Math 361, Problem set 7
Due 10/25/10
1. (1.9.6) Let the random variable X have E[X ] = , E[(X )2 ] = 2 and
mgf M (t), h < t < h. Show that
E
X
= 0,
E
X
2
=1
and
E exp t
X
= et/ M
t
, h < t < h.
(Re
Math 361, Problem Set 2
September 17, 2010
Due: 9/13/10
1. (1.3.11) A bowl contains 16 chips, of which 6 are red, 7 are white and 3
are blue. If four chips are taken at random and without replacement,
Math 361, Problem set 3
Due 9/20/10
1. (1.4.21) Suppose a fair 6-sided die is rolled 6 independent times. A match
occurs if side i is observed during the ith trial, i = 1, . . . , 6.
(a) What is the p
Math 361, Problem set 4
Due 9/27/10
1. (1.4.26) Person A tosses a coin and then person B rolls a die. This is repeated independently until a head or one of the numbers 1, 2, 3, 4 appears,
at which tim
Midterm Exam I
Math 361 9/27/10 Name: Read all of the following information before starting the exam: READ EACH OF THE PROBLEMS OF THE EXAM CAREFULLY! Show all work, clearly and in order, if you want
Math 361, Problem set 5
Due 10/04/10
1. (1.6.8) Let X have the pmf p(x) = ( 1 )x , x = 1, 2, 3, . . . , and zero else2
where. Find the pmf of Y = X 3 .
1
Answer: Y has pmf p(x) = ( 2 )x/3 for x = 1, 8
Math 361, Problem set 6
Due 10/18/10
1. (1.8.3) Let X have pdf f (x) = (x + 2)/18 for 2 < x < 4, zero elsewhere.
Find E[X ], E[(X + 2)3 ] and E[6X 2(X + 2)3 ].
Answer:
E[X ] =
1
18
4
x2 + 2xdx =
2
4
4
Math 361, Problem set 11
Due 11/6/10
3
1. (3.4.32) Evaluate 2 exp(2(x 3)2 )dx - without a calculator. Use the
appendix table. Answer:
1
Note that if X has a N (3, 2 ) distribution then, X has pdf
2
fX
Math 361, Problem set 10
Due 11/6/10
1
1. Let p(x1 , x2 ) = 16 , x1 = 1, . . . , 4 and x2 = 1, . . . , 4, zero elsewhere, be the
joint pmf of X1 , . . . , X2 . Show that X1 and X2 are independent.
Ans
Math 361, Problem set 9
Due 11/8/10
1. (2.2.3) Let X1 and X2 have the joint pdf h(x1 , x2 ) = 2ex1 x2 , 0 < x1 <
x2 < , zero elsewhere. Find the joint pdf of Y1 = 2X1 and Y2 = X2 X1 .
Answer: We have
Math 361, Problem Set 2
November 4, 2010
Due: 11/1/10
1. (2.1.5) Given that the nonnegqative function g (x) has the property that
g (x)dx = 1, show that
0
f (x1 , x2 ) =
2g ( x2 + x2 )
1
2
x2 + x2
1
2