Detecting BS in Correlation Windows

 

Figuring out the sampling error of rolling correlation.

 

Financial theory requires correlation to be constant (or, at least, known and nonrandom). Nonrandom means predictable with waning sampling error over the period concerned. Ellipticality is a condition more necessary than thin tails, recall my Twitter fight with that non-probabilist Clifford Asness where I questioned not just his empirical claims and his real-life record, but his own theoretical rigor and the use by that idiot Antti Ilmanen of cartoon models to prove a point about tail hedging. Their entire business reposes on that ghost model of correlation-diversification from modern portfolio theory. The fight was interesting sociologically, but not technically. What is interesting technically is the thingy below.

How do we extract sampling error of a rolling correlation? My coauthor and I could not find it in the literature so we derive the test statistics. The result: it has less than \(10^{-17}\) odds of being sampling error.

The derivations are as follows:

Let \(X\) and \(Y\) be \(n\) independent Gaussian variables centered to a mean \(0\). Let \(\rho_n(.)\) be the operator.

\(\rho_n(\tau)= \frac{X_\tau Y_\tau+X_{\tau+1} Y_{\tau+1}\ldots +X_{\tau+n-1} Y_{\tau+n-1}}{\sqrt{(X_{\tau}^2+X_{\tau+1}^2\ldots +X_{\tau+n-1}^2)(Y_\tau^2+Y_{\tau+1}^2\ldots +Y_{\tau+n-1}^2)}}.\)\

 

 

First, we consider the distribution of the Pearson correlation for \(n\) observations of pairs assuming \(\mathbb{E}(\rho) \approx 0\) (the mean is of no relevance as we are focusing on the second moment):


\(f_n(\rho)=\frac{\left(1-\rho^2\right)^{\frac{n-4}{2}}}{B\left(\frac{1}{2},\frac{n-2}{2}\right)},\)


with characteristic function:

\(\chi_n(\omega)=2^{\frac{n-1}{2}-1} \omega ^{\frac{3-n}{2}} \Gamma \left(\frac{n}{2}-\frac{1}{2}\right) J_{\frac{n-3}{2}}(\omega ),\)

where \(J_{(.)}(.)\) is the Bessel J function.

We can assert that, for \(n\) sufficiently large: \(2^{\frac{n-1}{2}-1} \omega ^{\frac{3-n}{2}} \Gamma \left(\frac{n}{2}-\frac{1}{2}\right) J_{\frac{n-3}{2}}(\omega ) \approx e^{-\frac{\omega ^2}{2 (n-1)}},\) the corresponding characteristic function of the Gaussian.

Moments of order \(p\) become:


\(M(p)= \frac{\left( (-1)^p+1\right) \Gamma \left(\frac{n}{2}-1\right) \Gamma \left(\frac{p+1}{2}\right)}{2 \left(\frac{1}{2},\frac{n-2}{2}\right) \Gamma \left(\frac{1}{2} (n+p-1)\right)}\)

where \(B(.,.)\) is the Beta function. The standard deviation is \(\sigma_n=\sqrt{\frac{1}{n-1}}\) and the kurtosis \(\kappa_n=3-\frac{6}{n+1}\).

This allows us to treat the distribution of \(\rho\) as Gaussian, and given infinite divisibility, derive the variation of the component, sagain of \(O(\frac{1}{n^2})\) (hence simplify by using the second moment in place of the variance):.

\(\rho_n\sim \mathcal{N}\left(0,\sqrt{\frac{1}{n-1}}\right).\)

To test how the second moment of the sample coefficient compares to that of a random series, and thanks to the assumption of a mean of \(0\), define the squares for nonoverlapping correlations:

\(\Delta_{n,m}= \frac{1}{m} \sum_{i=1}^{\lfloor m/n\rfloor} \rho_n^2(i; n),\)

where \(m\) is the sample size and \(n\) is the correlation window. Now we can show that:

\(\Delta_{n,m}\sim \mathcal{G}\left(\frac{p}{2},\frac{2}{(n-1) p}\right),\)
where \(p=\lfloor m/n\rfloor\) and \(\mathcal{G}\) is the Gamma distribution with PDF:
\(f(\Delta)= \frac{2^{-\frac{p}{2}} \left(\frac{1}{(n-1) p}\right)^{-\frac{p}{2}} \Delta ^{\frac{p}{2}-1} e^{-\frac{1}{2} \Delta (n-1) p}}{\Gamma \left(\frac{p}{2}\right)},\)
and survival function:

\(S(\Delta)=Q\left(\frac{p}{2},\frac{1}{2} \Delta (n-1) p\right),\)
which allows us to obtain p-values below, using \(m=714\) observations (and using the leading order $O(.)$:


 

Such low p-values exclude any controversy as to their effectiveness cite{taleb2016meta}.

We can also compare rolling correlations using a Monte Carlo for the null with practically the same results (given the exceedingly low p-values). We simulate \(\Delta_{n,m}^o\) with overlapping observations:
\(\Delta_{n,m}^o= \frac{1}{m} \sum_{i=1}^{m-n-1} \rho_n^2(i),\)

Rolling windows have the same second moment, but a mildly more compressed distribution since the observations of \(\rho\) over overlapping windows of length \(n\) are autocorrelated (with, we note, an autocorrelation between two observations \(i\) orders apart of \(\approx 1-\frac{1}{n-i}\)). As shown in the figure below for \(n=36\) we get exceedingly low p-values of order \(10^{-17}\).

A nice Limit

From @infseriesbot, prove the identity: \(\gamma_n = \displaystyle\int_1^\infty \frac{\{x\}}{x^2} (n-\log(x)) \log^{n-1} \, dx \).

We have \(\{x\}=x-i \, \text{for} \, i \leq x \leq i+1\),

so

\(\text{rhs}=\displaystyle\sum _{i=1}^{\infty } \displaystyle\int_i^{i+1} \frac{(x-i) (n-\log (x)) \log ^{n-1}(x)}{x^2} \, dx,\)

and since \(\displaystyle\int \frac{(x-i) (n-x \log ) \left(x \log ^{n-1}\right)}{x^2} \, dx=\log ^n(x) \left(-\frac{i}{x}-\frac{\log (x)}{n+1}+1\right)\)

\(\displaystyle \int_i^{i+1} \frac{(x-i) (n-\log (x)) \log ^{n-1}(x)}{x^2} \, dx=\frac{(i+1) \log ^{n+1}(i)+(-(i+1) \log (i+1)+n+1) \log ^n(i+1)}{(i+1) (n+1)}\),

allora:

\(\text{rhs}=\underset{T\to \infty }{\text{lim}}\left(\frac{(n-(T+1) \log (T+1)+1) \log ^n(T+1)}{(n+1) (T+1)}+\sum _{i=2}^T \frac{\log ^n(i)}{i}\right)\),

and since \(\frac{\log ^n(T+1) (n-(T+1) \log (T+1)+1)}{(n+1) (T+1)}\overset{T}{\to}\frac{\log ^{n+1}(T)}{n+1}\),

From the series representation of the Stieltjes Gamma function, \(\gamma_n\):

\(\underset{T\to \infty }{\text{lim}}\left(\sum _{i=1}^T \frac{\log ^n i}{i}-\frac{\log ^{n+1}(T)}{n+1}\right)=\gamma _n\)

Maximum Ignorance Probability, with application to surgery’s error rates

Introduction and Result

A maximum entropy alternative to Bayesian methods for the estimation of independent Bernouilli sums.

Let \(X=\{x_1,x_2,\ldots, x_n\}\), where \(x_i \in \{0,1\}\) be a vector representing an n sample of independent Bernouilli distributed random variables \(\sim \mathcal{B}(p)\). We are interested in the estimation of the probability p.

We propose that the probablity that provides the best statistical overview, \(p_m\) (by reflecting the maximum ignorance point) is

\(p_m= 1-I_{\frac{1}{2}}^{-1}(n-m, m+1)\), (1)

where \(m=\sum_i^n x_i \) and \(I_.(.,.)\) is the beta regularized function.

Comparison to Alternative Methods

EMPIRICAL: The sample frequency corresponding to the “empirical” distribution \(p_s=\mathbb{E}(\frac{1}{n} \sum_i^n x_i)\), which clearly does not provide information for small samples.

BAYESIAN: The standard Bayesian approach is to start with, for prior, the parametrized Beta Distribution \(p \sim Beta(\alpha,\beta)\), which is not trivial: one is contrained by the fact that matching the mean and variance of the Beta distribution constrains the shape of the prior. Then it becomes convenient that the Beta, being a conjugate prior, updates into the same distribution with new parameters. Allora, with n samples and m realizations:

\(p_b \sim Beta(\alpha+m, \beta+n-m)\) (2)

with mean \(p_b = \frac{\alpha +m}{\alpha +\beta +n}\). We will see below how a low variance beta has too much impact on the result.

Derivations

Let \(F_p(x)\) be the CDF of the binomial \( \mathcal{B} in(n,p)\). We are interested in \(\{ p: F_p(x)=q \}\) the maximum entropy probability. First let us figure out the target value q.

To get the maximum entropy probability, we need to maximize \(H_q=-\left(\;q \; log(q) +(1-q)\; log (1-q)\right)\). This is a very standard result: taking the first derivative w.r. to q, \(\log (q)+\log (1-q)=0, 0\leq q\leq 1\) and since \(H_q\) is concave to q, we get \(q =\frac{1}{2}\).

Now we must find p by inverting the CDF. Allora for the general case,

\(p= 1-I_{\frac{1}{2}}^{-1}(n-x,x+1)\).

And note that as in the graph below (thanks to comments below by überstatistician Andrew Gelman), we can have a “confidence band” (sort of) with

\(p_\alpha= 1-I_{\alpha}^{-1}(n-x,x+1)\) ;

in the graph below the band is for values of: \(\alpha= \frac{1}{2}, \frac{3}{4}\).

Application: What can we say about a specific doctor or center’s error rate based on n observations?

Case (Real World): A thoraxic surgeon who does mostly cardiac and lung transplants (in addition to emergency bypass and aortic ruptures) operates in a business with around 5% perioperative mortality. So far in his new position in the U.S. he has done 60 surgeries with 0 mortality.

What can we reasonable say, statistically, about his error probability?

Note that there may be selection bias in his unit, which is no problem for our analysis: the probability we get is conditional on being selected to be operated on by that specific doctor in that specific unit.

Assuming independence, we are concerned with \(Y = 0, 1, \ldots,n\) a binomially distributed r.v. \(\sim \mathcal{B}(n,p)\) where n is the number of trials and \(p\) is the probability of failure per trial. Clearly, we have no idea what p and need to produce our best estimate conditional on, here, \(y=0\).

Here applying (1) with \(m=0\) and \(n=60\), we have \(p=0.01148\).

Why is this preferable to a Bayesian approach when, say, n is moderately large?

A Bayesian would start with a prior expectation of, say .05, and update based on information. But it is highly arbitrary. Since the mean is \(\frac{\alpha}{\alpha +\beta}\), we can eliminate one parameter. Let us say we start with \(Beta(\alpha, 19 \alpha )\) and have no idea of the variance. As we can see in the graph below there are a lot of shapes to the possible distribution: it becomes all in the parametrization.

APPENDIX: JAYNES’ BRANDEIS PROBLEM

When I worked on this problem, and posted the initial derivations, I wasn’t aware of Jaynes’ “Brandeis Problem”. It is not the same as mine as it ignores n and it led to a dead-end because the multinomial is unwieldy. But his approach would have easily let to more work if we had computational abilities then (maximization, as one can see below, can be seamless).

Acknowledgments

Thanks to Saar Wilf for useful discussions.

The Lognormal at high variance: a probability paradox

You have zero probability of making money. But it is a great trade.

One-tailed distributions entangle scale and skewness. When you increase the scale, their asymmetry pushes the mass to the right rather than bulge it in the middle. They also illustrate the difference between probability and expectation as well as the difference between various modes of convergence.

Consider a lognormal \(\mathcal{LN}\) with the following parametrization, \(\mathcal{LN}\left[\mu t-\frac{\sigma ^2 t}{2},\sigma \sqrt{t}\right]\) corresponding to the CDF \(F(K)=\frac{1}{2} \text{erfc}\left(\frac{-\log (K)+\mu t-\frac{\sigma ^2 t}{2}}{\sqrt{2} \sigma \sqrt{t}}\right) \).

The mean \(m= e^{\mu t}\), does not include the parameter \(\sigma\) thanks to the \(-\frac{\sigma ^2}{2} t\) adjustment in the first parameter. But the standard deviation does, as \(STD=e^{\mu t} \sqrt{e^{\sigma ^2 t}-1}\).

When \(\sigma\) goes to \(\infty\), the probability of exceeding any positive \(K\) goes to 0 while the expectation remains invariant. It is because it masses like a Dirac stick at \(0\) with an infinitesimal mass at infinity which gives it a constant expectation. For the lognormal belongs to the log-location-scale family.

\(\underset{\sigma \to \infty }{\text{lim}} F(K)= 1\)

Option traders experience an even worse paradox, see my Dynamic Hedging. As the volatility increases, the delta of the call goes to 1 while the probability of exceeding the strike, any strike, goes to \(0\).

More generally, a \(\mathcal{LN}[a,b]\) has for mean, STD, and CDF \(e^{a+\frac{b^2}{2}},\sqrt{e^{b^2}-1} e^{a+\frac{b^2}{2}},\frac{1}{2} \text{erfc}\left(\frac{a-\log (K)}{\sqrt{2} b}\right)\) respectively. We can find a parametrization producing weird behavior in time as \(t \to \infty\).

Thanks: Micah Warren who presented a similar paradox on Twitter.

A Peculiar Integral

Prove

\(I= \displaystyle\int_{-\infty }^{\infty}\sum_{n=0}^{\infty } \frac{\left(-x^2\right)^n }{n!^{2 s}}\; \mathrm{d}x= \pi^{1-s}\).

We can start as follows, by transforming it into a generalized hypergeometric function:

\(I=\displaystyle\int_{-\infty }^{\infty }\, _0F_{2 s-1} (\overbrace{1,1,1,…,1}^{2 s-1 \text{times}}; -x^2)\mathrm{d}x\), since, from the series expansion of the generalized hypergeometric function, \(\, _pF_q\left(a_1,a_p;b_1,b_q;z\right)=\sum_{k=0}^{\infty } \frac{\prod_{j=1}^p \left(a_j\right)_k z^k}{\prod_{j=1}^q k! \left(b_j\right)_k}\), where \((.)_k\) is the Pochhammer symbol \((a)_k=\frac{\Gamma (a+k)}{\Gamma (a)}\).

Now the integrand function does not appear to be convergent numerically, except for \(s= \frac{1}{2}\) where it becomes the Gaussian integral, and the case of \(s=1\) where it becomes a Bessel function. For \(s=\frac{3}{2}\) and \( x=10^{19}\), the integrand takes values of \(10^{1015852872356}\) (serious). Beyond that the computer starts to produce smoke. Yet it eventually converges as there is a closed form solution. It is like saying that it works in theory but not in practice!

For, it turns out, under the restriction that \(2 s\in \mathbb{Z}_{>\, 0}\), we can use the following result:

\(\int_0^{\infty } t^{\alpha -1} _pF_q \left(a_1,\ldots ,a_p;b_1,\ldots ,b_q;-t\right) \, dt=\frac{\Gamma (\alpha ) \prod {k=1}^p \Gamma \left(a_k-\alpha \right)}{\left(\prod {k=1}^p \Gamma \left(a_k\right)\right) \prod {k=1}^q \Gamma \left(b_k-\alpha \right)}\)

Allora, we can substitute \(x=\sqrt(u)\), and with \(\alpha =\frac{1}{2},p=0,b_k=1,q=2 s-1\), given that \(\Gamma(\frac{1}{2})=\sqrt(\pi)\),

\(I=\frac{\sqrt{\pi }}{\prod _{k=1}^{2 s-1} \sqrt{\pi }}=\pi ^{1-s}\).

So either the integrand eventually converges, or I am doing something wrong, or both. Perhaps neither.





On the Multivariate Lognormal with Application to Blood Pressure

Background: We’ve discussed blood pressure recently with the error of mistaking the average ratio of systolic over diastolic for the ratio of the average of systolic over diastolic. I thought that a natural distribution would be the gamma and cousins, but, using the Framingham data, it turns out that the lognormal works better. For one-tailed distribution, we do not have a lot of choise in handling higher dimensional vectors. There is some literature on the multivariate gamma but it is neither computationally convenient nor a particular good fit.

Well, it turns out that the Lognormal has some powerful properties. I’ve shown in a paper (now a chapter in The Statistical Consequences of Fat Tails) that, under some parametrization (high variance), it can be nearly as “fat-tailed” as the Cauchy. And, under low variance, it can be as tame as the Gaussian. These academic disputes on whether the data is lognormally or power law distributed are totally useless. Here we realize that by using the method of dual distribution, explained below, we can handle matrices rather easily. Simply, if \(Y_1, Y_2, \ldots Y_n\) are jointly lognormally distributed with a covariance matrix \(\Sigma_L\), then \(\log(Y_1), \log(Y_2), \ldots \log(Y_n)\) are normally distributed with a matrix \(\Sigma_N\). As to the transformation \(\Sigma_L \to \Sigma_N\), we will see the operation below.

Let \(X_1=x_{1,1},\ldots,x_{1,n}, X_2= x_{2,1},\dots x_{2,n}\) be joint distributed lognormal variables with means \(\left(e^{\mu _1+\frac{\sigma _1^2}{2}}, e^{\mu _2+\frac{\sigma _2^2}{2}}\right)\) and a covariance matrix

\(\Sigma_L=\left(\begin{array}{cc}\left(e^{\sigma _1^2}-1\right) e^{2 \mu _1+\sigma _1^2}&e^{\mu _1+\mu _2+\frac{\sigma _1^2}{2}+\frac{\sigma _2^2}{2}}\left(e^{\rho \sigma _1 \sigma _2}-1\right)\\ e^{\mu _1+\mu _2+\frac{\sigma _1^2}{2}+\frac{\sigma _2^2}{2}}\left(e^{\rho \sigma _1 \sigma _2}-1\right)&\left(e^{\sigma _2^2}-1\right) e^{2 \mu _2+\sigma _2^2}\end{array}\right)\)

allora \(\log(x_{1,1}),\ldots, \log(x_{1,n}), \log(x_{2,1}),\dots \log(x_{2,n})\) follow a normal distribution with means \((\mu_1, \mu_2)\) and covariance matrix

\(\Sigma_N=\left(\begin{array}{cc}\sigma _1^2&\rho \sigma _1 \sigma _2 \\ \rho \sigma _1 \sigma _2&\sigma _2^2 \\ \end{array}\right)\)

So we can fit one from the other. The pdf for the joint distribution for the lognormal variables becomes:

Bivariate Lognormal Distribution

\(f(x_1,x_2)= \frac{\exp \left(\frac{-2 \rho \sigma _2 \sigma _1 \left(\log \left(x_1\right)-\mu _1\right) \left(\log \left(x_2\right)-\mu _2\right)+\sigma _1^2 \left(\log \left(x_2\right)-\mu _2\right){}^2+\sigma _2^2 \left(\log \left(x_1\right)-\mu _1\right){}^2}{2 \left(\rho ^2-1\right) \sigma _1^2 \sigma _2^2}\right)}{2 \pi x_1 x_2 \sqrt{-\left(\left(\rho ^2-1\right) \sigma _1^2 \sigma _2^2\right)}}\)

Kernel Distribution

We have the data from the Framingham database for, using \(X_1\) for the systolic and \(X_2\) for the diastolic, with \(n=4040, Y_1= \log(X_1), Y_2=\log(X_2): {\mu_1=4.87,\mu_2=4.40, \sigma_1=0.1575, \sigma_2=0.141, \rho= 0.7814}\), which maps to: \({m_1=132.35, m_2= 82.89, s_1= 22.03, s_2=11.91}\).

Some (mis)Understanding of life Expectancy, with some good news.

The Lancet article: Maron, Barry J., and Paul D. Thompson. “Longevity in elite athletes: the first 4-min milers.” The Lancet 392, no. 10151 (2018): 913 contains an eggregious probabilistic mistake in handling “expectancy” a severely misunderstood –albeit basic– mathematical operator. It is the same mistake you read in the journalistic “evidence based” literature about ancient people having short lives (discussed in Fooled by Randomness), that they had a life expectancy (LE) of 40 years in the past and that we moderns are so much better thanks to cholesterol lowering pills. Something elementary: unconditional life expectancy at birth includes all people who are born. If only half die at birth, and the rest live 80 years, LE will be ~40 years. Now recompute with the assumption that 75% of children did not make it to their first decade and you will see that life expectancy is a statement of, mostly, child mortality. It is front-loaded. As child mortality has decreased in the last few decades, it is less front-loaded but it is cohort-significant.

The article (see the Table below) compares the life expectancy of athletes in a healthy cohort of healthy adults to the LE at birth of the country of origin. Their aim was to debunk the theory that while exercise is good, there is a nonlinear dose-response and extreme exercise backfires.

Something even more elementary missed in the Lancet article. If you are a nonsmoker, healthy enough to run a mile (at any speed), do not engage in mafia activities, do not do drugs, do not have metabolic syndrome, do not do amateur aviation, do not ride a motorcycle, do not engage in pro-Trump rioting on Capitol Hill, etc., then unconditional LE has nothing to do with you. Practically nothing.

Just consider that 17% of males today smoke (and perhaps twice as much at the time of the events in the “Date” column of the table). Smoking reduces your life expectancy by about 10 years. Also consider that a quarter or so of Americans over 18 and more than half of those over 50 have metabolic syndrome (depending on how it is defined).

Lindy and NonLindy

Now some math. What is the behavior of life expectancy over time?

Let \(X\) be a random variable that lives in \((0,\infty)\) and \(\mathbb{E}\) the expectation operator under “real world” (physical) distribution. By classical results, see the exact exposition in The Statistical Consequences of Fat Tails:

\(\lim_{K \to \infty} \frac{1}{K} \mathbb{E}(X|_{X>K})= \lambda\)


If \(\lambda=1\) , \(X\) is said to be in the thin tailed class \(\mathcal{D}_1\) and has a characteristic scale . It means life expectancy decreases with age, owing to senescence, or, more rigorously, an increase of the force of mortality/hazard rate over time.

If \(\lambda>1\) , \(X\) is said to be in the fat tailed regular variation class \(\mathcal{D}2\) and has no characteristic scale. This is the Lindy effect where life expectancy increases with age.

If \(\lim_{K \to \infty} \mathbb{E}(X|_{X>K})-K= \lambda\) where \(\lambda >0\), then \(X\) is in the borderline exponential class.

The next conversation will be about the controversy as to whether human centenarians, after aging is done, enter the third class, just like crocodiles observed in the wild, where LE is a flat number (but short) regardless of age. It may be around 2 years whether one is 100 or 120.