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?- love(math) is unrequited. true.


Apr
2
comment How can we bound $P\{X \ge t\}$ from below?
@J.D. Chebyshev's inequality is still an upper bound on probability (for a non-negative variable). It is where Markov's inequality is derived from, so it's likely that the poster already knew of this bound regardless. I suppose one can expand the absolute value usually used in it into two events, then use complements and multiply by -1 to get a statement of lower bounded probability, but it won't be in terms of a useful bound as it will always involve the other probability of the other "half-event" that you split out of the absolute value.
Mar
30
comment How can I calculate the CDF of this random variable?
When I do the computation by hand, I am able to easily perform the integral over the variable $x_{2}$, and then the resulting integral over $x_{3}$ is of the form $\int_{0}^{\infty}\frac{\alpha}{\epsilon + \beta x_{3}}\cdot{}\exp{(\frac{-x_{3}}{\Omega_{3}})}$, for constants $\alpha$, $\epsilon$, and $\beta$. According to Wolfram integrator, an integral of this type requires the Exponential Integral function Ei() to express the solution.
Mar
27
awarded  Tumbleweed
Mar
26
comment An example of a “pathological” power-spectral density function?
I don't like the 'avoid extended discussions in comments' rule, so I abstain from participating in chat-migrated things. I think it's most helpful if everything appears right here on the same page. But I will think much more about this and I'll post any follow-up result as an answer to avoid making the comments thread too long.
Mar
26
comment An example of a “pathological” power-spectral density function?
To clarify, you can construct functions with properties like $x^{2}\sin(1/x^{2})$ that have the symmetry and derivative properties you wanted. But the fact that PSD functions come specifically as Fourier transforms restricts what kinds of functions they can be. So it's a harder analysis problem to prove that you could't concoct a weird trig function with a singularity in its argument such that the singularity blows up for higher powers.
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.