All entries for Sunday 09 October 2011

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.

QC5:
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).

QC6:
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.


Christopher Midgley : 09 Oct 2011 18:52 |  Tags:    |  Comments (0) |  Report a problem

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