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Q: Probability - Random Variable* ( Answered 4 out of 5 stars,   0 Comments )
Subject: Probability - Random Variable*
Category: Science > Math
Asked by: alpa101473-ga
List Price: $3.50
Posted: 22 Sep 2003 20:18 PDT
Expires: 22 Oct 2003 20:18 PDT
Question ID: 259269
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Subject: Re: Probability - Random Variable*
Answered By: xargon-ga on 22 Sep 2003 21:34 PDT
Rated:4 out of 5 stars

Let Y = A*cos(w*t) + c, where A has mean m and variance sigma^2, and w
and c are constants.  Find the mean and variance of Y.

In statistics, the mean is often described as the "expectation value",
often written as E[X], where X is a statistical variable. 
(Techicialy, there is a distinction, but for our purposes we can say
this; the third link I have provided makes clear the distinction.)  In
your problem, Y is such a variable.  The variance of a varible is
commonly notated as V[X].

Looking at some general properties for expectation and variance:

if we let a variable Y = aX + b
then: E[Y] = a*E[X] + b
and: V[Y] = (a^2)*V[X]

This is known as a linear transformation on the variable X
(multiplying by a constant and adding a constant term).  The links I
have provided provide proofs of thse properties.

Returning to the specific question at hand, we can apply the rules for
a linear transformation.  Given that Y = A*cos(w*t) + c, we know that
E[A] = m and V[A] = sigma^2.  Therefore:

E[Y] = E[A*cos(w*t) + c] = E[A]*cos(w*t) + c = m*cos(w*t) + c

V[Y] = V[A*cos(w*t) + c] = V[A]*(cos(w*t))^2 = (sigma^2)*(cos(w*t))^2

The following sites provide references for basic statistical
identities as well as proofs:

Iowa State University: Expectation, Mean, Variance

The Aarhus School of Business: Introduction to Expectation and

University of Texas at Dallas: Expectation of functions of random

I hope that answers everything you needed to know.  Please request any
clarifications you may have.  Thank you for using Google Answers.


Search used:

variance expectation
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