Download PSTAT 120B Probability and Statistics - Week 3

PSTAT 120B Probability and Statistics - Week 3
Fang-I Chu
University of California, Santa Barbara
April 17, 2013
Fang-I Chu
PSTAT 120B Probability and Statistics
Announcement
Office hour: Tuesday 11:00AM-12:00PM (Current instructor
asks me to hold only 1 hour for office hour)
I am very reachable by email. Feel free to schedule
appointment with me when you need help!
email: [email protected]
Bring two blue books next week (Collection for your midterm
and final)
You should have received at least one group email from me by
now. Put your name and email address on the back of roster
sheet if you are not on the group email list.
Fang-I Chu
PSTAT 120B Probability and Statistics
Topic for review
Transformation methods
Cumulative distribution function approach
Probability density function formula approach
Exercise #6.23 (a),(c)
Exercise #6.34
Guideline/skill of doing a proof
Moment Generating Function
Exercise #6.46
Hint for homework problem 1.
Hint for homework problem 4.
Fang-I Chu
PSTAT 120B Probability and Statistics
Cumulative distribution function approach
The cumulative distribution function approach can only be
applied on continuous cases.
The obtained results by using CDF approach and
transformation formula above are identical.
The CDF approach begins with working out the desired CDF,
say FU (u), write in the form of CDF for Y . Obtain pdf of
FU (u) by taking differentiation.
Remember to take care of the range of variables.
Information can be found at page 304.
Fang-I Chu
PSTAT 120B Probability and Statistics
Probability density function formula approach
assumption: Let Y have probability density function fY (y ),
and h(y ) is either increasing or decreasing for all y such that
fY (y ) > 0.
content: Denote U = h(Y ), then we have density function for
U,
dh−1
fU (u) = fY (h−1 (u)) · |
|
du
−1
d [h−1 (u)]
note dhdu =
.
du
Information can be found at page 316.
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 6.23
6.23
In exercise 6.1, we considered a random variable Y with probability
density function given by
2(1 − y )
0≤y ≤1
fY (y ) =
0
elsewhere
and used the method of distribution functions to find the density
functions of
(a)U1 = 2Y − 1
(c)U3 = Y 2
Fang-I Chu
PSTAT 120B Probability and Statistics
#6.23(a)-CDF approach
#6.23(a)
U1 = 2Y − 1
Solution:
Known:
fY (y ) = 2(1 − y ) for 0 ≤ y ≤ 1
U1 = 2Y − 1
Goal: Use CDF approach find fU1 (u1 ).
Way to approach:
FU1 (u1 ) = P(U1 ≤ u) = P(2Y − 1 ≤ u1 ) = P(Y ≤ u12+1 ) =
FY ( u12+1 )
1
fU1 (u1 ) = dud 1 FY ( u12+1 ) = 21 fY ( u12+1 ) = 12 · 2(1 − u12+1 ) = 1−u
2
for −1 ≤ u ≤ 1
Fang-I Chu
PSTAT 120B Probability and Statistics
#6.23(a)-pdf formula approach
Solution:
Known:
fY (y ) = 2(1 − y ) for 0 ≤ y ≤ 1
U1 = 2Y − 1
Goal: Use pdf formula approach find fU1 (u1 ).
Way to approach:
dy
Y = U12+1 , du
= 12
1
Apply the formula: fU1 (u1 ) =
−1 ≤ u ≤ 1
Fang-I Chu
1
2
· 2(1 −
u1 +1
2 )
=
1−u1
2
PSTAT 120B Probability and Statistics
for
#6.23(c)-CDF approach
#6.23(c)
U3 = Y 2
Solution:
Known:
fY (y ) = 2(1 − y ) for 0 ≤ y ≤ 1
U3 = Y 2
Goal: Use CDF approach find fU3 (u3 ).
Way to approach:
√
FU3 (u3 ) = P(U3 ≤ u3 ) = P(Y 2 ≤ u3 ) = P(Y ≤ u3 ) =
√
FY ( u3 )
√
√
√
fU3 (u3 ) = dud 3 FY ( u3 ) = 2√1u3 fY ( u3 ) = √1u3 · 2(1 − u3 ) =
√
1− u3
√
u3
for 0 ≤ u ≤ 1
Fang-I Chu
PSTAT 120B Probability and Statistics
#6.23(c)-pdf formula approach
Solution:
Known:
fY (y ) = 2(1 − y ) for 0 ≤ y ≤ 1
U3 = Y 2
Goal: Use pdf formula approach find fU3 (u3 ).
Way to approach:
Y =
√
U3 ,
dy
du3
=
1
√
2 u3
Apply the formula: fU3 (u3 ) =
0≤u≤1
Fang-I Chu
1
√
2 u3
· 2(1 −
√
u3 ) =
√
1− u3
√
u3
PSTAT 120B Probability and Statistics
for
Exercise 6.34
6.34
A density function sometimes used by engineers to model lengths
of life of electronic component is the Rayleigh density, given by
(
−y 2
2y
θ
)e
(
y >0
θ
f (y ) =
0
elsewhere
(a) If Y has the Rayleigh density, find the probability density
function for U = Y 2 .
(b) Use the result of part (a) to find E (Y ) and V (Y ).
Fang-I Chu
PSTAT 120B Probability and Statistics
#6.34(a)-pdf formula approach
#6.34(a)
If Y has the Rayleigh density, find the probability density function
for U = Y 2 .
Solution:
Known:
fY (y ) = ( 2y
θ )e
U = Y2
−y 2
θ
for y > 0
Goal: Use pdf formula approach find fU (u).
Way to approach:
Y =
√
U,
dy
du
=
1
√
2 u
√
−u
Apply the formula: fU (u) = 2√1 u · ( 2 θ u )e θ = θ1 e
Recognize U follows exponential with mean θ
Fang-I Chu
−u
θ
PSTAT 120B Probability and Statistics
for u > 0
#6.34(b)
#6.34(b)
(b) Use the result of part (a) to find E (Y ) and V (Y ).
Solution:
Known:
U follows exponential with parameter θ
Goal: Find E (Y ) and V (Y )
Way to approach:
√
√
R∞ 1
u
E (Y ) = E ( U) = 0 u 2 θ1 e − θ du = 2πθ
E (Y 2 ) = E (U) = θ (use formula √
for exponential distribution)
2
2
V(Y ) = E (Y ) − E (Y ) = θ − ( 2πθ )2 = θ 1 − π4
Fang-I Chu
PSTAT 120B Probability and Statistics
guideline/skill of doing a proof
When to use this guideline?
Any question asks you to show/to argue or to prove a
statement
Layout your 4 steps
(1) What information we have on hand
write down known/given conditions
(2) What do we want to prove
write down your goal
(3) How does (2) connect to (1)
find possible clues to build up the bridge of known and goal
(4) Fine tune your (3)
You got your proof!
Fang-I Chu
PSTAT 120B Probability and Statistics
Moment generating function (MGF)
Definition: Let X be a random variable with cdf FX , the
moment generating function (mgf) of X , denoted by
MX (t) = Ee tX
Note: Method of MGF is always the method you should
attempt first when dealing with sums of random variables.
Useful/Special MGF functions:
n
Binomial: MX (t) = [pe t + (1 − p)]
pe t
Geometric: MX (t) = 1−(1−p)e
t
Poisson: MX (t) = exp [λ(e t − 1)]
p2
1
Chi squared(p): MX (t) = 1−2t
,t<
µt
1
2
e
Double exponential: MX (t) = 1−(σt)
2 , |t| <
1
1
Exponential: MX (t) = 1−βt , t < β
α
1
Gamma: MX (t) = 1−βt
h
i
2 2
Normal: MX (t) = exp µt + σ 2t
Fang-I Chu
1
σ
PSTAT 120B Probability and Statistics
Exercise 6.46
6.46
Suppose that Y has a gamma distribution with α = n2 for some
positive integer n and β equal to some specified value. Use the
method of moment-generating functions to show that W = 2Y
β has
a χ2 distribution with n degrees of freedom.
Proof:
1. Information:
definition of MGF: mW (t) = E (e Wt )
W = 2Y
β
2. Goal: Show W has a χ2 distribution with n degrees. i.e.
n
mW (t) = (1 − 2t)− 2
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 6.46
6.46
Suppose that Y has a gamma distribution with α = n2 for some
positive integer n and β equal to some specified value. Use the
method of moment-generating functions to show that W = 2Y
β has
2
a χ distribution with n degrees of freedom.
Solution:
3. Bridge:
2Y
E (e Wt ) = E (e ( β )t ) = mY ( 2t
β)
MGF for Y as gamma distribution with parameters ( n2 , β):
n2
1
mY (t) = 1−βt
4. Fine tune:
mY ( 2t
β)
=
1
1−β· 2t
β
n
2
n
= (1 − 2t)− 2 , same MGF
for χ2 distribution with n degrees!
Fang-I Chu
PSTAT 120B Probability and Statistics
#1
#1
Let X be a continuous random variable with pdf fX (x). Let
Y = X 2 . For y > 0, use the cdf method to prove that
√
√
1
f (y ) = √ (fX ( y ) + fX (− y )).
2 y
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1
Proof:
(1) Information:
X is continuous random variable with pdf fX (x) for
−∞ < x < ∞.
define Y = X 2
(2) Goal: Show f (y ) =
approach
(3) Bridge:
√
1
√
2 y (fX ( y )
√
+ fX (− y )) use CDF
FY (y ) = P(Y ≤ y ) = P(X 2 ≤ y ) = P(X ≤?)
Take derivative of FY (y ) to obtain fy (y )
(4) Fine tune: you could wrap it up on your own!
Note: the range of x: −∞ < x < ∞
Fang-I Chu
PSTAT 120B Probability and Statistics
#4
#4
Write out the entire answer (including both the part done in class
and the remaining cases that you complete on your own) for the
following problem begun in lecture on Tuesday April 9:
If X and Y have joint pdf
2(x + y )
0≤x ≤y ≤1
fX ,Y (x, y ) =
0
elsewhere
Find fZ (z) where Z = X + Y
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #4
Hint:
Known:
fX ,Y (x, y ) = 2(x + y ) for 0 ≤ x ≤ y
denote Z = X + Y
Goal: Find fZ (z)
Way to approach:
F (z) = P(Z ≤ z) = P(X + Y ≤ z) = P(X ≤ z − Y ) =
R Zb R d
f (x, y )dydx
a c X ,Y
determine a, b, c, d by drawing graph.
P(X ≤ z − Y ) is the area formed by x + y ≤ z, 0 ≤ x ≤ 1,
0 ≤ y ≤ 1 and x ≤ y
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #4
Note:
to evaluate the area, you could break the whole area into
subareas, and use the boundary points to compute the
sub-areas. The value of the integral is the sum of subareas.
you can check whether your obtained fZ (z) is correct or not
R2
by confirming 0 fZ (z)dz = 1. (0 ≤ z ≤ 2, why?)
Your final answer of fZ (z) should be a function of only z,
contains no x or y .
When write down double integral, you could either put dy or
dx as inner intergrator. The order of putting dy or dx
sometimes depends on the situation. The range for x and y
for these two cases would be different (if not symmetric
case).The evaluated results should agree.
Fang-I Chu
PSTAT 120B Probability and Statistics