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Life of batteries (Probability)

  1. Dec 8, 2007 #1
    1. The problem statement, all variables and given/known data
    The lifetime of batteries are independent random variables Xi, i= 1,2,... each having exponential distribution given by the density f(x) = 2e^-2x, x>0, 0 elsewhere.
    If a flashlight needs two batteries to work, then the time that the flashlight can operate is a random variable Y=min(X1,X2). Find the CDF and PDF of Y


    2. Relevant equations
    Xi, i= 1,2,...
    Xi ~ f(x) = 2e^-2x, x>0, 0 elsewhere
    Y=min(X1,X2)

    3. The attempt at a solution
    What i'm not sure about is the initial approach to this problem. I'm not sure if I should treat it as an order statistics problem in which case i'd be trying to use the equation:

    g(y) = n((1-Fx(y))^(n-1))f(y)

    though i'm not sure if Fx(y) and f(y) are just the respective cdf and pdf of X.
    I know if i have the pdf, I can get the cdf easily.
     
  2. jcsd
  3. Dec 8, 2007 #2

    Avodyne

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    Do you know what a Dirac delta function is? Given a PDF f(x1,x2) for two variables x1 and x2, the PDF g(y) for any function y=h(x1,x2) of these is given by
    [tex]g(y) = \int dx_1\,dx_2\,f(x_1,x_2)\delta(y-h(x_1,x_2))[/tex]
    where [itex]\delta(y)[/itex] is the Dirac delta function, and where x1 and x2 are integrated over their complete ranges (whatever they are).
     
  4. Dec 9, 2007 #3
    Thanks for the help. I don't recall covering the dirac delta at all but i'm trying to use it and we'll see how it works. I was thinking there might be a transform involved as well? Thanks again, I appreciate it.
     
  5. Dec 9, 2007 #4

    Avodyne

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    I was able to use this method to solve the problem, but it's tricky. There's probably (no pun intended) a method that avoids it; maybe someone else can help.
     
  6. Dec 9, 2007 #5

    EnumaElish

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    Since Y is min(X1, X2), and if Prob{Xi < x) = F(x), what is the probability Prob{min(X1,X2) < y), in terms of the F function?
     
  7. Dec 10, 2007 #6
    I just can't figure out how to deal with that "min". I know that if say Y=X1^2 then I would just follow:

    cdf = Fy(y) = P(Y<y) = P(X^2<y) = P(X<Y^1/2) etc, but I don't know how to represent that 'minimum' when i'm trying to solve the inequality for X.

    Thanks again.
     
  8. Dec 10, 2007 #7

    EnumaElish

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    If the minimum of X1 and X2 is less than y, then it cannot be the case that "both X1 > y and X2 > y."
     
  9. Dec 10, 2007 #8
    Would it correct to say then P(Y1>y) = P(X1>Y)*P(X2>Y) ?
    Thanks.
     
  10. Dec 10, 2007 #9

    EnumaElish

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    Exactly, or P(Y<y) = 1 - (1-P(X1<Y))*(1-P(X2<Y)).
     
  11. Dec 11, 2007 #10
    Ok, i'm coming up with a pdf of:

    (4e^-2y)*(1+e^-2y)

    OR

    (4e^-2y)+(4e^-4y)

    Thoughts?
     
  12. Dec 12, 2007 #11

    Avodyne

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    Not what I got. And I got the same answer with my fancy delta-function method and EmulaFish's simpler method.

    Can you show your work?
     
  13. Dec 12, 2007 #12
    well after talking to my professor, he said I should be using the order statistic technique

    pdf of Y=min(X1,X2...Xn) = g(y) = n*[(1-Fx(y))^n-1]*fx(y)

    where n=2 for this case
    Fx(y) is cdf of X evaluated at y = -e^-2y
    fx(y) is pdf of X evaluated at y = 2e^-2y

    I put those into g(y) and that's how I came up with that.
    Thanks again.
     
  14. Dec 13, 2007 #13
    Any ideas on what I might be doing wrong? Thanks.
     
  15. Dec 13, 2007 #14
    I figured out one thing i've been doing wrong. My new answer for pdf of Y=min(X1,X2) is

    4e^-4y

    I'm pretty confident about that one.
     
  16. Dec 13, 2007 #15

    Avodyne

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    That's what I got!
     
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