# 02-24-14 -- Group 1 -- Group Quiz

## Problem 1

The distribution $\displaystyle p_X(x)$ :

## Problem 2

$\displaystyle \mu = \mathbb{E}(x) = \int_{-\infty}^{\infty} x p(x) ~ dx = \int_{0}^{2} x (1 - \frac{x}{2}) ~ dx = \frac{2}{3}$

$\displaystyle \mathbb{E}(x^2) = \int_{-\infty}^{\infty} x^2 p(x) ~ dx = \int_{0}^{2} x^2 (1 - \frac{x}{2}) ~ dx = \frac{2}{3}$

$\displaystyle Var(x) = \mathbb{E}(x^2) - (\mathbb{E}(x))^2 = \frac{2}{3} - \left( \frac{2}{3} \right)^2 = \frac{6}{9} - \frac{4}{9} = \frac{2}{9}$

$\displaystyle Std(x) = \sqrt{Var(x)} = \sqrt {\left( \frac{2}{9} \right)} \approx 0.471$

## Problem 3

The CDF is the integral of the PDF.

$\displaystyle x - \frac{x^2}{4}, 0 < x < 2$

$\displaystyle 0, x \leq 0$

$\displaystyle 1, x \geq 2$

## Problem 4

#### Code


% Function to draw random deviates from p(x) given a value drawn from
% uniform random
invCDF = @(p) (2-2*sqrt(1-p));

% Visualization
hist(invCDF(rand(100000, 1)))



#### Visualization

Here is an example of deviates drawn using the function above. As expected, it resembles the pdf of x.

## Problem 5

Normal distribution with mean $\displaystyle \mu = \frac{2}{3} N$ and variance $\displaystyle \sigma^2 = \frac{2}{9} N$ :

$\displaystyle S \sim \mathcal{N} \left( \mu, \sigma^2 \right) = \frac{1}{\sqrt{2 \pi \sigma^2}} exp \left( -\frac{(x - \mu)^2}{2 \sigma^2} \right) = \frac{1}{\sqrt{\frac{4 \pi}{9} N}} exp \left( {-\frac{x - \frac{2}{3} N}{\frac{4}{9} N}} \right)$

## Problem 6

#### Code

Here is code showing pvalues calculated for 2 different datasets: one that is uniform(0,5) and the other that is p(x) from problem 1, using the null hypothesis that the data was drawn from p(x)

% Function to draw random deviates from p(x) given a value drawn from
% uniform random
invCDF = @(p) (2-2*sqrt(1-p));

n = 28;
mu = (2/3) * n;
sigma = (2/9) * n;

% Random deviates drawn from p(x) and uniform(0,5)
randDraws =  sum(5 * rand(n, 1));
pDraws = sum(invCDF(rand(n,1)));

% Pvalue for test statistic for a dataset of random deviates uniform (0,5)
if randDraws < mu
randPval = 2 * normcdf(randDraws, mu, sigma);
else
randPval = 2 * (1 - normcdf(randDraws, mu, sigma));
end

% Pvalue for test statistic for a dataset of deviates drawn from p(x)
if pDraws < mu
pPval = 2 * normcdf(pDraws, mu, sigma);
else
pPval = 2 * (1 - normcdf(pDraws, mu, sigma));
end

randPval
pPval


#### Sample Result


randPval =

9.5479e-15

pPval =

0.6586



In general (i.e. even over multiple runs), the null hypothesis can be rejected at the 5% significance level (possibly even a smaller significance level) for the dataset drawn from uniform(0,5). The null hypothesis is not ruled out for the dataset drawn from p(x).

## Problem 7

Solution
Our approach is similar to problem 6. We perform p-value test for each of the known hypothesis, we add multiple hypothesis correction with cutoff of $\displaystyle \alpha' = \alpha/10$ . The hypotheses that fail the p-value test will be ruled out and we won't be able to say anything conclusively about the remaining hypotheses.

## Problem 8

For each hypothesis $\displaystyle H_j$ we compute the probability given the data as

  $\displaystyle P_j = Pr(Data | H_j)\cdot P(H_j) = \displaystyle\prod_{i=1}^{28} P_{X}^{(j)} (x_i) \cdot P(H_j)$


  $\displaystyle arg max_{j} (P_j)$



We pick the hypothesis that maximizes the probability given the data.

## Problem 9

Solution
We would multiply the characteristic functions and take the inverse fourier transform