Search results “Product of dependent random variables”

Intuition for why the variance of both the sum and difference of two independent random variables is equal to the sum of their variances.
View more lessons or practice this subject at http://www.khanacademy.org/math/ap-statistics/random-variables-ap/combining-random-variables/v/variance-of-sum-and-difference-of-random-variables?utm_source=youtube&utm_medium=desc&utm_campaign=apstatistics
AP Statistics on Khan Academy: Meet one of our writers for AP¨_ Statistics, Jeff. A former high school teacher for 10 years in Kalamazoo, Michigan, Jeff taught Algebra 1, Geometry, Algebra 2, Introductory Statistics, and AP¨_ Statistics. Today he's hard at work creating new exercises and articles for AP¨_ Statistics.
Khan Academy is a nonprofit organization with the mission of providing a free, world-class education for anyone, anywhere. We offer quizzes, questions, instructional videos, and articles on a range of academic subjects, including math, biology, chemistry, physics, history, economics, finance, grammar, preschool learning, and more. We provide teachers with tools and data so they can help their students develop the skills, habits, and mindsets for success in school and beyond. Khan Academy has been translated into dozens of languages, and 15 million people around the globe learn on Khan Academy every month. As a 501(c)(3) nonprofit organization, we would love your help! Donate or volunteer today!
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Volunteer here: https://www.khanacademy.org/contribute?utm_source=youtube&utm_medium=desc

Views: 22451
Khan Academy

How to find the joint probability density function for two random variables given that one is dependent on the outcome of the other. Based on using the conditional probability formula. Also finding the covariance of said random variables, using conditional expectation (or iterated expectation).

Views: 631
ManyMiniMoose

MIT 6.041SC Probabilistic Systems Analysis and Applied Probability, Fall 2013
View the complete course: http://ocw.mit.edu/6-041SCF13
Instructor: Jagdish Ramakrishnan
License: Creative Commons BY-NC-SA
More information at http://ocw.mit.edu/terms
More courses at http://ocw.mit.edu

Views: 20246
MIT OpenCourseWare

Fx should be replaced by Fy in the start. Also, the title in start has a typo (maximum should be replaced by minimum)

Views: 2354
Iqbal Shahid

Classic problem of finding the probability density function of the sum of two random variables in terms of their joint density function. Find the density function of the sum random variable Z in terms of the joint density function of its two components X and Y that may be independent or dependent of each other. See also lecture slides at http://www.mhhe.com/engcs/electrical/papoulis/graphics/ppt/lect8a.pdf

Views: 35782
Probability, Stochastic Processes - Random Videos

Unizor - Creative Minds through Art of Mathematics - Math4Teens
Notes to a video lecture on http://www.unizor.com
Independent Random Variables
Expectation of Product
Our goal in this lecture is to prove that expectation of a product of two independent random variables equals to a product of their expectations.
First of all, intuitively, this fact should be obvious, at least, in some cases.
When an expectation of a random variable is a value, around which results of random experiments are concentrated (like a temperature of a healthy person), product of results of two different experiments (product of temperatures of two different healthy persons) tend to concentrate around product of their expectations.
In some other cases, when such a concentration does not take place (like flipping a coin), that same rule of multiplicative property of an expectation is still observed.
A very important detail, however, differentiates property of a sum of two random variables from their product. The expectation of a sum always equals to a sum of expectations of its component. With a product the analogous property is true only in case the components are INDEPENDENT random variables.
Let's approach this problem more formally and prove this theorem.
Consider the following two random experiments (sample spaces) and random variables defined on their elementary events.
Ω1=(e1,e2,...,eM )
with corresponding measure of probabilities of these elementary events
P=(p1,p2,...,pM )
(that is, P(ei )=pi - non-negative numbers with their sum equaled to 1)
and random variable ξ defined for each elementary event as
ξ(ei) = xi where i=1,2,...M
Ω2=(f1,f2,...,fN )
with corresponding measure of probabilities of these elementary events
Q=(q1,q2,...,qN )
(that is, Q(fj )=qj - non-negative numbers with their sum equaled to 1)
and random variable η defined for each elementary event as
η(fj) = yj where j=1,2,...N
Separately, the expectations of these random variables are:
E(ξ) = x1·p1+x2·p2+...+xM·pM
E(η) = y1·q1+y2·q2+...+yN·qN
To calculate the expectation of a product of these random variables, let's research what values and with what probabilities this product can take.
Since every value of ξ can be observed with every value of η, we can conclude that all the values of their product are described by all values xi·yj where index i runs from 1 to M and index j runs from 1 to N.
Let's examine the probabilistic meaning of a product of two random variables defined on two different sample spaces.
Any particular value xi·yj is taken by a new random variable ζ=ξ·η defined on a new combined sample space Ω=Ω1×Ω2 that consists of all pairs of elementary events (ei , fj ) with the corresponding combined measure of probabilities of these pairs equal to
R(ei , fj ) = rij
where index i runs from 1 to M and index j runs from 1 to N.
Thus, we have defined a new random variable ζ=ξ·η defined on a new sample space Ω of M·N pairs of elementary events from two old spaces Ω1 and Ω2 as follows
ζ(ei , fj ) = xi·yj
with probability ri j
Before going any further, let's examine very important properties of probabilities rij.
We have defined rij as a probability of a random experiment described by a sample space Ω1 resulting in elementary event ei and, simultaneously, a random experiment described by a sample space Ω2 resulting in elementary event fj.
Incidentally, if events from these two sample spaces are independent,
rij = pi·qj
because, for independent events, probability of their simultaneous occurrence equals to a product of probabilities of their separate individual occurrences.
Keeping in mind the above properties of probabilities rij, we can calculate the expectation of our new random variable ζ.
E(ζ) = E(ξ·η) =
= (x1·y1)·r11+...+(x1·yN)·r1N +
+ (x2·y1)·r21+...+(x2·yN)·r2N +
...
+ (xM·y1)·rM1+...+(xM·yN)·rMN
On the other hand, let's calculate the product of expectations of our random variable ξ and η:
E(ξ)·E(η) =
=(x1·p1+...+xM·pM)·
·(y1·q1+...+yN·qN) =
= (x1·y1)·p1q1+...+(x1·yN)·p1qN +
+ (x2·y1)·p2q1+...+(x2·yN)·p2qN +
...
+ (xM·y1)·pMq1+...+(xM·yN)·pMqN
Obviously, if random variables ξ and η are INDEPENDENT, probability rij of ξ to take value xi and, simultaneously, η to take value yj equals to a product of corresponding probabilities pi·qj. In this case expressions for E(ξ·η) and E(ξ)·E(η) are identical.
That proves that for INDEPENDENT random variables mathematical expectation of their product equals to a product of their mathematical expectations.
End of proof.

Views: 253
Zor Shekhtman

pdf of a difference as function of joint pdf

Views: 1357
Anish Turlapaty

This video explains what is meant by the covariance and correlation between two random variables, providing some intuition for their respective mathematical formulations. Check out https://ben-lambert.com/econometrics-course-problem-sets-and-data/ for course materials, and information regarding updates on each of the courses. Quite excitingly (for me at least), I am about to publish a whole series of new videos on Bayesian statistics on youtube. See here for information: https://ben-lambert.com/bayesian/ Accompanying this series, there will be a book: https://www.amazon.co.uk/gp/product/1473916364/ref=pe_3140701_247401851_em_1p_0_ti

Views: 235482
Ben Lambert

This video explains what is meant by the expectations and variance of a vector of random variables. Check out https://ben-lambert.com/econometrics-course-problem-sets-and-data/ for course materials, and information regarding updates on each of the courses. Quite excitingly (for me at least), I am about to publish a whole series of new videos on Bayesian statistics on youtube. See here for information: https://ben-lambert.com/bayesian/ Accompanying this series, there will be a book: https://www.amazon.co.uk/gp/product/1473916364/ref=pe_3140701_247401851_em_1p_0_ti

Views: 24387
Ben Lambert

Variables
-First Visit to Max; Sojourn Times; Independent Variables; Uncorrelated Variables: A Counterexample; Generating Function; Product of Gen Functions; A Simple Example; Gen Function for 2 Dice; Clever Loaded Dice; Well-Known Distributions
These lectures were offered as an online course at the Harvard Extension School
This online math course develops the mathematics needed to formulate and analyze probability models for idealized situations drawn from everyday life.
View complete course (Outline, Problem sets,etc) at: http://www.extension.harvard.edu/open-learning-initiative/sets-counting-probability

Views: 273
It's so blatant

calculating the expected values of ratio and product of two random variables

Views: 4078
Anish Turlapaty

We discuss joint, conditional, and marginal distributions (continuing from Lecture 18), the 2-D LOTUS, the fact that E(XY)=E(X)E(Y) if X and Y are independent, the expected distance between 2 random points, and the chicken-egg problem.

Views: 118065
Harvard University

Stochastic Structural Dynamics by Prof. C.S. Manohar ,Department of Civil Engineering, IISC Bangalore. For more details on NPTEL visit http://nptel.iitm.ac.in

Views: 2157
nptelhrd

This video we create he probability distribution table for the sum of two dice.

Views: 60462
Brian Veitch

Say we want to find an expectation of a product of random variables. Can we just to it as the product of the expectations? Watch and see.

Views: 4910
Phil Chan

This videos explains what is meant by a moment of a random variable. Check out https://ben-lambert.com/econometrics-course-problem-sets-and-data/ for course materials, and information regarding updates on each of the courses. Quite excitingly (for me at least), I am about to publish a whole series of new videos on Bayesian statistics on youtube. See here for information: https://ben-lambert.com/bayesian/ Accompanying this series, there will be a book: https://www.amazon.co.uk/gp/product/1473916364/ref=pe_3140701_247401851_em_1p_0_ti

Views: 46889
Ben Lambert

Online Private Tutoring at http://andreigalanchuk.nl
Follow me on Facebook: https://www.facebook.com/galanchuk/
Add me on Linkedin: https://www.linkedin.com/in/andreigalanchuk?trk=nav_responsive_tab_profile

Views: 497
Andrei Galanchuk

Finding the probability that the total of some random variables exceeds an amount by understanding the distribution of the sum of normally distributed variables.
View more lessons or practice this subject at http://www.khanacademy.org/math/ap-statistics/random-variables-ap/combining-random-variables/v/analyzing-distribution-of-sum-of-two-normally-distributed-random-variables?utm_source=youtube&utm_medium=desc&utm_campaign=apstatistics
AP Statistics on Khan Academy: Meet one of our writers for AP¨_ Statistics, Jeff. A former high school teacher for 10 years in Kalamazoo, Michigan, Jeff taught Algebra 1, Geometry, Algebra 2, Introductory Statistics, and AP¨_ Statistics. Today he's hard at work creating new exercises and articles for AP¨_ Statistics.
Khan Academy is a nonprofit organization with the mission of providing a free, world-class education for anyone, anywhere. We offer quizzes, questions, instructional videos, and articles on a range of academic subjects, including math, biology, chemistry, physics, history, economics, finance, grammar, preschool learning, and more. We provide teachers with tools and data so they can help their students develop the skills, habits, and mindsets for success in school and beyond. Khan Academy has been translated into dozens of languages, and 15 million people around the globe learn on Khan Academy every month. As a 501(c)(3) nonprofit organization, we would love your help! Donate or volunteer today!
Donate here: https://www.khanacademy.org/donate?utm_source=youtube&utm_medium=desc
Volunteer here: https://www.khanacademy.org/contribute?utm_source=youtube&utm_medium=desc

Views: 15886
Khan Academy

Expectation of the sum of two functions of a random variable

Views: 324
Lawrence Leemis

MIT 6.042J Mathematics for Computer Science, Spring 2015
View the complete course: http://ocw.mit.edu/6-042JS15
Instructor: Albert R. Meyer
License: Creative Commons BY-NC-SA
More information at http://ocw.mit.edu/terms
More courses at http://ocw.mit.edu

Views: 4520
MIT OpenCourseWare

On finding the probability density function of the "Ratio of Two Random Variables".

Practice this lesson yourself on KhanAcademy.org right now:
https://www.khanacademy.org/math/probability/independent-dependent-probability/dependent_probability/e/multiplying-dependent-probabilities?utm_source=YT&utm_medium=Desc&utm_campaign=ProbabilityandStatistics
Watch the next lesson: https://www.khanacademy.org/math/probability/independent-dependent-probability/dependent_probability/v/monty-hall-problem?utm_source=YT&utm_medium=Desc&utm_campaign=ProbabilityandStatistics
Missed the previous lesson?
https://www.khanacademy.org/math/probability/independent-dependent-probability/dependent_probability/v/analyzing-dependent-probability?utm_source=YT&utm_medium=Desc&utm_campaign=ProbabilityandStatistics
Probability and statistics on Khan Academy: We dare you to go through a day in which you never consider or use probability. Did you check the weather forecast? Busted! Did you decide to go through the drive through lane vs walk in? Busted again! We are constantly creating hypotheses, making predictions, testing, and analyzing. Our lives are full of probabilities! Statistics is related to probability because much of the data we use when determining probable outcomes comes from our understanding of statistics. In these tutorials, we will cover a range of topics, some which include: independent events, dependent probability, combinatorics, hypothesis testing, descriptive statistics, random variables, probability distributions, regression, and inferential statistics. So buckle up and hop on for a wild ride. We bet you're going to be challenged AND love it!
About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content.
For free. For everyone. Forever. #YouCanLearnAnything
Subscribe to KhanAcademy’s Probability and Statistics channel:
https://www.youtube.com/channel/UCRXuOXLW3LcQLWvxbZiIZ0w?sub_confirmation=1
Subscribe to KhanAcademy: https://www.youtube.com/subscription_center?add_user=khanacademy

Views: 552564
Khan Academy

Covariance, Variance and the Slope of the Regression Line
Watch the next lesson: https://www.khanacademy.org/math/probability/statistics-inferential/normal_distribution/v/introduction-to-the-normal-distribution?utm_source=YT&utm_medium=Desc&utm_campaign=ProbabilityandStatistics
Missed the previous lesson?
https://www.khanacademy.org/math/probability/regression/regression-correlation/v/calculating-r-squared?utm_source=YT&utm_medium=Desc&utm_campaign=ProbabilityandStatistics
Probability and statistics on Khan Academy: We dare you to go through a day in which you never consider or use probability. Did you check the weather forecast? Busted! Did you decide to go through the drive through lane vs walk in? Busted again! We are constantly creating hypotheses, making predictions, testing, and analyzing. Our lives are full of probabilities! Statistics is related to probability because much of the data we use when determining probable outcomes comes from our understanding of statistics. In these tutorials, we will cover a range of topics, some which include: independent events, dependent probability, combinatorics, hypothesis testing, descriptive statistics, random variables, probability distributions, regression, and inferential statistics. So buckle up and hop on for a wild ride. We bet you're going to be challenged AND love it!
About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content.
For free. For everyone. Forever. #YouCanLearnAnything
Subscribe to KhanAcademy’s Probability and Statistics channel:
https://www.youtube.com/channel/UCRXuOXLW3LcQLWvxbZiIZ0w?sub_confirmation=1
Subscribe to KhanAcademy: https://www.youtube.com/subscription_center?add_user=khanacademy

Views: 298444
Khan Academy

Probability Foundation for Electrical Engineers by Dr. Krishna Jagannathan,Department of Electrical Engineering,IIT Madras.For more details on NPTEL visit http://nptel.ac.in

Views: 2897
nptelhrd

This video explains some of the properties of the expectations and variance operators, particularly that of pre-multiplying by a constant. Check out https://ben-lambert.com/econometrics-course-problem-sets-and-data/ for course materials, and information regarding updates on each of the courses. Quite excitingly (for me at least), I am about to publish a whole series of new videos on Bayesian statistics on youtube. See here for information: https://ben-lambert.com/bayesian/ Accompanying this series, there will be a book: https://www.amazon.co.uk/gp/product/1473916364/ref=pe_3140701_247401851_em_1p_0_ti

Views: 32564
Ben Lambert

A lecture on determining if X and Y are independent random variables. We look at the joint density function and determine if it is the product of the marginal density functions.

Views: 5279
Rose-Hulman Online

You'll become familiar with the concept of independent events, or that one event in no way affects what happens in the second event. Keep in mind, too, that the sum of the probabilities of all the possible events should equal 1.
Practice this lesson yourself on KhanAcademy.org right now:
https://www.khanacademy.org/math/probability/independent-dependent-probability/independent_events/e/independent_probability?utm_source=YT&utm_medium=Desc&utm_campaign=ProbabilityandStatistics
Watch the next lesson: https://www.khanacademy.org/math/probability/independent-dependent-probability/independent_events/v/getting-at-least-one-heads?utm_source=YT&utm_medium=Desc&utm_campaign=ProbabilityandStatistics
Missed the previous lesson?
https://www.khanacademy.org/math/probability/independent-dependent-probability/addition_rule_probability/v/addition-rule-for-probability?utm_source=YT&utm_medium=Desc&utm_campaign=ProbabilityandStatistics
Probability and statistics on Khan Academy: We dare you to go through a day in which you never consider or use probability. Did you check the weather forecast? Busted! Did you decide to go through the drive through lane vs walk in? Busted again! We are constantly creating hypotheses, making predictions, testing, and analyzing. Our lives are full of probabilities! Statistics is related to probability because much of the data we use when determining probable outcomes comes from our understanding of statistics. In these tutorials, we will cover a range of topics, some which include: independent events, dependent probability, combinatorics, hypothesis testing, descriptive statistics, random variables, probability distributions, regression, and inferential statistics. So buckle up and hop on for a wild ride. We bet you're going to be challenged AND love it!
About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content.
For free. For everyone. Forever. #YouCanLearnAnything
Subscribe to KhanAcademy’s Probability and Statistics channel:
https://www.youtube.com/channel/UCRXuOXLW3LcQLWvxbZiIZ0w?sub_confirmation=1
Subscribe to KhanAcademy: https://www.youtube.com/subscription_center?add_user=khanacademy

Views: 820286
Khan Academy

In this video I will show you a simple method which you can use to determine if two variables are likely to be correlated. This is a good first port of call and typically works well for sociology and psychology.
The calculation of Pearson's r can easily be done with built in functions in excel and other programs like OpenOffice Calc (which is used here). A scientific calculator is also able to easily perform the calculation with a built in function.
However, it is likely to miss more complex relationships (i.e. logarithmic relations). Thus this method is good for showing that variables are likely to be correlated but not very good at proving rigorously that they uncorrelated.

Views: 849
The Complete Guide to Everything

-- Created using PowToon -- Free sign up at http://www.powtoon.com/youtube/ -- Create animated videos and animated presentations for free. PowToon is a free tool that allows you to develop cool animated clips and animated presentations for your website, office meeting, sales pitch, nonprofit fundraiser, product launch, video resume, or anything else you could use an animated explainer video. PowToon's animation templates help you create animated presentations and animated explainer videos from scratch. Anyone can produce awesome animations quickly with PowToon, without the cost or hassle other professional animation services require.

Views: 57
fernanda eva

definition of covariance and its relation to variance of sum
From www.statisticallearning.us

Views: 8989
Anish Turlapaty

We discuss functions of two discrete random variables. In particular, we discuss finding the PMF of a function of two random variables, when we have their joint PMF

Views: 5824
Probability Course

This video gives a formula for correlation in terms of covariances and variances. Then, I show why correlation is between -1 and 1, using a standard formula for variance of a linear combination of random variables.
The video is useful to show how to manipulate variances of linear combinations and how to do some of the essential algebra involved in econometrics.
This is part of my series of videos reviewing the mathematics and statistics prerequisites for econometrics.

Views: 8372
intromediateecon

Unizor - Creative Minds through Art of Mathematics - Math4Teens
Notes to a video lecture on http://www.unizor.com
Random Variables
Correlation
In this lecture we will talk about independent and dependent random variables and will introduce a numerical measure of dependency between random variables.
Assume a random variable ξ takes values
x1, x2,..., xM
with probabilities
p1, p2,..., pM.
Further, assume a random variable η takes values
y1, y2,..., yN
with probabilities
q1, q2,..., qN.
The known property of mathematical expectations for independent random variables is the basis of measuring the degree of dependency between any pair of random variables.
First of all, we introduce a concept of covariance of any two random variables:
Cov(ξ,η) =
= E[(ξ−E(ξ))·(η−E(η))]
Simple transformation by opening parenthesis converts it into an equivalent definition:
Cov(ξ,η) = E(ξ·η)−E(ξ)E(η)
Now we see that for independent random variables their covariance equals to zero (see property (c) above).
Incidentally, the covariance of a random variable with itself (kind of ultimate dependency) equal to its variance:
Cov(ξ,ξ) = E(ξ·ξ)−E(ξ)E(ξ) =
= E[(ξ−E(ξ))²] = Var(ξ)
Also notice that another example of very strong dependency, η = A·ξ, where A is a constant, leads to the following value of covariance:
Cov(ξ,Aξ) =
= E(ξ·Aξ)−E(ξ)E(Aξ) =
= A·E[(ξ−E(ξ))²] = A·Var(ξ)
This shows that, when coefficient A is positive (that is, positive change of ξ causes positive change of η=A·ξ), covariance between them is positive as well and proportional to coefficient A. If A is negative (that is, positive change of ξ causes negative change of η=A·ξ), covariance between them is negative as well and still proportional to coefficient A.
One more example.
Consider "half-dependency" between ξ and η, defined as follows.
Let ξ' be an independent random variable, identically distributed with ξ.
Let η = (ξ + ξ')/2.
So, η "borrows" its randomness from two independent identically distributed random variables ξ and ξ'.
Then covariance between ξ and η is:
Cov(ξ,η) = Cov(ξ,(ξ+ξ')/2) =
=E[ξ·(ξ+ξ')/2)]−
−E(ξ)·E((ξ+ξ')/2) =
=E(ξ²)/2+E(ξ·ξ')/2 −
−[E(ξ)]²/2−E(ξ)·E(ξ')/2
Since ξ and ξ' are independent, expectation of their product equals to a product of their expectations.
So, our expression can be transformed further: =E(ξ²)/2+E(ξ)·E(ξ')/2 −
−[E(ξ)]²/2−E(ξ)·E(ξ')/2 =
= Var(ξ)/2
As we see, covariance between "half-dependent" random variables ξ and η=(ξ+ξ')/2, where ξ and ξ' are independent identically distributed random variables, equals to half of the variance of ξ.
All the above manipulations with covariance led us to some formulas where the variance plays a significant role. If we want a kind of measure that reflects the dependency between random variables not related to variances, but always scaled in the interval [−1, 1], we have to scale the covariance by a factor that depends on variances, thus forming a coefficient of correlation: R(ξ,η) =
= Cov(ξ,η)/√(Var(ξ)·Var(η)
Let's examine this coefficient of correlation in cases we considered above as examples.
For independent random variables ξ and η the correlation is zero because their covariance is zero.
Correlation between a random variable and itself equals to 1: R(ξ,ξ) = Cov(ξ,ξ)/Var(ξ,ξ) = 1
Correlation between a random variables ξ and Aξ equals to 1 (for positive constant A) or −1 (for negative A):
R(ξ,Aξ) =
= Cov(ξ,Aξ)/√(Var(ξ)·Var(Aξ)
= A/|A|
which equals to 1 or −1, depending on a sign of A.
This seems to corresponds our intuitive understanding of rigid relationship between ξ and Aξ.
Correlation between "half-dependent" random variables, as introduced above, is:
R(ξ,(ξ+ξ')/2) = Cov(ξ,(ξ+ξ')/2) / √[Var(ξ)·Var((ξ+ξ')/2] = √2/2.
As we see, in all these examples the correlation is a number from an interval [−1,1] that is equal to zero for independent random variables, equals to 1 or −1 for rigidly dependent random variables and is inside this interval for partially dependent (like in our example of "half-dependent") random variables.
For those interested, it can be proved that this statement is true for any pair of random variables.
So, the coefficient of correlation is a good tool to measure the degree of dependency between two random variables.

Views: 64
Zor Shekhtman

wo events A and B are independent (often written as .
Why this defines independence is made clear by rewriting with conditional probabilities:
Thus, the occurrence of B does not affect the probability of A, and vice versa. Although the derived expressions may seem more intuitive, they are not the preferred definition, as the conditional probabilities may be undefined if P(A) or P(B) are 0. Furthermore, the preferred definition makes clear by symmetry that when A is independent of B, B is also independent of A.
More than two events[edit]
A finite set of events {Ai} is pairwise independent if every pair of events is independent[2]—that is, if and only if for all distinct pairs of indices m, k,
A finite set of events is mutually independent if every event is independent of any intersection of the other events[2]—that is, if and only if for every n-element subset {Ai},
This is called the multiplication rule for independent events. Note that it is not a single condition involving only the product of all the probabilities of all single events (see below for a counterexample); it must hold true for all subsets of events.
For more than two events, a mutually independent set of events is (by definition) pairwise independent; but the converse is not necessarily true (see below for a counterexample).
Two random variables X and Y are independent if and only if (iff) the elements of the π-system generated by them are independent; that is to say, for every a and b, the events {X a} and {Y b} are independent events (as defined above). That is, X and Y with cumulative distribution functions {\displaystyle F_{X}(x)} F_X(x) and {, are independent iff the combined random variable (X, Y) has a joint cumulative distribution function
or equivalently, if the joint density exists,
More than two random variables[edit]
A set of random variables is pairwise independent if and only if every pair of random variables is independent.
A set of random variables is mutually independent if and only if for any finite subset are mutually independent events (as defined above).
The measure-theoretically inclined may prefer to substitute events {X ∈ A} for events {X a} in the above definition, where A is any Borel set. That definition is exactly equivalent to the one above when the values of the random variables are real numbers. It has the advantage of working also for complex-valued random variables or for random variables taking values in any measurable space (which includes topological spaces endowed by appropriate σ-algebras).
Conditional independence[edit]
Main article: Conditional independence
Intuitively, two random variables X and Y are conditionally independent given Z if, once Z is known, the value of Y does not add any additional information about X. For instance, two measurements X and Y of the same underlying quantity Z are not independent, but they are conditionally independent given Z (unless the errors in the two measurements are somehow connected).
The formal definition of conditional independence is based on the idea of conditional distributions. If X, Y, and Z are discrete random variables, then we define X and Y to be conditionally independent given Z if
for all x, y and z such that P(Z = z) 0. On the other hand, if the random variables are continuous and have a joint probability density function p, then X and Y are conditionally independent given Z if
for all real numbers x, y and z such that pZ(z) 0.
If X and Y are conditionally independent given Z, then
for any x, y and z with P(Z = z) 0. That is, the conditional distribution for X given Y and Z is the same as that given Z alone. A similar equation holds for the conditional probability density functions in the continuous case.
Independence can be seen as a special kind of conditional independence, since probability can be seen as a kind of conditional probability given no events.
Independent σ-algebras[edit]
and an infinite family of σ-algebras is said to be independent if all its finite subfamilies are independent.
The new definition relates to the previous ones very directly:
Two events are independent (in the old sense) if and only if the σ-algebras that they generate are independent (in the new sense). The σ-algebra generated by an event
Two random variables X and Y defined over Ω are independent (in the old sense) if and only if the σ-algebras that they generate are independent (in the new sense). The σ-algebra generated by a random variable X taking values in some measurable space S consists, by definition, of all subsets of Ω of the form X−1(U), where U is any measurable subset of S.
Using this definition, it is easy to show that if X and Y are random variables and Y is constant, then X and Y are independent, since the σ-algebra generated by a constant random variable is the trivial σ-algebra {∅, Ω}. Probability zero events cannot affect independence so independence also holds if Y is only Pr-almost surely constant.

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maths gotserved

Venn diagrams and the addition rule for probability
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An introduction to a special class of random variables called binomial random variables.
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Conditional probability, Product rule, Concept of a Random variable

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Industrial Engineering

Probability density functions for continuous random variables.
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Statistics Lecture 4.4: The Multiplication Rule for "And" Probabilities.

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Professor Leonard

MIT RES.6-012 Introduction to Probability, Spring 2018
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MIT OpenCourseWare

Unizor - Creative Minds through Art of Mathematics - Math4Teens
Notes to a video lecture on http://www.unizor.com
Random Variables - Correlation
In this lecture we will talk about independent and dependent random variables and will introduce a numerical measure of dependency between random variables.
First of all, we introduce a concept of covariance of any two random variables:
Cov(ξ,η) = E[(ξ−E(ξ))·(η−E(η))]
Simple transformation by opening parenthesis converts it into an equivalent definition:
Cov(ξ,η) = E(ξ·η)−E(ξ)E(η)
Now we see that for independent random variables their covariance equals to zero.
Incidentally, the covariance of a random variable with itself (kind of ultimate dependency) equal to its variance:
Cov(ξ,ξ) = E(ξ·ξ)−E(ξ)E(ξ) =
= E[(ξ−E(ξ))²] = Var(ξ)
Also notice that another example of very strong dependency, η = A·ξ, where A is a constant, leads to the following value of covariance:
Cov(ξ,Aξ) =
= E(ξ·Aξ)−E(ξ)E(Aξ) =
= A·E[(ξ−E(ξ))²] = A·Var(ξ)
One more example.
Consider "half-dependency" between ξ and η, defined as follows.
Let ξ' be an independent random variable, identically distributed with ξ.
Let η = (ξ + ξ')/2.
So, η "borrows" its randomness from two independent identically distributed random variables ξ and ξ'.
Then covariance between ξ and η is:
Cov(ξ,η) = Cov(ξ,(ξ+ξ')/2) =
=E[ξ·(ξ+ξ')/2)]−
−E(ξ)·E((ξ+ξ')/2) =
=E(ξ²)/2+E(ξ·ξ')/2 −
−[E(ξ)]²/2−E(ξ)·E(ξ')/2
Since ξ and ξ' are independent, expectation of their product equals to a product of their expectations.
So, our expression can be transformed further: =E(ξ²)/2+E(ξ)·E(ξ')/2 −
−[E(ξ)]²/2−E(ξ)·E(ξ')/2 =
= Var(ξ)/2
As we see, covariance between "half-dependent" random variables ξ and η=(ξ+ξ')/2, where ξ and ξ' are independent identically distributed random variables, equals to half of the variance of ξ.
All the above manipulations with covariance led us to some formulas where the variance plays a significant role. If we want a kind of measure that reflects the dependency between random variables not related to variances, but always scaled in the interval [−1, 1], we have to scale the covariance by a factor that depends on variances, thus forming a coefficient of correlation: R(ξ,η) =
= Cov(ξ,η)/√[Var(ξ)·Var(η)]
Let's examine this coefficient of correlation in cases we considered above as examples.
For independent random variables ξ and η the correlation is zero because their covariance is zero.
Correlation between a random variable and itself equals to 1: R(ξ,ξ) = Cov(ξ,ξ)/Var(ξ,ξ) = 1
Correlation between a random variables ξ and Aξ equals to 1 (for positive constant A) or −1 (for negative A):
R(ξ,Aξ) =
= Cov(ξ,Aξ)/√[Var(ξ)·Var(Aξ)]
= A/|A|
which equals to 1 or −1, depending on a sign of A.
This seems to corresponds our intuitive understanding of rigid relationship between ξ and Aξ.
Correlation between "half-dependent" random variables, as introduced above, is:
R(ξ,(ξ+ξ')/2) =
= Cov(ξ,(ξ+ξ')/2)/
/√[Var(ξ)·Var((ξ+ξ')/2)]
= √2/2.
As we see, in all these examples the correlation is a number from an interval [−1,1] that is equal to zero for independent random variables, equals to 1 or −1 for rigidly dependent random variables and is inside this interval for partially dependent (like in our example of "half-dependent") random variables.
For those interested, it can be proven that this statement is true for any pair of random variables.
So, the coefficient of correlation is a good tool to measure the degree of dependency between two random variables.

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Zor Shekhtman

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