Can I Get A Squee For Math?

If I were to get another PhD, I think I'd want it in Awesomeness.

So today has been an absolute blast, and it’s all because of math.

Seriously.

I know, I know, you’re probably rolling your eyes and thinking Gillis has finally lost his gourd. But no, that is not the case. Well, it might be the case, but that’s a story for another time.

So why was today all-that-and-a-bag-of-chips thank you very much mathematics? Well, for at least 2 reasons:

  • Today my friend Kim defended her Ph.D. dissertation. Her talk was fantastic and I couldn’t help but smile as I watched her explain some rather complicated and complex mathematics in a way that was understandable – even to those who had no mathematics training. This was due to Kim’s obvious mastery of the subject, and the obvious passion she has for it. She is exceptionally talented and has the very desirable and sought after ability to translate high-brow math-speak to something that everyone can grasp.
  • My whiteboard at the end of the day

    I also spent the afternoon playing around with some mathematics1 for a paper that I’m coauthoring with my friend Almost Doctor Lorna. You see, she has spent the last little while working on a rather incredible simulation study built on a statistical model that is intended to understand disease spread in animal populations. Her work is pretty phenomenal, and I can’t wait until the day she too gets to defend. Anyway, she noticed some interesting results and we are now trying to determine the mathematical justification for what she saw. That, or identify that it might just be coincidence – that is, no pattern actually exists; we just think it’s there.

Anyway, I’m currently trying to show the following (generally speaking – or possibly on average):

\ln\displaystyle{\prod_{t}}\displaystyle{\prod_{i}}p(x|\bar\theta)\displaystyle{\prod_{j}}(1-p(x|\bar\theta))>\ln\displaystyle{\prod_{t}}\frac{1}{m}\displaystyle{\sum_{l=1}^{m}}\displaystyle{\prod_{i}}p(x|\theta^{(l)})\displaystyle{\prod_{j}}(1-p(x|\theta^{(l)})),

where

p(x|\mathbf{\theta})=1-\exp\left\{-f(\mathbf{\theta}, x)\right\},

and

\bar\theta=\displaystyle{\sum_{l=1}^{m}}\frac{\theta^{(l)}}{m},

but I’m stuck. This could be a side effect of being tired. Or perhaps I’ve been staring at this problem too long and the answer is actually obvious. Or it could be that my brain may or may not be enjoying the soothing aromas and flavours of a scotch-y beverage2. It really is a mystery.


1 Squee!

2 If you know me, you’ll know the answer to this riddle3.

3 It’s so not a riddle. Ha!


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