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Probability · Lecture 19 of 76 · 15:26
PMF of a Function of a Random Variable
Study guide
What this lecture covers
This recitation problem gives a random variable X taking values -3, -2, -1, 1, 2, 3 with PMF p_X(x) = x^2/a for an unknown constant a, and asks you to find a, then to find the PMF of Z = X^2, a function of X. It is a practice problem for manipulating PMFs and applying the normalization axiom that a PMF must sum to 1.
After watching, you should be able to use the "PMF values sum to 1" property to solve for an unknown normalizing constant in a PMF, and understand how to approach finding the PMF of a function of a random variable by combining the probability mass of X-values that map to the same output.
Key ideas
- PMF definition:
p_X(k)is the probability that random variableXequalsk; it can be thought of as a function that returns the probability mass sitting over a given value. - Normalization axiom: summing a PMF over all possible values of
Xmust equal 1, which gives an equation you can solve for an unknown parameter. - Symmetry: because
p_X(x) = x^2/adepends only onx^2, the PMF is symmetric, sop_X(-3) = p_X(3),p_X(-2) = p_X(2), andp_X(-1) = p_X(1), which simplifies the summation. - Solving for the unknown constant: summing
2(9/a) + 2(4/a) + 2(1/a) = 28/aand setting it equal to 1 givesa = 28. - PMF of a function of a random variable: since
Z = X^2, several values ofXcan map to the same value ofZ(for example, bothX=2andX=-2giveZ=4), so findingp_Zrequires combining the probability mass of allX-values that produce eachZ-value.
Walkthrough
Setting up the problem and reviewing what a PMF means (0:01)
The problem gives p_X(x) = x^2/a for x in {-3,-2,-1,1,2,3} and 0 elsewhere, with a > 0 unknown. The lecture reviews that p_X(k) denotes the probability mass lying over the value k, setting up the two tasks: solve for a, then find the PMF of Z = X^2.
Solving for the unknown constant a (2:08)
Using the axiom that a PMF must sum to 1 over all possible values, the lecture plots the PMF (noting its symmetry around 0), sums the six probability masses as 2(9/a) + 2(4/a) + 2(1/a) = 28/a, sets this equal to 1, and solves to get a = 28, showing this is the only value that produces a valid PMF.
Finding the PMF of Z = X^2 (5:10)
The lecture moves on to the second part of the problem, deriving the PMF of Z, which requires identifying which values of X map to each possible value of Z and combining their probabilities, following the chapter progression toward representing p_Z algebraically.
Before you watch
- Review the definition of a probability mass function and the axiom that probabilities of all outcomes in a sample space sum to 1, from earlier lectures in this course.
- Be comfortable with the expected value rule / functions of random variables, since this problem's second part depends on relating
XandZ = X^2.
Check your understanding
- Why must a PMF sum to 1 over all possible values, and how does that let you solve for an unknown parameter?
- Why does
p_X(-3)equalp_X(3)in this problem? - When computing the PMF of
Z = X^2, why might more than one value ofXcontribute to the same value ofZ?
Chapters
- 0:00 <Untitled Chapter 1>
- 0:15 Problem Statement
- 1:57 What Does a Pmf Really Tell You
- 3:07 Axioms of Probability
- 7:32 Definition of Pmf
- 13:35 Represent the Pmf of Z Algebraically
From the YouTube description
MIT 6.041SC Probabilistic Systems Analysis and Applied Probability, Fall 2013
View the complete course: http://ocw.mit.edu/6-041SCF13
Instructor: Katie Szeto
License: Creative Commons BY-NC-SA
More information at http://ocw.mit.edu/terms
More courses at http://ocw.mit.edu
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