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Exponential distribution

  1. Nov 7, 2007 #1
    1. The problem statement, all variables and given/known data

    component has lifetime X that is exponentialy distibuted with parameter gamma.

    a) if the cost per unit time is a constant, c, what is the expectec cost of its lifetime?
    b) if c is not constant, and=c(1-.5e^(a*x) such that a<0. (aka it costs more over time) what is its cost inrespect to its lifetime?

    2. Relevant equations

    f(x)=gamma*e^(-gamma*x)
    E(X)=1/gamma
    Var(X)=1/gamma^2


    3. The attempt at a solution


    i thitnk its just

    a) c/gamma
    b) (c(1-.5e^(ax)))/gamma

    but that seems to easy...
     
    Last edited: Nov 7, 2007
  2. jcsd
  3. Nov 7, 2007 #2

    EnumaElish

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    To find E[cost] you need to integrate the relevant unit cost (per unit time) weighted by f(x) over the entire domain of X.
     
  4. Nov 8, 2007 #3
    what do you mean "weighted by f(x)"?

    so integrate c...so cx

    and integrate (c(1-.5e^(ax)))/gamma
    which is cx-(.5/a)e^(ax)

    where x is time

    but dunno what to do now
     
  5. Nov 8, 2007 #4

    HallsofIvy

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    Well, you didn't actually integrate "over the entire domain of X"! You found the anti-derivative which is not quite the same thing. For one thing, the "expected value" does not depend upon x. "Weighted by f(x)" means, here, "multiplied by f(x)" before you integrate which is apparently what you did. What is the entire domain of X?
     
  6. Nov 8, 2007 #5
    since x is t then 0 to infinity?
     
  7. Nov 8, 2007 #6
    f(x)=gamma*e^(-gamma*x)

    so would i go

    c*gamma*e^(-gamma*x)
    and
    (c(1-.5e^(ax)))/gamma* gamma*e^(-gamma*x)

    and then integrate form x=0 to infinity?
     
  8. Nov 8, 2007 #7

    EnumaElish

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    Correct.
     
  9. Nov 8, 2007 #8
    c*gamma*e^(-gamma*x)

    becomes (-C*gamma^2)*e^(-gamma*x) from 0 to infin
    so at x=infin that equals 0
    at x=0, it is -c*gamma^2

    so c*gamma^2 is my first answer?

    and then (c(1-.5e^(ax)))/gamma* gamma*e^(-gamma*x)
    becomes c(1-.5e^(ax)))*e^(-gamma*x)
    so C(e^(-gamma*x)-e^(-gamma*x+ax))
    which becomes Ce^(-gamma*x)-Ce^(-gamma*x+ax)
    and again...Ce^(-gamma*x)-Ce^(x(a-gamma))
    so integrating that i get

    (-C/gamma)e^(-gamma*x)-(C/(a-gamma)e^(x(a-gamma))) from 0 to infin
    again, at infin it equals 0
    at 0, the first expression is -C/gamma and the 2nd is C/(a-gamma)
    so that means 0-(-C/gamma -C/(a-gamma) )
    so my answer is C/gamma +C/(a-gamma)?????
     
  10. Nov 8, 2007 #9
    just checked over my mtath and i did them wrong...the first answer is just C

    and the 2nd would be C(1-(.5gamma/a-gamma)

    NOTE i made a mistake real early...its (c(1-.5e^(ax))) not (c(1-.5e^(ax)))/gamma
    and changed that for this above calc
     
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