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=) ...


11

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

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

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

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 ...


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, ...


6

What does 'simulate a covariance matrix' mean? If the question means, generate an arbitrary correlation matrix for 1000 stocks, then we can choose any symmetric matrix with all 1s down the diagonal, so long as every element is between -1 and 1 and the matrix is positive semi-definite. The large size of the matrix means that putting random values in every ...


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

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

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)...


5

If you assume that a financial asset price has a change that is a wiener process then you can view the future value of that asset as the initial value plus the sum of the independent daily changes (for equity or returns based then you would need log version of this): $$ S_t = S_0 + \sum \Delta S_i $$ where $\Delta S_i = S_i - S_{i-1} $ is a wiener process. ...


4

Mutual information measures how much knowing one variable reduces uncertainty about another variable. It considers any type of dependency (linear or non-linear), it's measured in bits, and it is widely used in machine learning, computer vision NLP and other fields.


4

The Newey-West procedure is meant to adjust the covariance matrix of the parameters to account for autocorrelation and heteroskedasticity. It is typically used in financial applications when one estimates the alpha (a parameter in a regression model) of a portfolio or strategy. One would adjust the standard errors using the Newey-West procedure in order to ...


4

The clearest and most intuitive article I have seen so far is Kritzman et al., Regime Shifts: Implications for Dynamic Strategies in FAJ (May / June 2012) It not only shows how you can use HMM for financial modelling but it also goes through the actual estimation algorithm (Baum-Welch) step-by-step and even gives full Matlab-code. From the abstract: ...


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) \...


4

The solution provided can be derived using the CAPM. For asset $A$ you have: $$R_A-R_f = \alpha_A +\beta_A(R_M-R_f)+\epsilon_A$$ Similarly for asset B: $$R_B-R_f = \alpha_B +\beta_B(R_M-R_f)+\epsilon_B$$ Calculate the covariance: $$\text{Cov}(R_A, R_B) = \text{Cov}(\beta_AR_M, \beta_BR_M)$$ Here I have dispensed with all the constant terms, and also ...


4

Multivariate volatility models for replacing the sample covariance matrix with in the mean-variance portfolio selection model: RiskMetrics 1996 EWMA (Exponentially weighted moving average) covariance matrix RiskMetrics 2006 EWMA covariance matrix Multivariate DCC-GARCH covariance matrix Jon Danielsson "Financial risk forecasting" has EWMA and GARCH for R ...


4

Just to expand on Alex answer. Empirically it is simply not true. Focusing on the diagonal of the variance-covariance matrix, we know that there is a large variance risk premium. Take a look at table 3 from Carr and Wu (2009). Regarding covariances we do not have much evidence, because there are no options on every single pair of stocks. However, we do know ...


3

You can obtain the covariance between 2 portfolios by multiplying the row vector, containing the weights of portfolio A with the variance-covariance matrix of the assets and then multiplying with the column vector, containing the weights of assets in portfolio B. Equally you can set up a new portfolio A+B by creating a new column vector that contains the ...


3

I thought I would answer the question of "what am I using." All shrinkage estimators map to a Bayesian estimator that differs only in the prior distributions. In other words, you get a point estimate that is indistinguishable from a Bayesian estimate except that the calculation rule determines the prior distribution. Stein estimators for the Gaussian are ...


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