 2y, ...) are split up in (at first) abstract factors like $$ \Delta R_i = \sum_{j=1}^n e_{i,j} f_j $$ where $\Delta R_i$ is the change in the rate $i$ and $f_j$ is factor $j$ and $e_{i,j}$ is the (factor loading=) ...


12

Unlike the tangency portfolio on the efficient frontier (which represents the most efficient portfolio in terms of max expected sharp ratio), min var portfolios have no ex ante theory that suggests it should outperform a cap weighted market portfolio. The same can be said about other risk-weighted portfolio construction schemes, including equal risk ...


9

The minimum variance optimization framework does not guarantee positive return whatsoever. As a matter of fact what you are trying to do is something close to the following: $$\underset{w}{\arg \min} \quad w' Q w \quad \text{s.t} \quad Aw \leq b,\quad \sum_i w_i=1$$ The fact that you get positive return is a nice result that you get from your backtest (i....


9

If $X \sim N(\mu, V)$ is multivariate gaussian, you can write $X = \mu + C Y$ where $ Y \sim N(0,1) $ is a standard Gaussian and $C$ is the lower-triangular Choleski matrix of $V$. You can then express $ v = \sum_{i=1}^n (X_i - S/n)^2 $, where $ S = \sum_{i=1}^n X_i $, in terms of $Y$ and $C$. (I do not reproduce the computations: they are straightforward.) ...


9

Have a look at this classic paper: Honey, I Shrunk the Sample Covariance Matrix by O. Ledoit and M. Wolf The abstract answers your question already: The central message of this article is that no one should use the sample covariance matrix for portfolio optimization. It is subject to estimation error of the kind most likely to perturb a mean-...


8

To clarify notation, you have an universe of $n=2000 \space$ stocks and two portfolio vectors $\mathbf{a},\mathbf{b}\in\mathbb{R}^{n}$ with $\left\|\mathbf{a}\right\|_{1}=\left\|\mathbf{b}\right\|_{1}=1$. Further, you have Estimators for the true Variance $\operatorname{Var}\left[\mathbf{a}\right]$ resp. $\operatorname{Var}\left[\mathbf{b}\right]$ and the ...


8

The estimation of a covariance matrix is unstable unless the number of historical observations $T$ is greater than the number of securities $N$ (5000 in your example). Consider that 10 years of data represents only 120 monthly observations and about 2500 daily observations. Depending on the application, using data dating farther back than 10 years may be ...


7

Transaction costs - even for banks, funds etc, every trade has an associated cost, so if you would be buying a small number of shares, it's probably cheaper to carry the risk and not make those small trades. The source data is imperfect, and contains noise. A lot of the smaller components are simply artefacts of that noise so it would be both an unnecessary ...


7

I personally use the simple Garch(1,1) for volatility filtering in the risk management area. In fact in most cases I don't even estimate the parameters, I stick 0.94 for mean reversion, 0.04 for the squared error and I get the constant by matching the series variance. My experience is that there is no point pretending to finetune parameters when vol is ...


7

If you really believed the CAPM's prediction that $\alpha=0$, then imposing $\alpha=0$ in your estimation would indeed lead to your 2nd formula. The problems? The CAPM doesn't work so imposing a false restriction during estimation is problematic. More generally, taking factor models extremely seriously and imposing $\alpha=0$ in estimation to gain ...


6

The standard estimator of the covariance matrix is: $$\widehat{ \mathrm{cov}}(X) = \frac 1 {n-1} \sum_{i=1}^n (X_i-\bar X)(X_i-\bar X)^T,$$ where $X_i$ is the column vector containing the $i$th observation of all the observables. Each summand is an outer product of a vector with itself, i.e., a square matrix having rank at most one. Therefore $$\mathrm{rk\;...


6

I think an extremely interesting strand of research on this topic is represented by extensions of vine copulas with time-varying parameters. For vine copulas in general have a look at this site from the Technische Universität München: Vine Copula Models One of their research projects, which is the most relevant in this context, is:Time varying vine copula ...


6

I am not sure if I understood your question correctly but I will try to answer it anyway. If you have a standard normal random vector $z \sim N(\mathbb{0},I_n)$ (where $z,0 \in \mathbb{R}^{n\times1}$ and $I_n \in \mathbb{R}^{n\times n}$ is the identity matrix) and you want to transform it into a multivariate normal $x \sim N(\mu,\Sigma)$ you do it the ...


6

Let $s$ be a $N\times1$ vector of standard deviations and $C$ be an $N\times N$ correlation matrix. The covariance matrix is equal to $$\Sigma=\text{diag}(s) \ C \ \text{diag}(s)$$ where $\text{diag}(x)$ is a function that takes an $N\times1$ vector and puts it on the diagonal of a $N\times N$ matrix. If you get some better standard deviation estimates, ...


5

Any explanations? Yes. Within each asset category we find that stocks may be: Unattractively underperforming the category norm Attractive as they meet the expected norm Unsustainable as their returns exceed the category norm and may suffer mean reversion By focusing on low variance, we exclude type (3) stocks that damage portfolio performance through high ...


5

Minimizing risk alone would not imply a positive expected return, except for the following: The assets that are being included have positive expected returns. If you took a portfolio of assets that had a negative expected return, and minimized their risks, you would probably still end up with a portfolio that has a negative expected return. Most of these ...


5

The given matrix can not represent a covariance matrix since it would imply that asset 1 is negatively correlated to asset 2 and asset 3. But asset 2 is negatively correlated to asset 3 which contradicts the first statement. In general a covariance matrix has to be positive semi-definite and symmetric, and conversely every positive semi-definite symmetric ...


5

Go ahead and compute a sample covariance matrix with 5,000 stocks on a few years (or less) of daily or monthly returns data. This can be done almost instantly on a modern computer. There is a very good chance that this matrix will not be a covariance matrix. You can check by inspecting the eigenvalues. If any are negative then you don't have a covariance ...


5

If $\Sigma$ is the covariance matrix of all assets and $w$ is the column vector of weightings of the asset in a certain portfolio. Then $$ w^T \Sigma w = VAR $$ is the variance of the portfolio. The contribution to volatility of asset $i$ is given by $$ w_i (\Sigma w)_i/\sqrt{VAR}, $$ where $(\Sigma w)_i$ is the $i_{th}$ entry in the vector $\Sigma w$. Note ...


5

You're not going to get an analytic formula except in special cases of function $\rho(x)$. And you're probably going to want $\rho$ convex. If $\rho$ is convex, the problem is a convex optimization problem and can be efficiently solved numerically. If $\rho$ isn't convex, the optimization problem may be difficult to solve. If $\rho(x) = |x|$ you basically ...


5

By: bilinearity of covariance, independence of Brownian increments, and Itô's isometry, we obtain: $$\begin{align} & \text{Cov}\left(\int^{t_1}_0\sigma(t)dW_t,\int^{t_2}_0\sigma(t)dW_t\right) \\[6pt] & \qquad = \text{Cov}\left(\int^{t_1\wedge t_2}_0\sigma(t)dW_t,\int^{t_1\wedge t_2}_0\sigma(t)dW_t +\int^{t_1\vee t_2}_{t_1\wedge t_2}\sigma(t)...


4

You can use the Exponentially Weighted Average directly aswell, finding the covariances and then normalizing back to the correlations: $ \sigma_{t+1,jk} = (1-\lambda) \sum_{n=0}^\infty \lambda^{n} r_{j,t-n} r_{k,t-n} $ (this assumes average returns 0 etc etc. More general versions can be derived)


4

Is the covariance of the raw return forecasts a good forecaster of the covariance of market returns? As you suggest, the covariance of the raw return forecasts is a lousy forecast of the covariance of market returns. Grinold & Kahn explain why quite eloquently in Active Portfolio Management, 2nd edition (pg. 275). It might be tempting to augment the ...


4

For the stationary multivariate normal case, the expected returns vector does not matter. This is because the cross-sectional mean is subtracted out before calculating the standard deviation. The cross-sectional mean can be more conveniently thought of as like the return on an equally weighted portfolio. Similarly, I would argue that the expected cross-...


4

The formula is $$ \mu = \lambda CX $$ in your notation. You find it in many places, e.g. here. The assumption is that you know $\lambda$ which is a strong assumption. Furthermore it only holds if investors are unconstrained (long/short not long only). It is intuitive as it says that given the weighting the return expectation increases with risk aversion ...


4

I think you're looking for multivariate GARCH models of which this is an overview paper. Multivariate GARCH models have one big drawback: they are pretty hard to estimate due to the number of correlations. This paper by Caporin and McAleer might be of interest in that regard.


4

$\sigma_p=\sqrt{\omega_a^2 \sigma_a^2+(1-\omega_a)^2 \sigma_b^2+2 \omega_a (1-\omega_a) \rho_{ab} \sigma_a \sigma_b}$ with $\rho_{ab}=-1$ the term under the square root simplifies to $(\omega_a \sigma_a-(1-\omega_a) \sigma_b)^2$ which is equivalent to $(-\omega_a \sigma_a+(1-\omega_a) \sigma_b)^2$ therefore $\sigma_p=\omega_a \sigma_a-(1-\omega_a) \...


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