General Random Variables 2 Tutorial Overview
Learn about extending the notion of Probability Density Function (PDF) to multiple random variables, including jointly continuous PDF, marginal PDF, and conditional PDF. Understand the definitions and concepts of continuous PDF, jointly continuous PDF, and the importance of joint density function. Discover how to compute probabilities using joint density function for events defined in terms of random variables.
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Tutorial 6: General Random Variables 2 Yitong Meng March 6, 2016 1
Outline Extend the notion of PDF to the case of multiple random variables Jointly continuous PDF Marginal PDF Conditional PDF Exercises 2
Continuous PDF Recall that ? is continuous if there is a function ?(?) (the density) such that ? ?(? ?) = ??? ?? We generalize this to two random variables. 3
Jointly continuous PDF Definition Two random variables ? and ? are jointly continuous if there is a function ??,??,? on ?2, called the joint probability density function, such that ? ? ?,? ? = ??,??,? ???? ? ?,? ? 4
Jointly continuous PDF The integral is over {(?,?) ? ?,? ?}. We can also write the integral as ? ? ? ? ?,? ? = ??,??,? ?? ?? ? ? = ??,??,? ?? ?? 5
Jointly continuous PDF In order for a function ?(?,?) to be a joint density it must satisfy ?(?,?) 0 ??,??,? ???? = 1 6
Jointly continuous PDF Just as with one random variable, the joint density function contains all the information about the underlying probability measure if we only look at the random variables ? and ? . In particular, we can compute the probability of any event defined in terms of ? and ? just using ?(?,?). 7
Jointly continuous PDF Here are some events defined in terms of ? and ? : ? ? {?2+ ?2 1} {1 ? 4,? 0} They can all be written in the form {(?,?) ?} for some subset ? of ?2. 8
Jointly continuous PDF Proposition. ??? ? ?2, ? ?,? ? = ??,??,? ???? (?,?) ? The two-dimensional integral is over the subset B of ?2 9
Jointly continuous PDF Typically, when we want to actually compute this integral we have to write it as an iterated integral. It is a good idea to draw a picture of ? to help do this. Many sample observations are shown from a joint probability distribution. The marginal densities are shown as well. 10
Exercise 1 6? + ? , ?? 1 ? 2 ??? 4 ? 5 0, ? ?,? = ?? ?????? What it the probability of 1 ? 1.5 ??? 4.5 ? 5 ? ? ? 11
Solution ?(1 ? 1.5 ,4.5 ? 5) 1.5 4.5 5??,??,? ???? = 1 1.5 4.5 5 1 = 1 6? + ? ???? =0.25 12
Marginal PDF Suppose we know the joint density ??,??,? of ? and ? . How do we find their individual densities ??? , ??? . These are called marginal densities. The CDF of ? is ??? = ? ? ? = ? < ? ?, < ? < ? ??,??,? ?? ?? = 13
Marginal PDF Differentiate this with respect to ? and we get ??? = ??,??,? ?? In words, we get the marginal density of X by integrating y from to in the joint density. 14
Marginal PDF Proposition. If ? and ? are jointly continuous with joint density ??,??,? , then the marginal densities are given by ??? = ??,??,? ?? ??? = ??,??,? ?? 15
Conditional PDF Definition. Suppose X and Y are continuous random variables with joint probability density function f(x,y) and marginal probability density functions ??(?) and ??(?), respectively. Then, the conditional PDF of ?given ? = ? is defined as: (?|?) = ?(?,?)/??(?) provided ??(?) > 0. 16
Exercise Suppose the continuous random variables X and Y have the following joint probability density function: ?(?,?) = 3/2 for ?2 ? 1 and 0 < ? < 1. What is the conditional distribution of Y given X = x? 17
Solution We can use the formula: (?|?) = ?(?,?)/?X(x) to find the conditional PDF of Y given X. 18
So, we basically have a plane, shaped like the support, floating at a constant 3/2 units above the xy-plane. To find ??(?) then, we have to integrate: ?(?,?) = 3/2 over the support x2 y 1. 19
That is: ??? = ?2? ?,? ?? = ?2 = 3 3 2?? 2 (1- ?2) for 0 < x < 1. 20
Now, we can use the joint PDF ?(?,?) that we were given and the marginal PDF ??(?) that we just calculated to get the conditional PDF of Y given X = x: 3 2 ? ? =? ?,? = 1 1 ?2, = 3 2(1 ?2) ?? ? 0 < ? < 1,?2 ? 1 21
That is, given x, the continuous random variable Y is uniform on the interval (x2, 1). For example, if x = , then the conditional PDF of Y is: 2 1 4 (?|1/4) = 11 = 1/(15/16) = 16/15 ??? 1/16 ? 1. 22
And, if x = , then the conditional PDF of Y is: 2 1 2 (?|1/2) = 11 = 1/(1 1/4) = 4/3 ??? 1/4 ? 1. 23
