Example of distribution of two continuous random variables
Let the joint p.d.f. of X and Y be
3
f ( x, y ) x 2 (1 | y |), - 1 x 1. - 1 y 1.
2
(a)
(b)
(c)
(d)
What is F(0.5,0.5)=P(X0.5,Y0.5)?
What are the marginal p.d.f. of X and Y?
What is the conditional
Example of distribution of two continuous random variables
Letthejointp.d.f. ofXand Ybe f(x,y) =%x2(1h|yl), -1<x<1. -1<y<1.
(a) What is F(0.5,0.5)=P(XSO.5,YSO.5)?
(b) What are the marginal p.d.f. of X and Y?
(c) What is the conditional probability densi
Supplementary for 19/10/2016
1. For Uniform distribution, it is certainly much easier and more recommended to find the moments
by the following direct evaluation:
b
1 n
1
E ( X )
x dx
(b n 1 a n 1 )
b a
(b a )( n 1)
a
n
Therefore, for example,
1
a b
E(
Explanation of the main result in section 5.2 of lecture notes:
Look at the diagram below, we match the following small probabilities:
Prcfw_a A a da, b B b db f A, B (a, b)dadb
and
Prcfw_( X , Y ) Blue _ area f X ,Y ( x, y ) area _ of _ blue _ area
To fi
4.12
(6) Independent, because fx(w)fy(y) = f(1',y)-
4-18 Probability of men who ptefet to date non-smokers (P(NS) = 0.63
Probability of men who prefer to date smokers (P(S) = 0.13
Probability of men who don't can: (P(DC) = 0.24
(a) Jomt pmf of 1' and 11
An interesting example of modeling using conditional distribution:
Q: An insect lays a large number of eggs, each surviving with probability p. On the average,
how many eggs will survive?
We assume the number of eggs laid is a random variable following Po
STAT2001 Tutorial 2 Solution
Sep. 23th, 2016
1. Sample Space = cfw_ HT, TH, HH, TT
A = cfw_ HT, HH
B= cfw_ HH, TH
C = cfw_ HH, TT
A B = A C = B C = cfw_ HH
A B C = cfw_ HH
P( A) = P( B) = P(C ) =
1
2
P( A B) =
1
= P( A) P( B)
4
P( A C ) =
1
= P( A)
Chapter 3: Continuous Distributions
STAT2001
2016 Term I
Outline
1. Continuous random variables
2. The Uniform distribution and Exponential distribution
3. The Gamma distribution and Chi-square distribution
4. The Normal distribution
(Textbook chapters: 3
Assignment 4 Solution
1(a)
1 1
ce
x y
dydx 1
0 x
1
1
ce x e y dydx 1
0
x
1
ce x (e e x )dx 1
0
e2 1
1
] 1 c 2
0.677394
e
1
2 2
e
2
2
c[e 2 e
(b) No. Because the support is not rectangular.
(c)
1
f X ( x) ce x e y dy ce x (e e x ) c (e x 1 e 2 x ),0
In last class, we have introduced COVARIANCE and CORRELATION COEFFICIENT (or
simply CORRELATION, usually denoted as ) as measures for LINEAR relationship
between 2 random variables. I would like to give more remarks here:
1. A proof of -11:
Consider 2 ran
STAT 2001A / 2001B
Basic Concepts in Statistics and Probability I
2016 Term I
1.
Description:
This course is designed to study the basic concepts of probability and statistics. Topics include elementary
probability, Bayes theorem, random variables, distri
Chapter 5: Distributions of functions of random variables
STAT2001
2016 Term I
Outline
1. Functions of one random variable
2. Transformations of two random variables
3. Several independent random variables
4. Random functions associated with Normal distri
Moments and moment generating function are completely new to most of you. Therefore it is
worthwhile to spend more time and use more examples to understand them.
1. Moments:
Firstly, we explained mean, variance, skewness, are important quantities that can
1.1-2 Probability of insuring exactly 1 our. HA) = 0.]
Probability of mnunng more than I eat. H3) = 0.9
Pxobnbilily of insuring a sperm cu. P(C') = 0.25
P(B n C) = 0.10
P(AnC')=P(C)-P(Bn() =0.l = P(A)
PM A 0') = o.
1.1-4 (a) 82 cfw_1111111111. HHHHT. HHHT
STAT2001 Assignment 4
Do all 6 questions. Show your steps clearly.
Deadline for this assignment is 18th Nov. 5:00p.m. You can submit to the assignment locker
(next to LSB 125) or to your Tutors.
Q1. The random variables X and Y have the joint probability
A summary of the special distributions discussed:
For discrete distributions, we have learnt Binomial distribution, Hypergeometric
distribution, Poisson distribution, Geometric distribution, Negative Binomial distribution.
For continuous distributions, we
1. Example 3 on P.13 of chapter 5 notes can also be solved using distribution-function technique:
Y1
X1
, Y2 X 2 ;
X2
f X1 , X 2 ( x1 , x2 ) 2,0 x1 x2 1,
FY1 ,Y2 ( y1 , y 2 )
P (Y1 y1 , Y2 y2 )
P(
X1
y1 , X 2 y2 )
X2
P ( X 1 y1 X 2 , X 2 y2 )
y2 y1x2
5.12 Herew=\/;(7,$=Iand0<w<oomapsont00<y<oo. Thus
2f
g(y)= J37
e y/Q, 0<y<oo.
2ng=
5.1-8 2: = (9W7
1: 3
.17 $03) "(213)
e", 0<z<oo
3/7
W'(%)(%) W
510/7 9 y
a,
A
1.
ll
Q
A
Q
v
II
, 0<y<oo.
5.110 Since-1<z<3.wehavc0$y<9.
When0<y<Lthen
l
xl_. y, _2y-' $2.
Solution for STAT2001 2016 Midterm
1. (a) Let X1 denote the number of babies born during next 5 hours, then
X1 P oisson(1 ), 1 =
5
3 = 0.625
24
So,
P (X1 2) = P (X1 = 0) + P (X1 = 1) + P (X1 = 2)
= e0.625 +
e0.625 0.625 e0.625 0.6252
+
1!
2!
= 0.9743.
(b
1, Applying Expected Value calculation in a decision making problem:
The manager of a bakery knows that the number of chocolate cakes he can sell on any given
day is a random variable with probability mass function
f ( x)
1
6
for x 0,1,2,3,4,5.
He also k
In some settings, random variables can be intuitively considered independent. For example, you roll a
dice twice and let X to be the result of the first throw and Y to be the result of the second throw. In
some modelling exercises, it is also common to as
STAT2001 Tutorial 5 solution
Oct. 18, 2016
Exercise 1
Y B(2000, 0.001)
For n is large and p is small, we could use Poisson distribution to approximate
Binomial distribution. Thus = 2000 0.001 = 2, Y P oisson(2).
P (Y 4) = P (Y = 0) + P (Y = 1) + . + P (Y
STAT2001 Tutorial 4 Solution
Oct. 11, 2016
Exercise 1
(a)
X
P (X = k) = 1
k1
c e2
X e2 2k
k1
c e2
X e2 2k
k!
k0
k!
=1
2 0
e
2
=1
e2
0!
c(e2 1) = 1 c =
e2
1
1
(b)
M (t) = E (exp(tx) = c
X (2et )k
k1
= c
X (2et )k
k!
k0
t
= ce2e
1
X e2et (2et )k
k0
k!
k!
STAT2001 Assignment 5
Deadline for this assignment is 2nd Dec. 5:00p.m. You can submit to the assignment
locker (next to LSB 125), to Blackboard system or to your TAs.
1.
Find the pdf of Y=exp(X) when X follows a normal distribution with mean
and standar
Solution to assignment 1
1
(a) There are 8! = 40320 ways.
(b) There are 2 x 7! = 10080 ways.
(c) There are 2 x 4! x 4! = 1152 ways
(d) There are 4! x 5! = 2880 ways
(e) There are 4! x 2^4 = 384 ways
2.
(a) There are 2 acceptable answers for this part:
(i)
The derivations for example 17 on the lecture notes involve more mathematical steps. Therefore I type
this note as supplementary. There are two methods to find the m.g.f.:
Method 1:
Make use of the following mathematical result:
k 1 k r
w , 1 w 1.
(1 w)
STA2001 Tutorial 6
Oct. 28, 2010
Attention: Question 3 is modied to avoid ambiguity. Students who have
attended the tutorials, especially at the rst time slot, please refer to the
explanation in the solution.
1. Trac on a road is passing a point at rate o