Engineering Probability Exam 3 solution  Tues 20180508
Name, RCSID: W. Randolph Franklin, frankwr
OK to give the formulas w/o working them out.
Rules:
 You have 80 minutes.
 You may bring three 2sided 8.5"x11" papers with notes.
 You may bring a calculator.
 You may not share material with each other during the exam.
 No collaboration or communication (except with the staff) is allowed.
 Check that your copy of this test has all nine pages.
 Each part of a question is worth 5 points.
 You may cross out three question parts, which will not be graded.
 When answering a question, don't just state your answer, prove it.
Questions:

You toss two coins. Each comes up heads half of the time. However, for some funny reason, they both come up heads together, or both come up tails together. Intuitively, they not independent. This question is to prove that from the definition of independence.
P[HT] = 0. However P[A=H] = P[B=T] = 1/2, so P[HT] != P[A=H]P[B=T]. That's the def of not independent.

This time, you toss three coins, A, B, and C. These are the probabilities:
P[TTT] = P[THH] = P[HTH] = P[HHT] = 0
P[TTH] = P[THT] = P[HTT] = P[HHH] = 1/4
My notation is that TTH means that coin A is tails, coin B tails, and coin C heads. Etc.

Are the individual coins fair (i.e., heads half the time)?
P[A=H] = 0+0+1/4+1/4 = 1/2 so fair. Ditto B and C.

Are coins A and B independent?
P[A=H,B=H] = 1/4, good. P[A=H,B=T]=1/4, good. P[TH] = 1/4, good. P[TT] = 1/4, good. Yes independent.

Are all 3 coins independent?
P[HHH] = 1/4 != P[A=H]P[B=H]P[C=H] = 1/8. Not independent.


This question is about a continuous probability distribution on 2 variables.
$$f_{XY}(x,y) = \begin{cases} c x y & \text{ if } (0\le x) \ \& \ (0\le y)\ \& \ (0\le x+y \le 1) \\ 0 & \text{ otherwise}\end{cases}$$
The nonzero region is the triangle with vertices (0,0), (1,0) and (0,1).
c is some constant, but I didn't tell you what it is.

What is c?
$\int_0^1\int_0^{1x} xy\ dy\ dx = 1/24$ so $c=24$

What is $F_{XY}(x,y)$?
$$F_{XY}(x,y)=\begin{cases} 0 & \text{ if } x\le0 \cup y\le0 \\ 1 & \text{ if } x\ge 1 \cap y\ge1 \\ 6x^2y^2 & \text{ if } 0\le x \cap 0\le y \cap x+y\le1 \\ (\int_0^x\int_0^{1x} + \int_0^{1y}\int_{1x}^y + \int_{1y}^x\int_{1x}^{1x_0}) (12x_0y_0 dy_0dx_0) & \text{ otherwise}\end{cases}$$
The last case above splits the nonzero integration region into two rectangles and a triangle.
It's also acceptable to draw a figure and say something intelligent w/o being explicit about all the details.

What is $f_X(x)$?
$f_X(x)= \int_0^{1x}f_{XY}(x,y) dy = 12x(1x)^2$
Note that $\int_0^1 f_X(x)=1$, which is correct.

Are X and Y independent?
$f_X(x)=12x(1x)^2,f_Y(y)=12y(1y)^2,f_X(x)f_Y(y)\ne f_{XY}(x,y)$
No.

What is $P[X\le Y]$ ?
$\int_0^1\int_0^{\min(x,1x)} f_{XY}(x,y) dy\ dx$. However, since $f_{XY}(x,y) = f_{XY}(y,x)$, $P[X\le Y]=1/2$

Define a new random variable $Z=X+Y$. What is $F_Z(z)$?
$f_Z(z) = \int_0^z f_{XY}(x,zx) dx = 24\int_0^z x(zx)dx$ for $0\le z\le 1$
$F_Z(z) = \int f_Z(z)dz$

What is $E[X]$?
$\int_0^1 xf_X(x)dx = 2/5$

What is $COV[X,Y]$?
E[XY] = $\int_0^1\int_0^{1x}xyf_{XY}dx dy=8\int_0^1x^2(1x)^4 dx, E[X]=E[Y]=2/5$, COV[X,Y]=E[XY]E[X]E[Y]. You don't have to work through the math.

What is $\rho_{X,Y}$?
$\sigma_X=\sigma_Y= E[X^2]E[X]^2$
$\rho_{X,Y}=COV[X,Y]/(\sigma_X\sigma_Y)$

What is $f_Y(yx)$?
$f_Y(yx)=f(x,y)/f(x) = 12xy/(4x^3) = 2\frac{y}{x^2}$

What is $E[Yx]$?
Integrate the above over $y$ to get $x^{2}$


Compute $$\int_0^\infty e^{x^2} dx$$ .
Consider $\sigma=1/\sqrt{2}$. Working a little, this will give $\int_0^\infty e^{x^2} dx=\sqrt{\pi}/2=0.886$. It was also ok just to use a calculator.

What is the legal range for a correlation coefficient?
1 to 1

What is the variance of the sum of 100 independent variables, each of which is N(0,1)?
100.

You have 10 independent random variables. Each is uniform on [0,1]. What is the expected value of the max?
Let $W=\max(X_i). F_W(w)=w^{10}. f_W(w)=10w^9.E[W]=10/11.$

You toss 10 independent fair coins, one after the other.

What is the expected total number of heads, given that the first 5 coins came up heads?
The last 5 coins do not depend on the first 5. So the expectation is 7.5.

What is the probability of the total number of heads being 10, given that the first 5 coins came up heads?
1/32

End of exam 3, total 90 points (considering that 3 questions aren't graded).