1,591 reputation
615
bio website suitdummy.blogspot.com
location Cambridge, MA
age
visits member for 2 years, 4 months
seen Jul 8 at 19:04

I once launched swi-prolog and asked it a question:

ely:~/home$ prolog
Welcome to SWI-Prolog (Multi-threaded, 64 bits, Version 5.10.1)
Copyright (c) 1990-2010 University of Amsterdam, VU Amsterdam
SWI-Prolog comes with ABSOLUTELY NO WARRANTY. This is free software,
and you are welcome to redistribute it under certain conditions.
Please visit http://www.swi-prolog.org for details.

For help, use ?- help(Topic). or ?- apropos(Word).

?- love(math) is unrequited.
true.

Mar
26
comment An example of a “pathological” power-spectral density function?
This makes the assumption that it is Riemann integrable, but there are surely many pathological functions that should be considered here which aren't. Things like $x^{2}\sin(1/x^{2})$ and the sort. Many of these satisfy the properties of a PSD, but I haven't been able to find one yet where raising it to higher powers makes it less integrable. The fact that we can assume the function has no singularities makes it seem like you are correct, but I can't quite convince myself that it's airtight.
Mar
26
awarded  Scholar
Mar
26
comment Computing the derivative of a quadratic form and matrix chain rule
Thank you for the clarifying remarks. After I had gone back and did some debugging, I found a different reason why the code wasn't converging. In the end, with the derivatives I had calculated, I did get it to converge.
Mar
26
accepted Computing the derivative of a quadratic form and matrix chain rule
Mar
26
comment An example of a “pathological” power-spectral density function?
As an aside, I used to work as a radar engineer, and once an analyst kept grumbling about why his radar simulation was giving him nonsense $\infty$ values. I looked at his code and it was because he was using a non-integrable function for the antenna gain pattern (hence infinite energy radar). When I tried to explain this, he got angry that I was using the word "integrable" (a concept he apparently thought he could forget about after college).
Mar
26
comment An example of a “pathological” power-spectral density function?
Consider the function $\frac{1}{\sqrt{|f|}}$. The singularity is just barely integrable, but if you square it, then it's no longer integrable. Really, you're looking for function that are in $L^{k}(\mathbb{R})$ for some $k<n$ but not for $L^{n}(\mathbb{R})$. I'm sure some of the basic examples from functional analysis textbooks would satisfy what you're looking for.
Mar
26
comment An example of a “pathological” power-spectral density function?
I'm confused, are you asking us to find a definite integral that does not evaluate to a constant? There doesn't appear to be additional variables of which this integral could be a function.
Mar
26
comment On the integral $\int_{b'}^{b''}f(x)\ln(1+a x)\mathrm{d}x$
As an aside, I had a fun time assuming that $X$ was exponentially distributed with $\lambda=1$ and then computing the mean of $\log(1+X)$. It turns out to be Gompertz constant; I wonder if this coincidence has been noted before.
Mar
26
comment On the integral $\int_{b'}^{b''}f(x)\ln(1+a x)\mathrm{d}x$
Yes, I agree with that last statement but only for $x\in[0,\infty]$ Given that $X$ has finite mean over $[0,\infty]$ then the transformation of variables $X\to Y=\log(1+aX)$ should also have a finite mean on the same region. This won't be true over $(-1,\infty]$ because $|\log(1+x)|$ on that region can be an arbitrarily large.
Mar
26
comment On the integral $\int_{b'}^{b''}f(x)\ln(1+a x)\mathrm{d}x$
The space of functions that is integrable with respect to a given probability measure $\nu=f(x)dx$ is $L^{1}(\mathbb{R},\nu)$. The function $\log(1+x)$ doesn't decay towards $0$ as $x\to\infty$. This is a problem because then it's much harder to determine if $\log(1+x)\in L^{1}(\mathbb{R},\nu)$. If $f(x)$ does not die off to zero sufficiently quickly, then this function won't be integrable. You should look up some standard results in integrability theory to see why your stated conditions are not sufficient to determine the integrability of $\log(1+ax)$.
Mar
24
comment On hitting time of Brownian motion and Ito's lemma
This is a good approach. More generally, you could look at the standard proofs for hitting times of the interval $(a,b)$ and then just let $a\to-\infty$ in the arguments.
Mar
24
comment On the integral $\int_{b'}^{b''}f(x)\ln(1+a x)\mathrm{d}x$
Even if $b'>-1$, that doesn't mean $\log(1+ab')$ is well-behaved. I think you have a backward inequality as well. I'm not sure that $\alpha$-stability buys you anything, because you're only ever working with one copy of the distribution. You're not using any central limit theorem type properties, and so all of the suggestions above about how there can be arbitrary probability density on $[b',b'']$ still applies. There's just not much you can possibly say without stronger assumptions.
Mar
24
answered On the integral $\int_{b'}^{b''}f(x)\ln(1+a x)\mathrm{d}x$
Mar
24
answered Likelihood Ratio Test for Linear Regression
Mar
23
comment Error of Pearson Correlation Coefficient
Just as an aside, there are many ways to evaluate fits. No single scoring method is ever "best", and in order to provide informative, truthful results to the readers of your work, you need to usually apply a wide array of model checks and fit scores. One place to start reading is Chapter 6 of the excellent resource book "Bayesian Data Analysis" by Gelman, Carlin, Stern, and Rubin.
Mar
23
awarded  Enlightened
Mar
22
awarded  Nice Answer
Mar
22
revised Gradient of a Mahalanobis distance
added 2 characters in body
Mar
22
revised Gradient of a Mahalanobis distance
deleted 117 characters in body
Mar
22
revised Gradient of a Mahalanobis distance
edited body