# Tag Info

14

it doesn't require normality. What it requires is that the investor's decisions are determined by mean and variance. A normal distribution is determined by mean and variance, so if you assume joint normality then there is no point in the investor being interested in anything else. (we try to discuss assumptions thoroughly in our book, Introduction to ...

8

The Markowitz mean-variance model is the basis for many extensions and portfolio solutions that have been discovered over the years: The standard model (Markowitz, 1952, 1959) originally only considered: Constrained model where short sales are forbidden Only risky assets considered for investment (no risk-free asset) Scenarios that the mean-variance model ...

7

Two separate cases were identified by R.C. Merton in 1972: In the economically more relevant case, where $r_f < b/c$, efficient portfolios are combinations of a long position in [the tangency] portfolio M and lending or borrowing at the risk–free rate. In the case where $r_f > b/c$, efficient portfolios are generated by short (or zero) positions in ...

6

In the early days of Portfolio Theory there were different views about short positions. Some authors modeled short positions as negative and required all weights to add up to 1 (first equation), others (including Markowitz himself) thought this was not realistic (he thought if you have 1 dollar you cannot both buy 1 dollar worth of stock and also short 1 ...

5

The weak EMH states that it is impossible to earn an excess return given publicly known information such as past prices. Clearly, different securities have different expected returns. For example: the bond and the stock of one company or a security that generates twice the return of another one. This difference in expected return is explained by a ...

5

To solve this constraint minimization problem, first form the Lagrangian Function \begin{align} L(w,\lambda_1,\lambda_2)=w'\Sigma w + \lambda_1(w'\boldsymbol{\mu}-m) + \lambda_2 (w'\boldsymbol{1}-1). \end{align} The first order conditions for a minimum are then given by \begin{align} \frac{\delta L(w,\lambda_1,\lambda_2)}{\delta w}&=2 \Sigma w + \...

5

I prefer to interpret the mean-variance frontier as a consequence of linear algebra as developed in Hansen and Richard (1987) and discussed in Cochrane (2005). In brief: The space of returns is a hyperplane in the vector space of payoffs. The set of returns on the mean-variance frontier is a line in the space of returns. Any two distinct points on a line ...

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

The Efficient Market Hypothesis (EMH) states that you cannot beat the market on a risk-adjusted basis by looking at past prices. You can certainly earn higher returns than the market if you take on more risk (by leveraging, for example). Modern Portfolio Theory allows you to construct portfolios that are efficient. According to this theory, you still cannot ...

4

The problem lies in the definition of risk. It seems that in the cited paper, the authors treat risk as a concept connected with the uncertainty of the out-of-sample performance of the portfolio. In that way portfolios constructed using the proposed robust estimators would be what they call minimum-risk portfolios. Contrasted with minimum-variance ...

4

You can use the package quadprog and define everything yourself. Code can look like this: library(quadprog) Sigma = cov(data) mu = mean(data) Amat_in # define constraints here bvec_in # define rhs of constraints here solve.QP( Dmat = 2*Sigma, dvec = mu, meq=0,Amat=Amat_in,bvec=bvec_in) EDIT: Yes, and reading the documentation we see that portfolio.optim(...

4

In a linear regression approach you do the following: $$(X \beta - y)^2 \rightarrow Min$$ thus you try to predict something. Your objective is quadratic. You usually add constraints on $\sum \beta_i^2$ or $\sum |\beta_i|$. Without constraints the estimator is: $$\hat{\beta} = (X^T X)^{-1} X^T y,$$ where $X^T y$ has to do with the covariance of $X$ and $... 4 This is easier to explain with an example. Let's say there is no risk free rate as your question seems to imply. Also assume there are two portfolios$\phi_p$and$\phi_q$with the following characteristics: Both lie on the frontier; The first one has an expected return of 0.04 and standard deviation of 0.2 The second one has an expected return of 0.10 ... 4 You cannot eliminate the dependence of a solution on the risk aversion parameter (which this author confusingly calls$\lambda$). Perhaps a source of confusion? Typically$\lambda$is used to denote a Lagrange multiplier in Lagrangian optimization, but the author is using$\lambda$as a risk tolerance parameter. (In your other linked question,$\lambda$... 4 No, even if returns were perfectly normal (it really doesn't matter whether mean is zero and standard deviation is 1 - they can be anything), it wouldn't ensure that markowitz would perform well out of sample. The reason is because even if data is normally distributed it is hard to estimate means of returns. The standard error for an estimate of a mean like ... 3 It is surprising. What I think is: Markowitz became interested in the general problem when there are constraints (including inequality constraints) on the portfolio weights (in addition to the standard$\sum w_i = 1$constraint). Once he devised a computer algorithm [the Critical Line Method] for solving this problem (he was a math programming whiz) he seems ... 3 Risk is a broader concept than variance. That paper is specifically focused on robust estimators (i.e., estimators that are less sensitive to outliers) of dispersion. A robust estimator of dispersion is not the same thing as variance (which may be a dispersion parameter for some classes of distributions). Nevertheless, these robust estimators could be used ... 3 There are many papers on this subject (try googling portfolio optimization skewness kurtosis) that can describe the assumptions of including skewness and kurtosis in a utility function (if that's what you're interested in). I would highlight two main points. Mean-variance optimization does not make an assumption of normality. Assume returns are distributed ... 3 You should have a look at chapter 8 (p. 261ff.) of Hedge Fund Market Wizards by Jack D. Schwager Excerpt from there (but it is much more detailed in the book): Perhaps the most potent risk control Platt employs in BlueCrest’s discretionary strategy is maintaining an extremely tight rein on what a trader can lose before capital is withdrawn. A mere 3 ... 3 There is a recent a paper recently using a population test of all CRSP data from 1925-2013 as a test of whether a mean and a variance exist versus they do not exist. It overwhelmingly excluded mean-variance finance as not possible. It is also a population study so for mean-variance to be valid, there would have to be radically different behavior before and ... 3 Broadly speaking, as you probably already know, there are 2 approaches to estimating large covariance matrices: 1) Shrinkage Methods like Ledoit-Wolf that try to reduce the noise in a large matrix (N by N) that has been estimated using the conventional method. 2) Factor Models of Covariance as described in for example Connor Korajczik 2007 that assume that ... 3 Let's derive a possible approach from utility theory. Our investor is risk averse and exhibits CARA utility using an exponential utility function with risk aversion parameter$\gamma>0$(risk averse agent): $$u(x)=\frac{1-e^{-\gamma x}}{\gamma}$$ A 3rd order Taylor series expansion around$x=0yields \begin{align} u(x)\approx& x - \frac{1}{2}\gamma ... 2 Answering "No" to the title question, I'll mention that variance is a rather poor measure of risk, even if convinient and nicely behaving. Variance is not even a risk measure, with the standard deviation eventully being a deviation risk measure, while not necessarily for downside risk (see David Nawrocki-"A Brief History of Downside Risk Measures" for ... 2 EMH says that one can not earn excess return using some information. This is known as joint-hypothesis problem: to test for market efficiency one have to determine first what is "normal" market return, i.e. what type of information is normally priced by the market. Usually to test for EMH they use CAPM or 3-factor Fama-French model (which is a kind of CAPM-... 2 Lets_1 = r_1 -r_f$and$s_2 =r_2-r_f. Then, this is the maximization problem: \begin{align*} & \ \max_{w_1, w_2} SR = \frac{\mu_p}{\sigma_p}, \, \mbox{ subject to}\\ \mu_p = & \ w_1 s_1 + w_2 s_2,\\ \sigma_p^2 = & \ \sigma^2\big(w_1^2 + w_2^2 + 2 w_1 w_2 \rho\big),\\ 1 = & \ w_1+w_2. \end{align*} By certain substitution, we convert the ... 2 There is no generic solution. However, the KKT conditions are of the forms \begin{align*} \begin{cases} Qy + \lambda_1 \mu +\lambda_2 Py = 0,\\ \mu^T y = 1,\\ y^TPy \leq k^2 \sigma^2,\\ \lambda_2 \big( y^TPy - k^2 \sigma^2\big) = 0. \end{cases} \end{align*} Here, the condition $$\lambda_2 \big( y^TPy - k^2 \sigma^2\big) = 0$$ means that two cases need to ... 2 That's the way you apply. Usually you get the closest number of shares possible. However, if you use that strategy you are very likely to underperform the market. Check table 3 on this paper for the Out of sample performance of the Markowitz strategy. Over their sample the Sharpe Ratio is 0.07 whereas a simple naive strategy 1/N yielded 0.18. 2 First of all I’ll work with column vectors because I find it easier than with row vectors as you did. I guess it’s a little bit easier if we modify your first equation a little bit. Notice that is really the first order condition of the following lagrangian:$L(w, \lambda, \delta)= \frac{1}{2}{\bf w^TCw} - \lambda({\bf w^Tm} - \mu_v) - \delta({\bf w^Tu} - 1)... 2 They are the same. The maximum growth rate is achieved when the Sharpe ratio is maximized. For the proof, see here. 2 Out-of-sample is basically impossible to predict means. Second moments are much easier. You can take a look at this post: Estimating$\mu$- only increasing$T$improves estimate? Only with infinite$T$you would be able to correctly estimate$\mu\$. So theoretically your procedure could be correct if means are time-varying, but out of sample I bet your ...

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