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SDS 321: Introduction to Probability and
StatisticsLecture 19: Continuous random variables-Deriveddistributions, max of two independent r.v.'s
Purnamrita Sarkar
Department of Statistics and Data Science
The University of Texas at Austin
www.cs.cmu.edu/psarkar/teaching 1
Roadmap
I
Two random variables: joint distributions
I
Joint pdf3
IJoint pdf to a single pdf: Marginalization3
IConditional pdf
I
Conditioning on an event3
IConditioning on a continuous r.v3
ITotal probability rule for continuous r.v's3
IBayes theorem for continuous r.v's3
IConditional expectation and total expectation theorem3 I
Independence3
I
More than two random variables.3
I
Derived distributions
I
Linear functions
IMonotonic functions
2 Try it yourself: linear function of continuous random variables I LetXbe a random variable with PDFfX(x), and letY= 2X+ 3 I
Then the CDF ofYis given by:
F
Y(y) =P(2X+ 3y) =P
Xy32 =FXy32 I
Dierentiating, we get:
f
Y(y) =dFY(y)dy
=dFXy32 dy 12 fXy32 3 Try it yourself: linear function of continuous random variables I LetXbe a random variable with PDFfX(x), and letY=2X+ 3 I
Then the CDF ofYis given by:
F
Y(y) =P(2X+ 3y) =P
X3y2 = 1FX3y2 I
Dierentiating, we get:
f
Y(y) =dFY(y)dy
=dFX3y2 dy 12 fXy3(2) 4
Functions of continuous random variables
I LetXbe a random variable with PDFfX(x), and letY=aX+b I f
Y(y) =dFYdy
(y) =8 >:1a fXyba a>0 1a fXyba a<0
1jajfXyba
5
Linear functions of continuous random variables
I So, for any continuous random variableX, ifY=aX+b, then f
Y(y) =1jajfXyba
I Let's consider a normal random variable...XN(0;1), then what is the distribution ofY=aX+b?I
We know that
f
X(x) =1p2ex2=2I
So, f
Y(y) =1jajfXyba
=1jaj1p2expfyba 2 =2g
1p2jajexp
(yb)22a2!I This is just the PDF of a normal distribution with meanband variancea2!I
In fact, ifXN(;2), thenY=aX+bN(a+b;a22).6
Linear functions of continuous random variables
I So, for any continuous random variableX, ifY=aX+b, then f
Y(y) =1jajfXyba
I Let's consider a normal random variable...XN(0;1), then what is the distribution ofY=aX+b?I
We know that
f
X(x) =1p2ex2=2I
So, f
Y(y) =1jajfXyba
=1jaj1p2expfyba 2 =2g
1p2jajexp
(yb)22a2!I This is just the PDF of a normal distribution with meanband variancea2!I
In fact, ifXN(;2), thenY=aX+bN(a+b;a22).6
Linear functions of continuous random variables
Iquotesdbs_dbs2.pdfusesText_2
Linear Approximations - Illinois Institute of Technology