the alpha and omega : topic:mathstat

E[e^sX] when X is normally distributed

Mahalanobis — 2008-12-27T21:58:00Z

Recently, I asked myself the question: "Why is the expected value of e^sX (X ~ N(0,1)) equal to e^(0.5*s^2)?". You know that compounding a sum of money (A) at a continously compounded rate X for a period involves multiplying it by e^X. In case X is non-random, the expected value is A*e^X. In case X is random, calculating the expected value becomes quite difficult. The solution for a standard normally distributed X can be derived as follows:

1. Write down the equation (μ = E[e^sX] where X ~ N(0,1)):
2. Substitute y = x - s and you get:More general, if X ~ N(μ*, σ²), then X = μ* + σZ with Z ~N(0,1). It follows thatOf course, practitioners often know the parameters they are interested in (e.g. s = 0.3), so they can take a shortcut and run a monte carlo analysis:

s <- 0.3
n <- 10000
x <- rnorm(n)
answer <- mean(exp(s*x))

Or a bit more sophisticated and less arbitrary:

s <- 0.3
n <- 10000
i <- (1:n)/(n+1)
x <- qnorm(i)
answer <- mean(exp(s*x))

NB: The expected value can be very misleading.
Source: Stochastische Grundlagen der Finanzmathematik, Klaus Pötzelberger

Volatility hurts.

Mahalanobis — 2008-12-22T08:07:00Z

Ernest Chan (blog) writes in his book Quantitative Trading - How to Build Your Own Algorithmic Trading Business:

Here is a little puzzle that may stymie many a professional trader. Suppose a certain stock exhibits a true (geometric) random walk, by which I mean there is a 50-50 chance that the stock is going up 1 percent or down 1 percent every [day]. If you buy this stock, are you most likely--in the long run and ignoring financing costs--to make money, lose money, or be flat?
Most traders will blur out the answer "Flat!," and that is wrong. The correct answer is that you will lose money, at the rate of 0.005 percent (or 0.5 basis points) every [day]! This is because for a geometric random walk, the average compounded rate of return is not the return μ, but is g = μ - σ^2/2.

For this very reason, geometric Brownian Motion is often written as
where "μ - σ^2/2" is the expected return and "μ" is the return of the expected prices, i.e. ln(E[S_t]/E[S_t-1]).

If you lose 50 percent of your portfolio, you have to make 100 percent to get back to even... that's what everybody knows. But it's also interesting to see how mild volatiltiy hurts over time. Here are ten (random - no cherry picking) realizations of a geometric Brownian Motion with a daily volatility of 1% (i.e. a yearly volatility of 16% when having 252 trading days) over the period of 100 years:

R Development Core Team (2008). R: A language and environment for statistical computing.

PS: Hey, this looks promising:

The Skill versus Luck Paradox cont'd

Mahalanobis — 2007-05-15T19:50:00Z

A while ago, we adressed the following question: A portfolio manager knows that his strategy can, on average, outperform the benchmark index by 3% annually. His portfolio has an annual volatility (standard deviation) of 25% against the index's 15%. Assuming that the correlation between the returns of the portfolio and the returns of the index is 0.9, how many years would it take to outperform the index with 90% probability?

The correct answer is a whopping 300 years! (apply the Itô-Döblin formula)

Today somebody asked me if I could run a couple of simulations to get a better understanding of the result. What I did was plot 20 simulations of log(Portfolio/Index) for varying correlations. For ρ = 0.9 2 out of 20 (10%) are--as expected--below zero:

My Master's Thesis is Significant

Mahalanobis — 2007-04-12T16:11:00Z

By strolling through the library I figured out that the number of pages of my master's thesis is five standard deviations lower than the average number of pages my friends have written to graduate from the Vienna University of Economics and Business Administration:

Author	Pages	Title
Michael Sigmund	139	Anwendungsgebiete der Spieltheorie in den Sozialwissenschaften
Robert Ferstl	127	Werkzeuge zur Analyse räumlicher Daten - eine Softwareimplementation in EViews und MATLAB
Karin Doppelbauer	122	Analiz rʹinka mjasa kur v Rossii - Eine Analyse des russischen Geflügelfleischmarktes
Markus Pock	107	Untersuchungen zu Wachstumseffekten der WWU mittels Zeitreihenanalyse
Christian Kraxner	105	Using credit derivatives for managing corporate bond portfolios
Christian Balbier	104	Föderale Strukturen in den neuen Mitgliedsstaaten der Europäischen Union
Stefan Woytech	103	Harmonisierung von internem und externem Rechnungswesen auf Basis der IAS/IFRS
Anton Burger	92	Reasons for the U.S. Growth Experience in the Nineties: Non Keynesian Effects, the Capital Market and Technology
Michael Stastny	31	Economic Growth and Output Variability: An Empirical Analysis (pdf)

The outlier status of my thesis is confirmed by conventional tests for outliers:

How cool is that?

"almost sure" versus "sure"

Mahalanobis — 2007-04-02T20:37:00Z

The difference between an event being almost sure and sure is the same as the subtle difference between something happening with probability 1 and happening always.

If an event is sure, then it will always happen. No other event (even events with probability 0) can possibly occur. If an event is almost sure, then there are other events that could happen, but they happen almost never, that is with probability 0.

Cool Example: Throwing a dart

For example, imagine throwing a dart at a square, and imagine that this square is the only thing in the universe. There is physically nowhere else for the dart to land. Then, the event that "the dart hits the square" is a sure event. No other alternative is imaginable.

Next, consider the event that "the dart hits the diagonal of the square exactly". The probability that the dart lands on any subregion of the square is equal to the area of that subregion. But, since the area of the diagonal of the square is zero, the probability that the dart lands exactly on the diagonal is zero. So, the dart will almost surely not land on the diagonal, or indeed any other given line or point. Notice that even though there is zero probability that it will happen, it is still possible.

Source: Wikipedia

The waiting time paradox

Mahalanobis — 2007-03-28T01:42:00Z

There is an old conundrum in queueing theory that goes like this:

A passenger arrives at a bus-stop at some arbitrary point in time
Buses arrive according to a Poisson process
The mean interval between the buses is 10 min.

What is the mean waiting time until the next bus?

Answer: 10 min. This is an example of length-biased sampling. The explanation of the paradox lies therein that the passengers' probability to arrive during a long interarrival interval is greater than during a short interval. ] Here is a neat non-technical explanation (taken from this book). [

Given the interarrival interval, within that interval the arrival instant of the passanger is uniformly distributed and the expected waiting time is one half of the total duration of the interval. The point is that in the selection by the random instant the long intervals are more frequently represented than the short ones (with a weight proportional to the length of the interval).

Consider a long period of time t. The waiting time to the next bus arrival W(τ) as a function of the arrival instant τ of the passenger is represented by:
where the X_i are the interarrival intervals. The mean waiting time, W_bar, is the average value of this sawtooth curve:
Note that long interarrival intervals contribute much more than short ones to the average waiting time. As t grows, t/n -> X_bar, hence,
For exponential distribution (as the X_i are distributed),
Therefore,
Altogether,

Q.E.D.

Sources:
Advanced Course in Operating Systems (University of Haifa), Lecture 1 & 2

ZunZun :: Curve fitting on the web

Mahalanobis — 2007-03-27T02:50:00Z

New Economist writes: In a new post on the Statistical Modeling, Causal Inference, and Social Science blog, Aleks Jakulin at Columbia University points us to a great online tool, ZunZun. It lets you use 2 and 3 dimensional 'Function Finders' to 'help determine the best curve fit for your data'."

On William Greene's site I found a neat data set (Data Tables :: Table F6.1) for estimating a Cobb-Douglas production function. ZunZun comes up with the following suggestion:

Y = β₁( L^0.5K^0.5) + β₂(cos(L)K^1.5)

Surface Plot:

Contour Plot:

The R² reaches an unrealistic 0.968 (0.94 for the Cobb-Douglas specification). Textbook data... Here is a scatterplot of the logarithmized data (created with R):

Proof without Words: Geometric Series

Mahalanobis — 2007-03-24T02:16:00Z

The Viewpoints 2000 Group
Mathematics Magazine, Vol. 74, No. 4. (Oct., 2001), p. 320.

related items:
Neat Proofs, Mahalanobis

I'm Confused: How Could Information Equal Entropy?

Mahalanobis — 2007-03-22T01:14:00Z

Taken from the bionet.info-theory FAQ: If someone says that information = uncertainty = entropy, then they are confused, or something was not stated that should have been. Those equalities lead to a contradiction, since entropy of a system increases as the system becomes more disordered. So information corresponds to disorder according to this confusion.

If you always take information to be a decrease in uncertainty at the receiver and you will get straightened out:

where H is the Shannon uncertainty:

and p_i is the probability of the ith symbol. If you don't understand this, read the short Information Theory Primer.

Imagine that we are in communication and that we have agreed on an alphabet. Before I send you a bunch of characters, you are uncertain (H_before) as to what I'm about to send. After you receive a character, your uncertainty goes down (to H_after). H_after is never zero because of noise in the communication system. Your decrease in uncertainty is the information (I) that you gain.

Since H_before and H_after are state functions, this makes I a function of state. It allows you to lose information (it's called forgetting).

Many of the statements in the early literature assumed a noiseless channel, so the uncertainty after receipt is zero (H_after=0). This leads to the SPECIAL CASE where I = H_before. But H_before is NOT "the uncertainty", it is the uncertainty of the receiver BEFORE RECEIVING THE MESSAGE.

Brainiacs Succeed in Mapping 248-Dimensional Object

Mahalanobis — 2007-03-20T01:35:00Z

MIT news office: An international team of 18 mathematicians has mapped one of the largest and most complicated structures in mathematics. If written out on paper, the calculation describing this structure, known as E₈, would cover an area the size of Manhattan.

The work is important because it could lead to new discoveries in mathematics, physics and other fields. In addition, the innovative large-scale computing that was key to the work likely spells the future for how longstanding math problems will be solved in the 21st century. Click here (or here ) to continue.

related items:
American Institute of Mathematics: E₈
Lecture Slides: The Character Table for E8, or How We Wrote Down a 453,060 x 453,060 Matrix and Found Happiness, David Vogan, MIT

MARS :: Multivariate Adaptive Regression Splines

Mahalanobis — 2006-10-30T15:15:00Z

"Multivariate Adaptive Regression Splines (MARS) is an implementation of techniques popularized by Friedman (1991) for solving regression-type problems.

MARS is a nonparametric regression procedure that makes no assumption about the underlying functional relationship between the dependent and [explanatory] variables. Instead, MARS constructs this relation from a set of coefficients and basis functions that are entirely "driven" from the regression data. In a sense, the method is based on the "divide and conquer" strategy, which partitions the input space into regions, each with its own regression equation. This makes MARS particularly suitable for problems with higher input dimensions (i.e., with more than 2 variables), where the curse of dimensionality [see also blessing of dimensionality] would likely create problems for other techniques.

The MARSplines technique has become particularly popular in the area of data mining because it does not assume or impose any particular type or class of relationship (e.g., linear, logistic, etc.) between the predictor variables and the dependent (outcome) variable of interest. Instead, useful models (i.e., models that yield accurate predictions) can be derived even in situations where the relationship between the predictors and the dependent variables is non-monotone and difficult to approximate with parametric models." [Continue]

Note to self: Read

Multivariate Adaptive Regression Splines, Jerome H. Friedman, The Annals of Statistics, Vol. 19, No. 1. (Mar., 1991), pp. 1-67.
The Elements of Statistical Learning: Data Mining, Inference and Prediction, Hastie, T., Tibshirani, R. and Friedman, J.H. (2001), Springer Verlag, New York

Hat tip to Diethelm Würtz

Taylor Effect and Empirical Scaling Law :: Rmetrics

Mahalanobis — 2006-10-21T21:19:00Z

Taylor Effect: It is by now well established in the financial econometrics literature that high frequency time series of financial returns are often uncorrelated but not independent because there are non-linear transformations which are positively correlated. In 1986 Taylor observed that the empirical sample autocorrelations of absolute returns, |r|, are usually larger than those of squared returns, |r|^2. A similar phenomena is observed by Ding et al. (1993) who examined daily returns of the S&P 500 index and conclude that, for this particular series, the autocorrelations of absolute returns raised to the power of θ are maximized when θ is around 1, that is, the largest autocorrelations are found in the absolute returns. Granger and Ding (1995) denote this empirical property of financial returns as Taylor Effect. Therefore, if r_t, t = 1,...T, is the series of returns and ρ_θ(k) denotes the sample autocorrelation of order k of |r_t|^θ, θ > 0, the Taylor effect can be defined as follows:

ρ₁(k) > ρ_θ(k) for any θ ≠ 1.

However, Granger and Ding (1994, 1996) analyze several series of daily exchange rates and individual stock prices, and conclude that the maximum autocorrelation is not always obtained when θ = 1 but for smaller values of θ. Nevertheless, they point out that the autocorrelations of absolute returns are always larger than the autocorrelations of squares. [1] This can also be observed when looking at USDCHF High Frequency FX rates (1996-04-01 00:00:00 to 2001-03-30 23:30:00; 62,496 observations):

teffectPlot, k = 1,...,10:

Scaling Law: Some financial time series show a selfsimilar behavior under temporal aggregation. The 'empirical scaling law' relates the average of the unconditional volatility, measured as the absolute value of the return, r(t_i), over a time interval to the size of the time interval:

where the drift exponent 1/E is an estimated constant that Müller et al. (1990) find to be similar across different currencies and ΔT is a time constant that depends on the currency [2]. The Wiener process, a continuous Gaussian random walk, exhibits a scaling law with a drift exponent of 0.5 (slope of green line). The estimated drift component for the USDCHF series is 0.52, which is actually not statistically different from 0.5:

For more information see Fractals and Intrinsic Time - A Challenge to Econometricians.

Here is the official Rmetrics site.

[1]: see Stochastic Volatility Models and the Taylor Effect, Alberto Mora-Galán and Ana Pérez and Esther Ruiz
[2] see The Impact of News on Foreign Exchange Rates: Evidence from High Frequency Data, Dirk Eddelbuettel and Thomas H. McCurdy

Neat Proofs

Mahalanobis — 2006-10-12T22:37:00Z

Addendum:

by Rick Mabry
Mathematics Magazine, Vol. 72, No. 1. (Feb., 1999), p. 63.

Garch-in-mean :: Recursive Estimates

Mahalanobis — 2006-09-17T22:16:00Z

Finance theory suggests that an asset with a higher perceived risk would pay a higher return on average [Caution]. For example, let r_t denote the ex post rate of return on some asset minus the return on a safe alternative asset. Suppose that r_t is decomposed into a component anticipated by investors at date t-1 (denoted μ_t) and a component that was unanticipated (denoted u_t):
Then the theory suggests that the mean return (μ_t) would be related to the variance of the return. The GARCH(1,1)-in-mean, or GARCH(1,1)-M, regression model is characterized by

for ε i.i.d. with zero mean and unit variance. The effect that higher perceived variability of u_t has on the level of r_t is captured by the parameter γ (see Hamilton's TSA Bible).

A realization (n = 500) of a Garch(1,1)-M process with κ = 0.005, γ = 2, ω = 0.0001, α = 0.1, β = 0.8, and ε ~ N(0,1) looks as follows:

Unfortunately, I got a GARCH-in-mean effect (γ) of -34 instead of +2 after I estimated the process given above (n=500) with R (R uses the Ox package with "garchOxFit" command to estimate GARCH models. See here for "garchOxFit" installation instructions for Windows OS.) Estimating the GARCH(1,1)-M coefficient for n=500, 1000, 2000, 3000, ..., 10000 yields the following result:

This would mean that you shouldn't even think of estimating such a model if you don't have at least 4000 observations. So far I haven't seen any applied work where people used more than 1000 observations. How cool is that? Suggestions welcomed.

Helmeted cyclists in more peril on the road ?

Mahalanobis — 2006-09-13T12:24:00Z

timesonline: Cyclists who wear helmets are more likely to be knocked off their bicycles than those who do not, according to research.

Motorists give helmeted cyclists less leeway than bare-headed riders because they assume that they are more proficient. They give a wider berth to those they think do not look like proper cyclists, including women, than to kitted-out lycra-clad warriors.

Ian Walker, a traffic psychologist, was hit by a bus and a truck while recording 2,500 overtaking manoeuvres. On both occasions he was wearing a helmet.

During his research he measured the exact distance of passing traffic using a computer and sensor fitted to his bicycle.Half the time Dr Walker, of the University of Bath, was bare-headed. For the other half he wore a helmet and has the bruises to prove it.

He even wore a wig on some of his trips to see if drivers gave him more room if they thought he was a woman. They did.

He was unsure whether the protection of a helmet justified the higher risk of having a collision. We know helmets are useful in low-speed falls, and so are definitely good for children.

On average, drivers steered an extra 3.3 in away from those without helmets to those wearing the safety hats. Motorists were twice as likely to pass very close to the cyclist if he was wearing a helmet.

For an excellent discussion, see here. Here is a nice picture of a bike.