STAT 36-217, HW4, due Tuesday 10/11/2011, 11:50 AM
This HW is slightly harder than usual as you have about 2 weeks to complete. Start
earlier when possible and seek help during oce hours if you have problems.
1. (20). Consider two random random variables
36-217, Fall 2013
Homework 6 Solutions
1. (15 points)
a. Note that, after one week the expected price of the commodity is 1/2 + 4 1/2 = 5/2 dollars per ounce.
Since, 5/2 > 2, if I want to maximize the expected amount of money that I possess at the end of
36-217, Fall 2013
Homework 5 Solutions
1.
a. (5 points)
P (A|B C) =
(A and B are conditionally independent) =
P (A B|C)P (C)
P (B C)
P (A|C)P (B|C)
P (A|C)P (B|C)P (C)
=
= P (A|C)
P (B C)
P (B|C)
b. (10 points) Using Bayes rule and the law of total probab
36-217, Fall 2013
Homework 3 Solutions
1. (15 points) Consider an event A , and dene A B = C. Note that: B and A , C .
By Bayes rule, we can write:
P (A|B) =
P (A B)
P (C)
=
.
P (B)
P (B)
Since, P (.) is a probability measure in , we know that 0 P (C) 1 a
36-217, Fall 2013
Homework 4 Solutions
1. (20 Points) Let Ai denote the event that all the points lie in the semicircle beginning at the ith points and
going clockwise for 180deg, i = 1 . . . n. A is the event that all the points are contained in some sem
36-217, FALL 2013
Homework 1 Solutions
1.
a. (3 points) If we arbitrarily choose some element x from a non-empty set (A B c ), then x is
either in A or not B. Thus, x should be in A. This implies (A B) (A B c ) = (A B) A.
Therefore, (A B) A = A
b. (3 poin
36-217, Fall 2013
Homework 7 Solutions
1. (15 points)
a. First, lets show that E(X) = np.
n
E(X) =
xi P (X = xi ) where xi = 0, . . . , n
i=0
n
=
n xi
p (1 p)nxi
xi
xi
i=0
n
=
np
i=0
(n 1)!
pxi 1 (1 p)nxi
(n xi )!(xi 1)!
n
P (X = xi 1) = np 1 = np
= np
i=
36-217, Fall 2013
Homework 10 Solutions
1. (a.) [2 points] Since Y unif orm(a, b), cdf of Y is:
0 if x < a
FX (x) = xa if a x < b
ba
1 if x b
(b.) [5 points]
FX (x) = P (X x) = P (X x|Z = 1)P (Z = 1) + P (X x|Z = 0)P (Z = 0)
= P (X1 x)p + P (X2 x)(1 p)
=
36-217, Fall 2013
Homework 8 Solutions
1. (25 points) Bubble sort algortihm again
Using the fact that the expected value of a sum of random variables is equal to the sum of the expectations,
we see that
E[I] =
E[I(i, j)]
j
i<j
Now, for each i < j,
E[I(i,
36-217, Fall 2013
Homework 11 Solutions
1. (15 Points)
(a) Let E(X) = , E(Y ) = , to make it clearer.
Cov[X, Y ] = E[(X )(Y )]
= E[XY X Y + ]
= E[XY ] E(X) E(Y ) +
= E[XY ]
= E[XY ] E[X]E[Y ]
(b) We know that Cov[A,B+C]=Cov[A,C]+Cov[A,B] (p. 218), and C
36-217, Fall 2013
Homework 2 Solutions
1. (12 points) The information given by the problem can be organized in a Venn diagram as presented in Figure
1. Firstly, note that in the triple intersection of the three sets there are 18 students because there are