Continuous Random Variables Chapter 5

February 4, 2018 | Author: Anonymous | Category: Math, Statistics And Probability, Normal Distribution
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Continuous Random Variables Chapter 5 Nutan S. Mishra Department of Mathematics and Statistics University of South Alabama

Continuous Random Variable When random variable X takes values on an interval For example GPA of students X  [0, 4] High day temperature in Mobile X  ( 20,∞) Recall in case of discrete variables a simple event was described as (X = k) and then we can compute P(X = k) which is called probability mass function In case of continuous variable we make a change in the definition of an event.

Continuous Random Variable Let X  [0,4], then there are infinite number of values which x may take. If we assign probability to each value then P(X=k)  0 for a continuous variable In this case we define an event as (x-x ≤ X ≤ x+x ) where x is a very tiny increment in x. And thus we assign the probability to this event P(x-x ≤ X ≤ x+x ) = f(x) dx f(x) is called probability density function (pdf)

Properties of pdf f (x)  0 upper lim



f ( x ) dx  1

lower lim

(cumulative) Distribution Function The cumulative distribution function of a continuous random variable is a

F (a )  P ( X  a ) 



f ( x ) dx

lower lim it

Where f(x) is the probability density function of x. b

P (a  x  b)  P (a  x  b) 

 a

 F (b )  F ( a )

f ( x ) dx

Relation between f(x) and F(x) x

F ( x) 



f ( t ) dt



dF ( x ) dx

 f ( x)

Mean and Variance upper lim

 xf ( x ) dx

 

lower lim upper lim



2



 (x   )

2

f ( x ) dx

lower lim upper lim



2





x f ( x ) dx   2

lower lim

2

Exercise 5.2 To find the value of k 1

 kx  1 3

0 1

L . H .S .  k  x  k [ 3

0

k 4

Thus f(x) = 4x3 for 0
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