Problem:
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
http://clue.eng.iastate.edu/~zhengdao/teaching/ee322/spring03/notes/note7/l7.pdf
The Aarhus School of Business: Introduction to Expectation and
Variance
http://www.hha.dk/ifi/QEM/expvar.pdf
University of Texas at Dallas: Expectation of functions of random
variables
http://www.utdallas.edu/~ammann/cs3341/node24.html
I hope that answers everything you needed to know. Please request any
clarifications you may have. Thank you for using Google Answers.
xargon-ga
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