16: Continuous Joint DistributionsLisa Yan
May 11, 2020
1
Lisa Yan, CS109, 2020
Quick slide reference
2
3 Continuous joint distributions 16a_cont_joint
18 Joint CDFs 16b_joint_CDF
23 Independent continuous RVs 16c_indep_cont_rvs
28 Multivariate Gaussian RVs 16d_sum_normal
32 Exercises LIVE
59 Extra: Double integrals 16f_extra
Continuous joint distributions
3
16a_cont_joint
Lisa Yan, CS109, 2020
Remember target?
4
Good timesβ¦
Lisa Yan, CS109, 2020
CS109 logo with darts
5
Quick check: What is the probability that a dart
hits at (456.2344132343, 532.1865739012)?
The CS109 logo was created by throwing 500,000 darts according to a joint distribution.
If we throw another dart according to the same distribution, what is
P(dart hits within π pixels of center)?
Lisa Yan, CS109, 2020
CS109 logo with darts
6
1 pixel = 1 dart thrown
at screen
Possible dart counts (in 100x100 boxes)
P(dart hits within π pixels of center)?
Lisa Yan, CS109, 2020
CS109 logo with darts
7
Possible dart counts (in 50x50 boxes)
P(dart hits within π pixels of center)?
Lisa Yan, CS109, 2020
CS109 logo with darts
8
Possible dart counts(in infinitesimally small boxes)
P(dart hits within π pixels of center)?
Lisa Yan, CS109, 2020
Continuous joint probability density functions
If two random variables π and π are jointly continuous, then there exists a joint probability density function ππ,π defined over ββ < π₯, π¦ < β such that:
π π1 β€ π β€ π2, π1β€ π β€ π2 = ΰΆ±π1
π2
ΰΆ±π1
π2
ππ,π π₯, π¦ ππ¦ ππ₯
9
Lisa Yan, CS109, 2020
From one continuous RV to jointly continuous RVs
Single continuous RV π
β’ PDF ππ such that βββππ π₯ ππ₯ = 1
β’ Integrate to get probabilities
Jointly continuous RVs π and π
β’ PDF ππ,π such thatββββββππ,π π₯, π¦ ππ¦ ππ₯ = 1
β’ Double integrate to get probabilities
10
Probability for jointly continuous RVs is volume under a surface.
Probability = area
under curve
Lisa Yan, CS109, 2020
Double integrals without tears
Let π and π be two continuous random variables.
β’ Support: 0 β€ π β€ 1, 0 β€ π β€ 2.
Is π π₯, π¦ = π₯π¦ a valid joint PDF over π and π?
Write down the definite double integral that
must integrate to 1:
π€
Lisa Yan, CS109, 2020
Double integrals without tears
Let π and π be two continuous random variables.
β’ Support: 0 β€ π β€ 1, 0 β€ π β€ 2.
Is π π₯, π¦ = π₯π¦ a valid joint PDF over π and π?
Write down the definite double integral that
must integrate to 1:
ΰΆ±π¦=0
2
ΰΆ±π₯=0
1
π₯π¦ ππ₯ ππ¦ = 1 or ΰΆ±π₯=0
1
ΰΆ±π¦=0
2
π₯π¦ ππ¦ ππ₯ = 1
(used in next slide)
Lisa Yan, CS109, 2020
Double integrals without tears
Let π and π be two continuous random variables.
β’ Support: 0 β€ π β€ 1, 0 β€ π β€ 2.
Is π π₯, π¦ = π₯π¦ a valid joint PDF over π and π?
13
1 = ΰΆ±ββ
β
ΰΆ±ββ
β
π π₯, π¦ ππ₯ ππ¦ = ΰΆ±π¦=0
2
ΰΆ±π₯=0
1
π₯π¦ ππ₯ ππ¦
ΰΆ±π¦=0
2
ΰΆ±π₯=0
1
π₯π¦ ππ₯ ππ¦ = ΰΆ±π¦=0
2
π¦ ΰΆ±π₯=0
1
π₯ ππ₯ ππ¦ = ΰΆ±π¦=0
2
π¦π₯2
20
1
ππ¦ = ΰΆ±π¦=0
2
π¦1
2ππ¦
ΰΆ±π¦=0
2
π¦1
2ππ¦ =
π¦2
4π¦=0
2
= 1 β 0 = 1
1. Evaluate inside integral by treating π¦ as a constant:
2. Evaluate remaining (single) integral:
Yes, π π₯, π¦ is a valid joint PDF
because it integrates to 1.
0. Set up integral:
Lisa Yan, CS109, 2020
Marginal distributions
Suppose π and π are continuous randomvariables with joint PDF:
The marginal density functions (marginal PDFs) are therefore:
ππ π = ΰΆ±ββ
β
ππ,π π, π¦ ππ¦ ππ π = ΰΆ±ββ
β
ππ,π π₯, π πx
14
ππ π₯ΰΆ±ββ
β
ΰΆ±ββ
β
ππ,π π₯, π¦ ππ¦ ππ₯ = 1
Lisa Yan, CS109, 2020
Back to darts!
Match π and π to their respective marginal PDFs:
15
(top-down)
(side view)
π€
Lisa Yan, CS109, 2020
Back to darts!
Match π and π to their respective marginal PDFs:
16
pixel x pixel y
(top-down)
(side view)
Lisa Yan, CS109, 2020
Extra slides
If you want more practice with double integrals,
Iβve included two exercises at the end of this lecture.
17
Joint CDFs
18
16b_joint_cdfs
Lisa Yan, CS109, 2020
An observation: Connecting CDF to PDF
For a continuous random variable π with PDF π, the CDF (cumulative distribution function) is
πΉ π = π π β€ π = ΰΆ±ββ
π
π π₯ ππ₯
The density π is therefore the derivative of the CDF, πΉ:
π π =π
πππΉ π
19
(Fundamental Theorem
of Calculus)
Lisa Yan, CS109, 2020
Joint cumulative distribution function
For two random variables π and π, there can be a joint cumulative distribution function πΉπ,π:
πΉπ,π π, π = π π β€ π, π β€ π
20
For continuous π and π:
πΉπ,π π, π = ΰΆ±ββ
π
ΰΆ±ββ
π
ππ,π π₯, π¦ ππ¦ ππ₯
ππ,π π, π =π2
ππ πππΉπ,π π, π
For discrete π and π:
πΉπ,π π, π =
π₯β€π
π¦β€π
ππ,π(π₯, π¦)
Lisa Yan, CS109, 2020
Single variable CDF, graphically
21
limπ₯βββ
πΉX π₯ = 0
limπ₯β+β
πΉπ π₯ = 1
πΉπ π₯ = π π β€ π₯ππ π₯
Review
Lisa Yan, CS109, 2020
Joint CDF, graphically
22
limπ₯,π¦βββ
πΉπ,π π₯, π¦ = 0
limπ₯,π¦β+β
πΉπ,π π₯, π¦ = 1
πΉπ,π π₯, π¦ = π π β€ π₯, π β€ π¦ππ,π π₯, π¦
Independent Continuous RVs
23
16c_indep_cont_rvs
Lisa Yan, CS109, 2020
Independent continuous RVs
Two continuous random variables π and π are independent if:
π π β€ π₯, π β€ π¦ = π π β€ π₯ π π β€ π¦
Equivalently:
πΉπ,π π₯, π¦ = πΉπ π₯ πΉπ π¦ππ,π π₯, π¦ = ππ π₯ ππ π¦
Proof of PDF:
24
ππ,π π₯, π¦ =π2
ππ₯ ππ¦πΉπ,π π₯, π¦ =
π2
ππ₯ ππ¦πΉπ π₯ πΉπ π¦
=π
ππ₯
π
ππ¦πΉπ π₯ πΉπ π¦
= ππ π₯ ππ π¦
=π
ππ₯πΉπ π₯
π
ππ¦πΉπ π¦
βπ₯, π¦
βπ₯, π¦
Lisa Yan, CS109, 2020
Independent continuous RVs
Two continuous random variables π and π are independent if:
π π β€ π₯, π β€ π¦ = π π β€ π₯ π π β€ π¦
Equivalently:
πΉπ,π π₯, π¦ = πΉπ π₯ πΉπ π¦ππ,π π₯, π¦ = ππ π₯ ππ π¦
More generally, π and π are independent if joint density factors separately:
ππ,π π₯, π¦ = π π₯ β π¦ , where ββ < π₯, π¦ < β
25
Lisa Yan, CS109, 2020
Pop quiz! (just kidding)
Are π and π independent in the following cases?
26
independent
π and π
ππ,π π₯, π¦ = π π₯ β π¦ ,
where ββ < π₯, π¦ < β
1. ππ,π π₯, π¦ = 6πβ3π₯πβ2π¦
where 0 < π₯, π¦ < β
2. ππ,π π₯, π¦ = 4π₯π¦where 0 < π₯, π¦ < 1
3. ππ,π π₯, π¦ = 24π₯π¦where 0 < π₯ + π¦ < 1
π€
Lisa Yan, CS109, 2020
Pop quiz! (just kidding)
Are π and π independent in the following cases?
27
1. ππ,π π₯, π¦ = 6πβ3π₯πβ2π¦
where 0 < π₯, π¦ < β
2. ππ,π π₯, π¦ = 4π₯π¦where 0 < π₯, π¦ < 1
3. ππ,π π₯, π¦ = 24π₯π¦where 0 < π₯ + π¦ < 1
π π₯ = 3πβ3π₯
β π¦ = 2πβ2π¦
π π₯ = 2π₯β π¦ = 2π¦
Cannot capture constraint on π₯ + π¦into factorization!
β
β
β
independent
π and π
ππ,π π₯, π¦ = π π₯ β π¦ ,
where ββ < π₯, π¦ < β
Separable functions:
Separable functions:
If you can factor densities over all of the
support, you have independence.
Bivariate Normal Distribution
28
16d_bivariate_normal
Lisa Yan, CS109, 2020
Bivariate Normal Distribution
π1 and π2 follow a bivariate normal distribution if their joint PDF π is
π π₯1, π₯2 =1
2ππ1π2 1 β π2πβ
12 1βπ2
π₯1βπ12
π12 β
2π π₯1βπ1 π₯2β π2π1π2
+π₯2βπ2
2
π22
Can show that π1~π© π1, π12 , π2~π© π2, π2
2
Often written as: πΏ~π©(π, πΊ)β’ Vector πΏ = π1, π2
β’ Mean vector π = π1, π2 , Covariance matrix: πΊ =π12 ππ1π2
ππ1π2 π22
29
We will focus on understanding the
shape of a bivariate Normal RV.Recall correlation: π =
Cov π1,π2
π1π2
(Ross chapter 6, example 5d)
Lisa Yan, CS109, 2020
Back to darts
30
(top-down)
(side view)
These darts were actually thrown according
to a bivariate normal distribution:
π = 450, 600
πΊ =9002/4 0
0 9002/25π, π ~π© π, πΊ
pixel x pixel y
Marginal
PDFs:π~π© 450,
9002
4π~π© 600,
9002
25
Lisa Yan, CS109, 2020
A diagonal covariance matrix
Let πΏ = π1, π2 follow a bivariate normal distribution πΏ~π©(π, πΊ), where
π = π1, π2 , πΊ =π12 0
0 π22
Are π1 and π2 independent?
31
β
π π₯1, π₯2 =1
2ππ1π2 1 β π2πβ
12 1βπ2
π₯1βπ12
π12 β
2π π₯1βπ1 π₯2β π2π1π2
+π₯2βπ2
2
π22
=1
2ππ1π2πβ12
π₯1βπ12
π12 +
π₯2βπ22
π22
(Note covariance: ππ1π2 = 0)
=1
π1 2ππβ π₯1βπ1
2/2π12 1
π2 2ππβ π₯2βπ2
2/2π22
π1 and π2 are independent
with marginal distributions
π1~π© π1 π12 , π2~π©(π2 π2
2)
(live)16: Continuous Joint Distributions (I)Slides by Lisa Yan
July 24, 2020
32
Lisa Yan, CS109, 2020
Jointly continuous RVs
π and π are jointly continuous if they have a joint PDF:
ππ,π π₯, π¦ such that ΰΆ±ββ
β
ΰΆ±ββ
β
ππ,π π₯, π¦ ππ¦ ππ₯ = 1
Most things weβve learned about discrete joint distributions translate:
33
Marginal
distributionsππ π = ΰΆ±
ββ
β
ππ,π π, π¦ ππ¦
Independent RVs ππ,π π₯, π¦ = ππ π₯ ππ π¦ ππ,π π₯, π¦ = ππ π₯ ππ π¦
ππ π =
π¦
ππ,π π, π¦
Expectation
(e.g., LOTUS)πΈ π π, π =
π₯
π¦
π π₯, π¦ ππ,π π₯, π¦ πΈ π π, π = ΰΆ±ββ
β
ΰΆ±ββ
β
π π₯, π¦ ππ,π π₯, π¦ ππ¦ ππ₯
β¦etc.
Review
ThinkSlide 35 has a question to go over by yourself.
Post any clarifications here!
https://us.edstem.org/courses/109/discussion/60584
Think by yourself: 2 min
34
π€(by yourself)
Lisa Yan, CS109, 2020
Warmup exercise
π and π have the following joint PDF:
1. Are π and π independent?
2. What is the marginalPDF of π? Of π?
3. What is πΈ π + π ?
35
π€(by yourself)
ππ,π π₯, π¦ = 3πβ3π₯
where 0 < π₯ < β, 1 < π¦ < 2
Lisa Yan, CS109, 2020
Warmup exercise
π and π have the following joint PDF:
1. Are π and π independent?
2. What is the marginalPDF of π? Of π?
3. What is πΈ π + π ?
36
π π₯ = 3πΆπβ3π₯ , 0 < π₯ < ββ π¦ = 1/πΆ, 1 < π¦ < 2
πΆ is a
constantβ
ππ,π π₯, π¦ = 3πβ3π₯
where 0 < π₯ < β, 1 < π¦ < 2
π₯π¦
π π,ππ₯,π¦
Lisa Yan, CS109, 2020
Warmup exercise
π and π have the following joint PDF:
1. Are π and π independent?
2. What is the marginalPDF of π? Of π?
3. What is πΈ π + π ?
37
ππ,π π₯, π¦ = 3πβ3π₯
where 0 < π₯ < β, 1 < π¦ < 2
π π₯ = 3πΆπβ3π₯ , 0 < π₯ < ββ π¦ = 1/πΆ, 1 < π¦ < 2
πΆ is a
constantβ
Breakout Rooms
Check out the question on the next slide. Post any clarifications here!
https://us.edstem.org/courses/667/discussion/94992
Breakout rooms: 4 min. Introduce yourself!
39
π€
Lisa Yan, CS109, 2020
The joy of meetings
Two people set up a meeting time. Each arrives independently at a time uniformly distributed between 12pm and 12:30pm.
Define π = # minutes past 12pm that person 1 arrives. π~Unif 0, 30π = # minutes past 12pm that person 2 arrives. π~Unif 0, 30
What is the probability that the first to arrive waits >10 mins for the other?
Compute: π π + 10 < π + π π + 10 < π = 2π π + 10 < π
1. What is βsymmetryβ here?
2. How do we integrate to compute this probability?
40
π€
(by symmetry)
Lisa Yan, CS109, 2020
The joy of meetings
Two people set up a meeting time. Each arrives independently at a time uniformly distributed between 12pm and 12:30pm.
Define π = # minutes past 12pm that person 1 arrives. π~Unif 0, 30π = # minutes past 12pm that person 2 arrives. π~Unif 0, 30
What is the probability that the first to arrive waits >10 mins for the other?
Compute: π π + 10 < π + π π + 10 < π = 2π π + 10 < π
42
(by symmetry)
= 2 β ΰΆ΅
π₯+10<π¦
ππ,π π₯, π¦ ππ₯ππ¦ = 2 β ΰΆ΅
π₯+10<π¦,0β€π₯,π¦,β€30
1/30 2ππ₯ππ¦
=2
302ΰΆ±10
30
ΰΆ±0
π¦β10
ππ₯ππ¦
(independence)
=2
302ΰΆ±10
30
π¦ β 10 ππ¦ = β― =4
9
Interlude for jokes/announcements
43
Lisa Yan, CS109, 2020
Announcements
44
Problem Set 5 out tonight
Due: Monday 8/3 1pm
Covers: Up to Wedβs lecture (16)
Mid-quarter feedback form
Link (Stanford account login)
Open until: Wednesday
Problem Set 4 due soon
Due: Monday 7/27 1pm
Grades
Pset 3 by: Tonight
Midterm by: End of weekend
Lisa Yan, CS109, 2020
Even when you shift the probability far left or far
right, the opposing candidate still gets some wins.
That doesnβt mean a forecast was wrong. Thatβs just
randomness and uncertainty at play. The probability
estimates the percentage of times you get an
outcome if you were to do something multiple times.
Interesting probability news
46
https://flowingdata.com/2016/07/28/
what-that-election-probability-means/
What That Election Probability Means
Lisa Yan, CS109, 2020
Ethics in probability: Utilization and Fairness under Uncertainty
47
βSuppose there is a remote stretch of coastline with two
small villages, A and B, each with a small number of
houses.β
βIn any particular week, the probability distribution over the
number of houses C impacted by power outages in each
village is as follows:β
https://dl.acm.org/doi/abs/10.1145/3351095.3372847 ACM FAT* Best Paper Award 2020
Suppose you are trying to assign generators to these villages permanently.
Utilization: E[# of generators that are actually used]
Fairness: E[village A houses in need get generators] ~= E[village B houses in need get generators]
How do you choose an allocation that optimizes utilization subject to our fairness constraint?
=> With probability :-)
Lisa Yan, CS109, 2020
Bivariate Normal Distribution
The bivariate normal distribution of πΏ = π1, π2 :
πΏ~π©(π, πΊ)β’ Mean vector π = π1, π2
β’ Covariance matrix: πΊ =π12 ππ1π2
ππ1π2 π22
β’ Marginal distributions: π1~π© π1, π12 , π2~π© π2, π2
2
β’ For bivariate normals in particular, Cov π1, π2 = 0 implies π1, π2 independent.
48
We will focus on understanding the
shape of a bivariate Normal RV.
Cov π1, π2 = Cov π2, π1 = ππ1π2
Review
Breakout Rooms
Check out the question on the next slide. Post any clarifications here!
https://us.edstem.org/courses/667/discussion/94992
Breakout rooms: 3 min. Introduce yourself!
49
π€
Lisa Yan, CS109, 2020
π, π Matching (all have π = 0, 0 )
50
π€
x
y
PD
F
x
y
1.
x
y
PD
F
x
y
3.
x
y
PD
F
x
y
2.
x
y
PD
F
x
y
4.
A.1 00 1
B.1 00 2
C.1 0.50.5 1
D.1 β0.5
β0.5 1
Lisa Yan, CS109, 2020
π, π Matching (all have π = 0, 0 )
51
x
y
PD
F
x
y
1.
x
y
PD
F
x
y
3.
x
y
PD
F
x
y
2.
x
y
PD
F
x
y
4.
A.1 00 1
B.1 00 2
C.1 0.50.5 1
D.1 β0.5
β0.5 1
Lisa Yan, CS109, 2020
Note strict inequalities; these properties hold
for both discrete and continuous RVs.
Probabilities from joint CDFs
52
Recall for a single RV π with CDF πΉπ:
CDF: π π β€ π₯ = πΉπ π₯
π π < π β€ π = πΉπ π β πΉ(π)
For two RVs π and π with joint CDF πΉπ,π:
Joint CDF: π π β€ π₯, π β€ π¦ = πΉπ,π π₯, π¦
π π1 < π β€ π2, π1 < π β€ π2 =
πΉπ,π π2,π2 β πΉπ,π π1,π2 β πΉπ,π π2,π1 + πΉπ,π π1,π1
Lisa Yan, CS109, 2020
Probabilities from joint CDFs
53
π π1 < π β€ π2, π1 < π β€ π2 =
πΉπ,π π2,π2 β πΉπ,π π1,π2 β πΉπ,π π2,π1 + πΉπ,π π1,π1
π2π1
π2π1
π₯
π¦
π¦
π2
π1
π2π1π₯
Lisa Yan, CS109, 2020
Probabilities from joint CDFs
54
π π1 < π β€ π2, π1 < π β€ π2 =
πΉπ,π π2,π2 β πΉπ,π π1,π2 β πΉπ,π π2,π1 + πΉπ,π π1,π1
π¦
π2
π1
π2π1π₯
Lisa Yan, CS109, 2020
Probabilities from joint CDFs
55
π π1 < π β€ π2, π1 < π β€ π2 =
πΉπ,π π2,π2 β πΉπ,π π1,π2 β πΉπ,π π2,π1 + πΉπ,π π1,π1
π¦
π2
π1
π2π1π₯
Lisa Yan, CS109, 2020
Probabilities from joint CDFs
56
π π1 < π β€ π2, π1 < π β€ π2 =
πΉπ,π π2,π2 β πΉπ,π π1,π2 β πΉπ,π π2,π1 + πΉπ,π π1,π1
π¦
π2
π1
π2π1π₯
Lisa Yan, CS109, 2020
Probabilities from joint CDFs
57
π π1 < π β€ π2, π1 < π β€ π2 =
πΉπ,π π2,π2 β πΉπ,π π1,π2 β πΉπ,π π2,π1 + πΉπ,π π1,π1
π¦
π2
π1
π2π1π₯
Lisa Yan, CS109, 2020
Probabilities from joint CDFs
58
π π1 < π β€ π2, π1 < π β€ π2 =
πΉπ,π π2,π2 β πΉπ,π π1,π2 β πΉπ,π π2,π1 + πΉπ,π π1,π1
π¦
π2
π1
π2π1π₯
Lisa Yan, CS109, 2020
Probabilities from joint CDFs
59
π π1 < π β€ π2, π1 < π β€ π2 =
πΉπ,π π2,π2 β πΉπ,π π1,π2 β πΉπ,π π2,π1 + πΉπ,π π1,π1
π¦
π2
π1
π2π1π₯
Lisa Yan, CS109, 2020
Probability with Instagram!
60
In image processing, a Gaussian blur is the result of blurring an
image by a Gaussian function. It is a widely used effect in
graphics software, typically to reduce image noise.
π π1 < π β€ π2, π1 < π β€ π2 =
πΉπ,π π2,π2 β πΉπ,π π1,π2 β πΉπ,π π2,π1 + πΉπ,π π1,π1
(for next
time)
Lisa Yan, CS109, 2020
Gaussian blur
In a Gaussian blur, for every pixel:β’ Weight each pixel by the probability that π
and π are both within the pixel bounds
β’ The weighting function is a Bivariate Gaussian (Normal) standard deviation parameter π
Gaussian blurring with π = 3:
ππ,π π₯, π¦ =1
2π β 32πβ π₯2+π¦2 /2β 32
What is the weight of the center pixel?
Center pixel: (0, 0)
Pixel bounds:
β0.5 < π₯ β€ 0.5β0.5 < π¦ β€ 0.5
61
π π1 < π β€ π2, π1 < π β€ π2 =
πΉπ,π π2,π2 β πΉπ,π π1,π2 β πΉπ,π π2,π1 + πΉπ,π π1,π1
Weight matrix:
π β0.5 < π β€ 0.5,β0.5 < π β€ 0.5 =
= 0.206
β Independent π~π© 0, 32 , π~π© 0, 32
β Joint CDF: πΉπ,π π₯, π¦ = Ξ¦π₯
3Ξ¦
π¦
3
Next time: More Cont. Joint and Central Limit Theorem
63
Extra
64
16f_extra
Lisa Yan, CS109, 2020
1. Integral practice
Let π and π be two continuous random variables with joint PDF:
What is π π β€ π ?
65
π π β€ π = ΰΆ΅
π₯β€π¦,0β€π₯,π¦β€1
4π₯π¦ ππ₯ ππ¦ = ΰΆ±
π¦=0
1
ΰΆ±
π₯β€π¦
4π₯π¦ ππ₯ ππ¦ = ΰΆ±
π¦=0
1
ΰΆ±
π₯=0
π¦
4π₯π¦ ππ₯ ππ¦
= ΰΆ±
π¦=0
1
4π¦π₯2
20
π¦
ππ¦ = ΰΆ±
π¦=0
1
2π¦3ππ¦ =2
4π¦4
0
1
=1
2
π π₯, π¦ = α4π₯π¦ 0 β€ π₯, π¦ β€ 10 otherwise
Lisa Yan, CS109, 2020
2. How do you integrate over a circle?
66
π π₯2 + π¦2 β€ 102 = ΰΆ±ΰΆ±ππ,π π₯, π¦ ππ¦ ππ₯
π₯2 + π¦2 β€ 102
P(dart hits within π = 10 pixels of center)?
Letβs try an example that doesnβt
involve integrating a Normal RV
βΊ
Lisa Yan, CS109, 2020
2. Imperfection on Disk
You have a disk surface, a circle of radius π . Suppose you have a single point imperfection uniformly distributed on the disk.
What are the marginal distributions of π and π? Are π and π independent?
67
ππ,π π₯, π¦ = α1
ππ 2π₯2 + π¦2 β€ π 2
0 otherwise
ππ π₯ = ΰΆ±ββ
β
ππ,π π₯, π¦ ππ¦ =1
ππ 2ΰΆ±
π₯2+π¦2β€π 2
ππ¦
=1
ππ 2ΰΆ±
π¦=β π 2βπ₯2
π 2βπ₯2
ππ¦
where βπ β€ π₯ β€ π
=2 π 2 β π₯2
ππ 2
ππ π¦ =2 π 2 β π¦
ππ 2where βπ β€ π¦ β€ π , by symmetry
No, π and π are dependent.
ππ,π π₯, π¦ β ππ π₯ ππ π¦