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Exponential Distribution with Probability

  1. Dec 7, 2013 #1
    $$f(y) = \begin{cases} \int_0^y\frac1\beta e^{\frac {-t}\beta}dt = -e^{\frac {-y}\beta}+1 & \text{for } 0 ≤ y < ∞,\\ 0& \text{for } elsewhere\end{cases}$$

    P(Y>3) = 1 - P(Y ≤ 3) = 1 - (-e^{-3/beta}+1) = .1353

    When I take log to both sides, I get 3.453.
    When I take ln to both sides, I get 1.4998. When I plug it back into the equation, 1.4998 looks right...However, I puzzle why there is a difference?

    Before this, could anyone please make sure beta = 1.4998?

    (1) P(Y<0) = 0, right?

    (2) P(Y<1) = -e^{1/1.4998} + 1 = .4866 , right?
     
    Last edited: Dec 7, 2013
  2. jcsd
  3. Dec 7, 2013 #2

    mfb

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    ln(e^x)=x, but log10(e^x) is not x. I guess you did a calculation error somewhere.
     
  4. Dec 7, 2013 #3
    you mean beta is not 1.4998?

    Could you also check (1) and (2) for me?
     
  5. Dec 7, 2013 #4

    Ray Vickson

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    Your notation is against all the conventions. Almost 100% of the time we denote the density function by f and the (cumulative) distribution function by F, so
    [tex] f(t) = \begin{cases}r e^{-rt}, & t \geq 0\\ 0, & t < 0 \end{cases} \; \text{ and }\; F(t) = \int_{-\infty}^t f(s) \, ds = \begin{cases}1-e^{-rt} & t \geq 0\\ 0 & t < 0 \end{cases}
    [/tex]
    It is also much more convenient to write the exponential in terms of the rate parameter (r) instead of the mean (1/r); however, if you prefer to use 1/r that is OK too. It is also faster to remember the result that ##P\{Y > t\} = e^{-rt}##, but, of course, that is what you ended up doing.

    Your results look OK to me.
     
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