### Why does Kaggle use Log-loss?

If you're not familiar with Kaggle, it's an organization dedicated to data science competitions to both provide ways for companies to potentially do analytics at less cost, as well as to identify talented data scientists.

Competitions are scored using a variety of functions, and the most common for binary classification tasks with confidence is something called log-loss, which is essentially $$\sum_{i=1}^{n} \log(p_i)$$, where $$p_i$$ is your model's claimed confidence for test data point $$i$$'s correct label. Why does Kaggle use this scoring function? Here I'll follow Terry Tao's argument.

Ideally what we'd like is a scoring function $$f(x)$$ that yields the maximum expected score precisely when the claimed confidence $$x_i$$ in the correct label for $$i$$ is actually what the submitter believes is the true probability (or frequency) of that outcome. This means that we want $L(x)=p\cdot f(x) + (1-p)\cdot f(1-x)$ for fixed $$p$$ to be maximized when $$x=p$$. Differentiating, this means $L'(x) = p\cdot f'(x) - (1-p)\cdot f'(1-x) = 0$ when $$x=p$$, hence $$p\cdot f'(p) = (1-p)\cdot f'(1-p)$$ for all $$p$$. This will be satisfied by any admissible $$f(x)$$ with $$x\cdot f'(x)$$ symmetric around $$x=\frac{1}{2}$$, but if we extend our analysis to multinomial outcomes we get the stronger conclusion that in fact $$x\cdot f'(x) = c_0$$ for some constant $$c_0$$. This in turn implies $$f(x)=c_0\cdot \log(x)+c_1$$. If we want $$f(1/2)=0$$ and $$f(1)=1$$, we end up with $$f(x)={\log}_2(2x)$$ and the expected score is $L(x)=x\cdot {\log}_2(2x) + (1-x)\cdot {\log}_2(2(1-x)).$

### A Bayes' Solution to Monty Hall

For any problem involving conditional probabilities one of your greatest allies is Bayes' Theorem. Bayes' Theorem says that for two events A and B, the probability of A given B is related to the probability of B given A in a specific way.

Standard notation:

probability of A given B is written $$\Pr(A \mid B)$$
probability of B is written $$\Pr(B)$$

Bayes' Theorem:

Using the notation above, Bayes' Theorem can be written: $\Pr(A \mid B) = \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)}$Let's apply Bayes' Theorem to the Monty Hall problem. If you recall, we're told that behind three doors there are two goats and one car, all randomly placed. We initially choose a door, and then Monty, who knows what's behind the doors, always shows us a goat behind one of the remaining doors. He can always do this as there are two goats; if we chose the car initially, Monty picks one of the two doors with a goat behind it at random.

Assume we pick Door 1 and then Monty sho…

### Notes on Setting up a Titan V under Ubuntu 17.04

I recently purchased a Titan V GPU to use for machine and deep learning, and in the process of installing the latest Nvidia driver's hosed my Ubuntu 16.04 install. I was overdue for a fresh install of Linux, anyway, so I decided to upgrade some of my drives at the same time. Here are some of my notes for the process I went through to get the Titan V working perfectly with TensorFlow 1.5 under Ubuntu 17.04.

Old install:
Ubuntu 16.04
EVGA GeForce GTX Titan SuperClocked 6GB
2TB Seagate NAS HDD

New install:
Ubuntu 17.04
Titan V 12GB
/ partition on a 250GB Samsung 840 Pro SSD (had an extra around)
/home partition on a new 1TB Crucial MX500 SSD
New WD Blue 4TB HDD