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January 05, 2012

[citation needed]; the difficulty of finding things

Typing in “tower property” in Google, I find that the first result is the ever ubiquitous Wikipedia (whose mastery of SEO means it turns up, with an occasional irrelevant article, on whatever subject you could care to name) article on Expected Value, in this case. Actually typing in “tower property” returns the article on “law of total expectation” which is apparently the one of its myriad names that Wikipedia has decided is most common. Looking at the other results on Google, even adding a helpful “statistics”, I find that “tower property” doesn’t appear to return anything else relevant. In fact, the only other place I can find it called “tower property” is in my notes :)

For nameless results, I find my best bet is simply to type in the result itself. For example, that E[XY]=E[YE[X|Y]] is proven at the end of this pdf document, which is likely lecture notes. If something has a lot of roots or powers, this is somewhat less applicable.

As of yet, I’ve not been able to find anything on what my notes refer to as “Fisher’s theorem”. It’s a theorem named after a famous mathematician who had many theorems named after him (some with others), so we’re already off the a bad start trying to find it. The theorem reads:

Let X_i \sim N(\mu, \sigma^2) be indepedent random variables. Define \overline{X}=\sum_{i=1}^n X_i and s^2=\frac{1}{n-1}\sum_{i=1}^n (\overline{X}_n-X_i)^2. Then:
*\overline{X} \sim N(\mu,\frac{\sigma^2}{n})
*\overline{X} and s^2 are independent.
*\frac{(n-1)s^2}{\sigma^2} \sim \chi^2_{n-1}
*\frac{\overline{X}-\mu}{\sqrt{s^2/n}} \sim t_{n-1}

It looks like it has something to do with sample mean and variance, but I’m only taking the first module on this topic, so what its use is I can’t say.

October 30, 2011

Balls in boxes and other miscellaneous thoughts

It is a reasonably common statement in probability that problems can be simplified down to spinners or balls in boxes. So:
Let there be k balls that we wish to place in n boxes. If we can distinguish between the balls, then each ball can go in n different places, so the total is \textstyle n^k. If we cannot distinguish between the balls, this is answered by the occupancy theorem: the multiset formula for n and k.

What, then, if it is the boxes we cannot distinguish between? If the balls also cannot be distinguished between, this is equivalent to partitioning k into at most n parts. The OEIS implies that this has a closed-form solution: \textstyle \frac{1}{n!} \frac{d^n\left( \frac{1}{\prod_{i=1}^k(1-x^i)} \right)}{dx^n}(0), which is nice to know, even if it suggests there’s no ‘nice’ combinatorial solution.
If we can distinguish between the balls but not the boxes, I believe this would come to a sum of Stirling numbers of the second kind: \sum_{i=0}^n \left\{\begin{matrix} k \\ i \end{matrix}\right\}.

Other miscellany: Mario Micallef told us a few integration tricks this week: being able to cancel entire functions while integrating from 0 to 2pi (due to them containing sine), \textstyle\int_0^{2\pi}sin^2(x)dx=\pi by comparison with \cos^2(x) across the same interval. Massively time-saving tricks I’d never considered before. Great!

From combinatorics: wondered if there was a way to check transitivity of a relation expressed in matrix form quickly. Found nothing more than checking each row one step at a time, but I least I can do that fairly quickly now.

October 09, 2011

Mathematical Statistics A – Example Sheet 1

Thought I may as well put answers to the A & C sections here, for simple-ish reflection. And to notice when I can’t recall how to do things, at all.

QA1: We compute the cumulative distribution function by integrating the density function:
  \begin{align*} F_x &= \int_{-\infty}^x f_x(s)ds\\ &=\int_0^x e^{-s}ds\\ &=\left. -e^{-s}\right|_0^x\\ &= 1 - e^{-x} \end{align*}
We argue Y is discrete because…it takes a finite number of possible values? It’s a step function? I’m not sure. Its support is {0,2} (I hope), pmf and cdf are:
  f_Y(y) =\left{ {1-\frac{1}{e}} \quad (y=0) \\\frac{1}{e} \quad\quad (y=2)\\0 \quad\quad (y \notin \{0, 2\}) \right.
  F_Y(y) =\left{ 0 \quad\quad (y<0) \\\1-\frac{1}{e} \quad (0\leq y<2)\\1 \quad\quad (y \geq 2) \right.
QA2 is simple infinite sum fiddling combined with the identity for e^k.

  Y = g(X)\\ F_Y(y) = (g(X) \leq y)\\ \begin{align*} F_Y(y) &= (X \leq g^{-1}(y))\\ &=F_X(g^{-1}(y)) \end{align*} \\ \begin{align*} f_y(y)&=f_x(g^{-1}(y))\times(g^{-1}(y))'\\ &= \frac{f_x(g^{-1}(y))}{g'(g^{-1}(y))} \end{align*}
By the inverse function theorem (differentiation). Y is continuous as the functions that make up Y are continuous – additionally, as g is an increasing bijection it is continuous and differentiable, and its derivative is always positive, so the pdf of Y is not undefined (g’ always nonzero).

For the last part, simply plug in g(X) = aX + b to what you’ve already worked out, and find Y is the Gaussian (given X is the standard Gaussian).

Find the mgf of the exponential in the normal way; use the same trick as Q4 to find the mgf of S_n. Find the mgf of S_n using the pdf; as they are equal use uniqueness to conclude the distribution is correct.

Integrate F_{S_n} from t to infinity; receive sum from 0 to n. Observe that as the number of photons detected must be an integer, P(N=k)=P(N<k+1)-P(N<k). Plug in your previously calculated values for the answer.

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