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Exponential Distribution and median

  1. Apr 2, 2010 #1
    1. The problem statement, all variables and given/known data

    IF X has an exponential distribution with parameter [itex]\lambda[/itex], derive a general expression for the (100p)th percentile of the distribution. Then specialize to obtain the median.

    2. Relevant equations

    [tex]X.exp(\lambda)=\lambda e^{-\lambda x} for x>0[/tex]

    [tex]p=F(\eta (p)) = \int_{-\infty}^{\eta (p)} f(y)dy[/tex]

    3. The attempt at a solution
    First, can I verify that this antiderivative is correct?
    Since F'(X) = f(X) [tex]F(X)=-e^{-\lambda x}[/tex]
    Finding the 100pth percentile is equivalent to finding the cumulative density function (the antiderivative) from 0-->p correct?

    And F(.5) = [itex]\eta[/itex] for the mean...

    I'm really confused on how to set this up and what i'm looking for. Any help would be great.
     
  2. jcsd
  3. Apr 2, 2010 #2

    rock.freak667

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    Homework Helper

    Your limits of integration are correct, you are just missing one term in the expression for P(X≤x)


    [tex]P(X \leq x) = \int _0 ^x \lambda e^{-\lamda t} dt[/tex]


    re-integrate it and you'll see.

    Also I think you want to find the median 'm' such that F(m)=1/2 or in the integral form

    [tex]\int _0 ^m \lambda e^{-\lamda t} dt =1/2[/tex]
     
  4. Apr 2, 2010 #3
    I don't follow at all. Can you try to explain it to me in english rather than math? I'm not a math major and the language confuses me more than anything else.
     
  5. Apr 2, 2010 #4
    When you write cumulative density function you probably mean cumulative distribution function. As far as I know cumulative density function has no meaning in probability theory and Wikipedia writes:
    However from you usage it seems clear that you intend for it to be cumulative distribution function (cdf).

    This is the right antiderivative for non-negative numbers however it may not be exactly the one you're looking for (remember there are infinitely many that can be obtained by adding constants). Remember that f(x)=0 for x<0 so the antiderivative is 0 if x < 0.
    [tex]G(x) = \begin{cases} -e^{-\lambda x} & 0 \leq x \\ 0 & x < 0 \end{cases}[/tex]
    is probably what you meant.

    Now as you wrote earlier the cdf is:
    [tex]F(x) = \int_{-\infty}^x f(y) dy = G(x) - \lim_{y \to -\infty}G(y)[/tex]
    It's not enough for it be an antiderivative of f(y), it must be the right one (given by the equation above).
    From that you should be able to compute the correct cdf (and it's not equal to G(x)).

    I'm not sure exactly what finding the cdf from 0->p means, but I don't think it's the right idea. By definition we say that a value x is the 100pth percentile if F(x)=p. Thus you need to solve this equation for x to find an expression for the 100pth percentile.

    The mean is just a fancy word for the 50th percentile. That is we call m the mean if F(m)=.5
     
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