Probability Theory, Math 170b, Winter 2013, Tonci Antunovic - Homework 2

From the textbook solve the problems 17, 18, 19, 22, 23 and 24 from the Chapter 4.Solve the problems 21, 22, 24, 30 from the Chapter 4 additional exercises at

http://www.athenasc.com/prob-supp.html

And also the problems below:

Problem 1. Show that for random variables X, Y and Z we have

E[E[E[X|Y ]|Z]] = E[X].

Apply this formula to the following problem: Roll a far 6-sided die and observe the number Zthat came up. Then toss a fair coin Z times and observe the number of heads Y . Then let Xbe a number uniformly chosen in the interval [0, Y ]. Find E[X].

Solution: We apply Law of repeated expectation twice to get

E[E[E[X|Y ]|Z]] = E[E[X|Y ]] = E[X].

In the example given E[X|Y ] = Y/2, E[E[X|Y ]|Z] = E[Y/2|Z] = Z/4 and

E[X] = E[E[E[X|Y ]|Z]] = E[Z/4] =1 + · · ·+ 6

4 · 6=

5

8.

Problem 2. A machine prints out the number 1, 2, or 3 with equal probabilities. A machinestarts printing numbers independently one after another and stops the first time it prints 1. LetX denote the number of 2s it printed. Compute E[X] and var(X).

Solution: Let Y be the first time the machine prints 1, so Y is geometric with parameter1/3 and E[Y ] = 3 and var(Y ) = (1 − 1/3)/(1/3)2 = 6. Given the value Y we have Y − 1preceding tosses each of which can be either 2 or 3 with equal probabilities. Thus, given Y therandom variable X is just binomial with parameters Y − 1 and 1/2 and so E[X|Y ] = (Y − 1)/2and var(X|Y ) = (Y − 1)/4.

ThenE[X] = E[E[X|Y ]] = E[(Y − 1)/2] = (3− 1)/2 = 1

var(X) = E[var(X|Y )] + var(E[X|Y ]) = E[(Y − 1)/4] + var((Y − 1)/2)

= (3− 1)/4 + var(Y )/4 = 1/2 + 6/4 = 2.

Problem 3. Break a stick of length 1 at a uniform location and then choose either of the twoparts with equal probabilities. If X denotes the length of the part you choose, compute E[X]and var(X).

Solution: If Y is the location of the break point, then Y is uniform in the interval [0, 1].Given Y we either choose Y or 1− Y each with probability 1/2 and so

E[X|Y ] = Y/2 + (1− Y )/2 = 1/2,

1

E[X2|Y ] = Y 2/2 + (1− Y )2/2 = Y 2 − Y + 1/2

⇒ var(X|Y ) = E[X2|Y ]− (E[X|Y ])2 = Y 2 − Y + 1/2− 1/4 = Y 2 − Y + 1/4.

and

var(X) = E[var(X|Y )]+var(E[X|Y ]) = E[Y 2−Y +1/4]+var(1/2) =

∫ 1

0t2− t+1/4 dt = 1/12,

since var(1/2) = 0.

Problem 4. Consider the triangle with vertices (0, 0), (1, 0) and (0, 1). Let Z be a uniformrandom variable in the interval [0, 1]. Draw a vertical line that intersects the x axis at Z. Thisline divides the triangle in two pieces. Select a point (X,Y ) uniformly at random from the rightpiece. Find the expectation E[X] of the x coordinate of the selected point.

Hint: You might have to integrate an ugly looking function. Factor the numerator, cancelstuff and simplify the function.

Solution: Given the value of Z the expected x coordinate of chosen point is

E[X|Z] =2

(1− Z)2

∫ 1

Z

∫ 1−x

0x dydx =

1/3− Z2 + 2Z3

3

(1− Z)2.

Now factor 1/3− Z2 + 2Z3

3 = (1− Z)2(1/3 + 2Z/3) and so

E[X|Z] = 1/3 + 2Z/3.

ThenE[E[X|Z]] = E[1/3 + 2Z/3] = 2/3.

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