lecture17.pdf

Lecture 17

Zhihua (Sophia) Su

University of Florida

Feb 17, 2015

STA 4321/5325 Introduction to Probability 1

Agenda

Expected Values of Continuous Random VariablesTchebysheff’s Inequality for Continuous Random VariablesExamples

Reading assignment: Chapter 4: 4.3


Expected Values of Continuous RandomVariables

DefinitionIf X is a continuous random variable with density fX , then theexpected value of X is defined as

E(X) =

ˆ ∞−∞

xfX(x)dx.

(Assuming that´∞−∞ | x | fX(x)dx <∞.)

Notice that it parallels with discrete random variables. The sumis replaced by an integral, and the probability mass function pXis replaced by the probability density function fX .



ResultIf X is a continuous random variable with density fX , then forany function g: R→ R,

E[g(X)] =

ˆ ∞−∞

g(x)fX(x)dx.

(Assuming that´∞−∞ | g(x) | fX(x)dx <∞.)



The variance of a continuous random variable is defined in asimilar fashion as a discrete random variable.

DefinitionIf X is a continuous random variable with probability densityfunction fX , then the variance of X is given by

V (X) = E[(X − E(X))2]

=

ˆ ∞−∞

(x− E(X))2fX(x)dx.


Expected Values of Continuous RandomVariablesLet us look at an example to understand these definitions andresults.

Example: For a given teller in a bank, let X denote theproportion of time, out of a 40-hour work week, that he isdirectly serving the customers. Suppose that X has aprobability density function given by

fX(x) =

{3x2 if 0 ≤ x ≤ 1,

0 otherwise.

(a) Find the mean proportion of time during a 40-hour workweek the teller directly serves customers.



Example (cont’d)(b) Find the variance of the proportion of time during a 40-hourwork week the teller directly serves customers.



There is another way to determine expectations of non-negativecontinuous random variables directly from their distributionfunctions.

ResultIf X is a non-negative continuous random variable, then

E(X) =

ˆ ∞0

P (X > x)dx =

ˆ ∞0

[1− FX(x)]dx.


Tchebysheff’s Inequality for ContinuousRandom Variables

The Tchebysheff’s theorem holds for continuous randomvariables in the same way as for discrete random variables.

ResultIf X is a continuous random variable with E(X) = µX andSD(X) = σX , then for any k ≥ 0,

P (| X − µX |< kσX) ≥ 1− 1

k2.


Examples

Example: The distribution function of the random variable X,the time (in years) from the time a machine is serviced until itbreaks down, is as follows:

FX(x) =

{1− e−4x if x > 0,

0 otherwise.

(a) Find E(X), V (X).


Examples

(b) Find an interval such that the probability that X lies in theinterval is at least 75%.


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