Dan's Segment 8

From Computational Statistics (CSE383M and CS395T)
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To Calculate

1. For a lognormal prior <math>f_X(x;\mu,\sigma) = \frac{1}{x \sigma \sqrt{2 \pi}}\, e^{-\frac{(\ln x - \mu)^2}{2\sigma^2}}</math> we can take the limit where <math>\sigma</math> goes to infinity. The exponential term falls off faster to zero than the fraction term, and sigma dominates over x, leaving us with <math>\frac 1{\sqrt{2\pi} \sigma} </math>

3. For a Student distribution <math> p(t) = \frac{\Gamma(\frac12[\nu+1])}{\Gamma(\frac12\nu)\sqrt{\nu\pi}\sigma} \left( 1+\frac1{\nu}\left[\frac{t-\mu}{\sigma}\right]^2\right)^{-\frac12(\nu+1)} </math> With <math>\nu</math> going to infinity, I can not figure out how it's supposed to converge on a gaussian. The gamma term should converge to <math>\frac1{\sqrt{2\pi}\sigma} </math> and it's pretty easy to see that, but getting the second term to converge to a normal doesn't follow for me. I keep thinking it should converge to 1, but obviously that isn't correct. Interestingly I noticed that wikipedia has the final result totally lacking sigma altogether, which is not the answer anyone on their personal wiki gave.

To Think About

1. Isn't this symbolic math? I don't know how to generalize it. Slide 9 shows a specific case where the inverse cdf had already been determined, presumably by hand, and just hardcoded in. This seems like the best practical solution to me.

2. If we assume random decay, then the amount of decay at any given time is directly proportional to the amount of material present. When there is a lot of material, there is a high rate of decay and vice-versa. Mathematically, that looks like this: <math>\frac{dN}{dt} = -\lambda N</math> and the solution to that differential equation is an exponential <math>N = N_0 e^{-\lambda t}, \,</math>