There are eight pages with eight questions. Answer any six questions. (If you answer more, then we will grade the first ones.)

We record the order in which you hand in your exam the better to keep track of the exams, and sometimes to plot scatterplots of order vs grade. (There is never any significant correlation.) Your order does not affect your grade.

• This exam is closed book but a calculator and two 3-sided letter-paper-size note sheets are allowed.
• You may not use other computers or communication devices, or share material with other students.
• For full points, your answers must be simplified as much as reasonable.
• Since this conflict exam is earlier than the regular exam, please turn in any notes that you made.

1. [_____/10] Evaluate and give me the answer as an expression perhaps involving {$\pi$} and {$\sqrt{\cdot}$}. Show your work. {$${\cal A} = \int_{-\infty}^\infty e^{-2x^2} dx$$}
{$\left( \sqrt{ \frac{2}{\pi} } \right) e^{-2x^2}$} is a Gaussian pdf with {$\mu=0, \sigma=1/2$} and integrates to 1. So, {$$\int_{-\infty}^\infty e^{-2x^2} dx = \sqrt{ \frac{\pi}{2} }$$}
2. [_____/10]
• Random variable X is continuous U[0,1].
• Random variable Y is continuous U[0,2].
• Random variable Z is continuous U[0,3].
• Random variable W {$\overset{\Delta}{=}$} min(X,Y,Z).
Compute P[0<W<1].
You could do it the detailed way. Or, you could notice that min(X,Y,Z) <=1 and so P[0<W<1]=1.
3. [_____/10] You are throwing darts uniformly at a circular dartboard with diameter 1 foot.
1. Compute the pdf {$f_{XY}(x,y)$}
{$$f_{XY}(x,y) = \left\lbrace \begin{matrix} 4/\pi & \text{if } x^2+y^2\le\frac{1}{4} \\ 0 & \text{otherwise} \end{matrix} \right.$$} The {$\Large \frac{4}{\pi}$} is chosen to make {$\int \int f_{XY}(x,y) dx dy =1$} .
2. Transform to fathoms: {$W \overset{\Delta}{=} X/6, Z \overset{\Delta}{=} Y/6$}. Compute the pdf {$f_{WZ}(w,z)$}.
{$$f_{WZ}(w,z) = \left\lbrace \begin{matrix} 144/\pi & \text{if } w^2+z^2\le9 \\ 0 & \text{otherwise} \end{matrix} \right.$$}
3. Transform {$(X,Y)$} to {$(R,\Theta)$} and compute {$f_{R\Theta}(\rho,\theta)$}.
{$X=R \cos\Theta, \,\, Y=R\sin\Theta$} so the Jacobian is {$R$} so {$dx dy = R d\rho d\theta$} so {$$f_{R\Theta}(\rho,\theta) = \left\lbrace \begin{matrix} 4r/\pi & \text{if } 0\le r\le\frac{1}{2} \\ 0 & \text{otherwise} \end{matrix} \right.$$}
(Not required) Check: {$$\int_0^{\frac{1}{2} } \int_0^{2\pi} f_{R\Theta}(\rho,\theta) d\theta d\rho = 1$$}
4. [_____/10]
• Random variable X is continuous U[-1,1].
• Random variable Y is defined as {$Y \overset{\Delta}{=} X^2$}.
Compute {$f_X(x), F_X(x), E[X], VAR[X], f_Y(y), F_Y(y), E[Y], VAR[Y], COV[X,Y], \rho_{XY}$}.
{$$f_X(x) = \left\lbrace \begin{matrix} 1/2 & \text{if} -1\le x\le 1 \\ 0 & \text{otherwise} \end{matrix} \right.$$} {$$F_X(x) = \left\lbrace \begin{matrix} 0 & \text{if } x\le -1 \\ \frac{x+1}{2} & \text{if } -1 \le x\le 1 \\ 1 & \text{if } x\ge 1 \end{matrix} \right.$$} {$$E[X] = 0$$} {$$VAR[X] = \int_{-1}^1 x^2 f_X(x) dx = 1/3$$} . {$F_Y(y)$} and {$F_Y(y)$} are tricky because 2 values of {$x$} map to each {$y$}. It might be easiest to start with {$F_Y(y)$}. The interesting values are {$0 \le y \le 1$}. {$$F_Y(y) = \left\lbrace \begin{matrix} 0 & \text{if } y\le 0 \\ \text{something} & \text{if } 0\le y\le 1 \\ 1 & \text{if } y\ge 1 \end{matrix} \right.$$} For {$0 \le y \le 1$} {$$F_Y(y) = P[Y\le y] = P[X^2 \le y] = P[|X| \le \sqrt{y} ]$$} {$$= F_X(\sqrt{y} ) - F_X(-\sqrt{y} ) = \sqrt{y}$$} Putting the pieces together: {$$F_Y(y) = \left\lbrace \begin{matrix} 0 & \text{if } y\le 0 \\ \sqrt{y} & \text{if } 0\le y\le 1 \\ 1 & \text{if } y\ge 1 \end{matrix} \right.$$} {$$f_Y(y) = \frac{dF_Y(y)}{dy} = \left\lbrace \begin{matrix} { \Large \frac{1}{ 2\sqrt{y} } } & \text{if } 0\le y\le 1 \\ 0 & \text{otherwise} \end{matrix} \right.$$} {$$E[Y] = 1/3$$} {$$VAR[Y] = E[Y^2] - E[Y]^2 = 1/5 - 1/9 = 4/45$$} {$$E[XY] = \int_{-1}^{1} xy f_X(x) dx = 1/2 \int_{-1}^{1} x^3 dx =0$$} {$$COV[X,Y] = E[XY] - E[X]E[Y] = 0$$}
{$$\rho = 0$$} That is, even though X completely determines Y, so they are definitely dependent, they are uncorrelated, since the dependency is nonlinear.
5. [_____/10] You toss a fair coin 100 times.
1. Compute the probability that the number of heads is between 60 and 90 inclusive using the central limit theorem. Express your answer using the Z function.
{$$\mu = np = 50, \,\, \sigma = \sqrt{npq} = 5.$$}
{$$60 = \mu+2\sigma, \,\, 90 = \mu+8\sigma$$}
Answer = Z(8) - Z(2) ( {$\approx 0.02$} , but you don't need to show that.)
2. You observe 55 heads from your experiment. You know that the mean number of heads is 50.
If you repeated this experiment (of 100 tosses) many times, what's the probability that you would see a number of heads at least this far from the mean in either direction? Express your answer using Z.
That's a 2-tail probability. {$55=\mu+\sigma$}. Answer = 2Z(1) ( {$\approx 2/3$} ).
6. [_____/10] Assume that a bicycle tire's probability of failing in the next hour is 0.1 and does not depend on its age. (That is appropriate if it fails from road hazards and not from wearing thin.)
1. What's the name of the appropriate probability distribution? Exponential.
2. You have 2 tires and no spares. What's the probability of neither failing in the next 4 hours?
Let {$p$} be the probability of one tire living for one hour. {$p=0.9$} . So, the probability of two tires living for 4 hours is {$p^8 = 0.9^8 = 0.43$}. You don't need to explicitly use the exponential distribution.
3. Now you're stocking supplies for a team of 10 bicyclists, so 20 tires are in use at any time. They race for 10 hours each. Compute the mean and standard deviation for the number of tires you expect to wear out.
The expected number of tires wearing out is 20. The distribution of the number of tires wearing out is Poisson with {$\mu=20$} and so {$\sigma=\sqrt{20}$}.
7. [_____/10] You toss 2 4-sided dice, a red one and a blue one.
• Each has faces numbered 1 to 4.
• The red one is fair.
• The blue one has P[i] = i/10.
1. What's the probability that your total is at least 6?
You can enumerate the possibilities. They are (2,4), (3,3), (3,4), (4,2), (4,3), (4,4). The respective probabilities are {$$\frac{1}{4}\frac{4}{10}, \frac{1}{4}\frac{3}{10}, \frac{1}{4}\frac{4}{10}, \frac{1}{4}\frac{2}{10}, \frac{1}{4}\frac{3}{10}, \frac{1}{4}\frac{4}{10}$$}. They sum to 1/2.
2. What's the probability that your total is at least 6 given that the red die shows 3 or 4?
P[red die shows 3 or 4] = 1/2.
P[red die shows 3 or 4 and total is at least 6] = 2/5.
P[your total is at least 6 given that the red die shows 3 or 4] {$= \frac{2/5}{1/2} = 4/5$}
3. You win an amount equal to the absolute value of the difference between the two dice. What's the expected value of your win?
  {$$0 \frac{1}{4}\frac{1}{10} + 1 \frac{1}{4}\frac{2}{10} + 2 \frac{1}{4}\frac{3}{10} + 3 \frac{1}{4}\frac{4}{10} + 1 \frac{1}{4}\frac{1}{10} + 0 \frac{1}{4}\frac{2}{10} + 1 \frac{1}{4}\frac{3}{10} + 2 \frac{1}{4}\frac{4}{10}$$} {$$+ 2 \frac{1}{4}\frac{1}{10} + 1 \frac{1}{4}\frac{2}{10} + 0 \frac{1}{4}\frac{3}{10} + 1 \frac{1}{4}\frac{4}{10}+3 \frac{1}{4}\frac{1a}{10} +2 \frac{1}{4}\frac{2}{10} +1 \frac{1}{4}\frac{3}{10} +0 \frac{1}{4}\frac{4}{10}$$} {$$= \frac{6}{5}$$}

1. [_____/10] A signal X is transmitted over a noisy channel. X is Gaussian with mean 0 and standard deviation 10. The channel adds noise N. N is Gaussian with mean 0 and standard deviation 20. The received signal is {$Y \overset{\Delta}{=}X+N$}.
What is the maximum a posteriori value for X given Y=10.
This is done on page 282. x=2.

(end of exam /WRF)