STAT 9715 / DR Gross
HW Assignment # 4:
Read from chapter 4: Sections 4.5 - end of chapter.
The examples are not nearly as onerous as some you have seen. Make your own selection.
Do problems: 4.17
Due: Thursday, sept 29.
4.19
4.25
4.31
4.45
4.47
4.60

3 y
1[e
( 3 y )0 3 y ( 3 y )1 3 y ( 3 y )2 3 y ( 3 y )3
+e
+e
+e
]
0!
1!
2!
3!
1( e3 y +3 y e3 y +
3 y
9 2
y e
2
3 y
+
9 3
y e
2
)
e
3 y
1[ 3 y +3 y e
y
9
+ y 2 e3 y ( 1+ y ) ]
2
y
Hence x e
3 3 x
0
(4)
34 3 3 x
dx= 4
x e dx
3 0 ( 4)
e
(4)
9
4 [1(

Dr. Gross
STAT 9715
Ross 8th Edition
HW #10
1) Problems from chapter 7
7.26
X1Xn ~ U(0,1), iid
Y = max(X1,Xn)
then 0<Y<1
(i) Use cdf method to find the distribution of Y for 0 < y< 1
FY ( y )=P [ Y y ] =P [ X 1 y , , X n y ]=by independenceof the X ' s
P

-1xy
dx
2
1
1
2
x dx +
(e)
0
1
0
6
f Y ( y )= f XY ( x , y ) dx=
7
0
It is clear that the f Y ( y ) 0 on [0,2]that it integrates 1.
2
2
3
6
6
6 2 6 2 8
2
ydy+ y dy =
+
=
21
28 0
21 2 28 3 7
2
E ( Y )=
0
6
xy
x + ) x+ y
(
)(
7
2
2
=
=
for x ( 0,1 )any

7.36
n rolls of a dice, X = # 1s Y=#2 s
Cov ( X , Y )=E ( XY ) E ( X ) E (Y )
n
E ( X )= =E (Y )
6
E( XY )=E [ XE ( Y | X ) ]
Because
and
1 nx
=
5
5
,
E (Y X=x)=E
(
Bin nx , p=
E ( XY )=E
[
)
1
E(Y X )= (n X)
5
]
(
2
1
n
1
n
1 1
1 5 n
X ( nX ) = E ( X )

11/1/15
STAT 9715
Dr. Gross
Cov ( X , Y )=
XY =
7
7 1 2 1
1
x = = ( 2120 )=
30
30 3 3 90
90
y
Cov ( X ,Y )
=
xy
1
90
( 451 )
2
=
45 1
=
90 2
6.23
f XY ( x , y )=12 xy ( 1x ) I ( y ( 0,1 ) ) I ( x ( 0,1 ) )
2 y I ( y ( 0,1 ) 6 x ( 1x ) I ( x ( 0,1 ) )
C

-1-
(a) Are X, Y independent? That is: Is f XY =f X f Y
y
?
x
e dy = x e I (0< x< )
x e( x+ y ) dy= x ex
0
f X ( x )=
0
= 1, because it is the integral of the Exp(1) pdf over its
range.
So X Gamma( =2, =1)
x
y
x e dx= e
( 2 ) =e
y
I ( y >0 )
( x+ y )
x

-1And for fixed
f Y |X =x ( y )=
x (0,1) and any
f XY ( x , y )
=
f X ( x)
y ( 0,1x)
24 xy
2y
=
0< y <1x
2
2
( 1x ) (1x )
24 x
2
Check f Y |X =x is a pdf on (0, 1-x):
1x
1x
f Y X =x ( y ) dy=
0
0
1x
So:
E ( Y |X =x ) =
0
2
2 ( 1x )
2y
1
dy=
=1
2
2
( 1x

1
=1
2
1
exp
Var
=1)=
because
and
Cov ( X , Y )=E ( XY ) E ( X ) E ( Y )=21=1
PART 2
(i)
f XY ( x , y )=C 1 x 3 e3 x C 2 e3 y for x >0 y >0
We can see that X, Y are independent and marginally
Y exp ( =3 ) so f Y X =x ( y )=3 e3 y I ( y> 0)
X Gamma( =4,

-1-
Answer Set #9
(Chapter 6 Bivariate, Continuous)
This answer set was recently typed. If you find any typos or mis-calculations, please report them
to me. Thank you.
y
6.9
2
6
xy
f XY ( x , y )= x 2+
I (0< x< 1,0< y< 2)
7
2
(
)
1
(a) To verify this is a

The entire assignment is due on Tuesday November 19 and
will be collected.
Please retain a copy.
HW # 7 - TRANSFORMATIONS
Read section 5.7. Use the CDF method we have already (11.11.14)
done in class to find the distributions and densities in the
followin

HW # 10
Dr Gross / STAT 9715
This homework covers:
1. Expectations and Conditional Expectations from Chapter 7 and our discrete and
continuous Bivariate notes.
2. Independence (Section 6.2 w/o ex 2d 2e, 2i, 2j; 2g optional) (Parts of Section 6.3)
3. Momen

HW # 6 - Continuous distributions
DUE THUSDAY 10.30.2014
Read Sections 5.1 - 5.6 as you do this homework assignment. This chapter covers a fair number of
continuous distributions that are all essential to using probability theory in applications.
Introduc

DR Gross / STA 9715 Ross 8th edition
HW 9 : Bivariate continuous distributions.
The following problems are concerned with continuous bivariate distributions from
chapter 6, omitting section 6.6 on order statistics, and not including section 6.7:
6.9
Add t

Review/Preview
1) "Bernoulli" Trials (Independent repeated trials)
Can result in A, B, or C (neither A nor B)
eg.
Throw two dice repeatedly
A = sum of 7
B = sum of 11
C = neither A nor B
P(A) = 6/36
P(B) = 2/36
P( C) = 1 - P(A) - P(B) = 28/36
Question: P[

HW # 6 - Continuous distributions
DUE THUSDAY Nov 3, 2016
Read Sections 5.1 - 5.6 as you do this homework assignment. This chapter covers a fair number of
continuous distributions that are all essential to using probability theory in applications.
Introdu

NOV 13, 2014 STAT 9715
Assignment # 8: Joint discrete Distributions Ross 8th ed. The answer set
will be discussed and posted 11/20/13 as this material will be covered by
the take-home midterm you will receive on Nov 25 to be submitted Dec
3.
_
Please read

STAT 9715 HW Assignment # 3:
Read from chapter 4: Sections 4.1-4.4.
The examples are not nearly as onerous as some you have seen. Make your own selection. Do
problems: 4.1 4.2
(4.7 4.8) together
For all problems compute E(X).
Dr Gross
4.13 4.14

Advanced Data Mining
Kamiar Rahnama Rad
lecture 1
chapter 1
statistical learning:
supervised: the presence of the outcome variable guides
the learning process, such as regression and classification.
unsupervised: we observe only the features and have n

Dr Gross / 15.08.28
Assignment # 1. Chapter 1 and 2 of Ross, 8th
edition.
In chapter 1: Read the chapter. You may skip the introduction to the counting problem in
proposition 6, but we shall be doing propositions 6.1 6.2, and the examples that follow.
Exa

1
th
Assignments # 2. Chapter 3 of Ross, 8 edition.
DR GROSS / Sept 15-17, 2015
Chapter 3: This chapter is particularly rich in examples. I shall try to indicate which
ones are less essential for you at this time. The examples are of course numbered as in

Dr. Gross
The jumps of F are at b = 1, 2, 3
(4.17)
a)
STA 9715
HW #4
P(X = 1) = F(1) F(1) =
=
P(X = 2) = F(2) F(2) =
=
=
P(X = 3) = F(3) F(3) = 1
but P(X = 1) + P(X = 2) + P(X = 3) =
+
+
=
1 because F is not purely discrete
b)
(4.19)
P(
) = F( ) F(
< X

1
DR GROSS / STAT 9715/ROSS 8th Ed
Answer set # 7 :
fX
U (-1, 1)
37. X
(a) . P
(|X|> 12 )
= P
( X < 12 )
(x) =
+P
1
2
for
1<x<1
( X > 12 )
=+ =
(b) . Let Y = |X|
support: 0 < y <1
2y
2
FY (y) = P (Y y) = P (|X| y ) = P ( y < X < y =
F'Y (y) = 1 for 0 < y

Ch. 4
Discrete r.v.
STA 9715
Ross 8th ed.
HW #3
Please report typing and other mistakes!
(4.1)
8W 1$
4B +2$
2O 0$
n = 2 draws w/o replacement
X = total winnings
2 X 4
Player gets +$2 for each black ball and -$1 for each white ball.
P(X = 4) = P(2B) =
P(

Dr. Gross
(4.71)
STA 9715
HW #5
Roulette wheel: #s 136, 0, double 0
Smith bet: # between 112
F = # not in 112
a)
P[FFFFF] = P(F)5 by independence
= (26/38)5 = 0.1500
b)
P(X = 4) = q3p
with p = 12/38;
= (26/38)3(12/38) = 0.1012
X = # bets to obtain 1st win