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Exponential Distribution memory loss

  1. Apr 11, 2013 #1

    Mentallic

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    1. The problem statement, all variables and given/known data

    Show that the exponential distribution has the memory loss property.


    2. Relevant equations

    [tex]f_T(t) = \frac{1}{\beta}e^{-t/\beta}[/tex]

    The memory loss property exists if we can show that

    [tex]P(X>s_1+s_2|X>s_1) = P(X>s_2)[/tex]

    Where

    [tex]P(X>s_1+s_2|X>s_1)=\frac{P(X>s_1+s_2,X>s_1)}{P(X>s_1)}[/tex]



    3. The attempt at a solution

    For [itex]P(X>s_2)[/itex] I computed the integral

    [tex]1-\int_0^{s_2}f_T(t)dt[/tex]

    And similarly I found [itex]P(X>s_1)[/itex] but I'm unsure how to find [itex]P(X>s_1+s_2,X>s_1)[/itex] to complete the problem.
     
  2. jcsd
  3. Apr 11, 2013 #2

    HallsofIvy

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    This is miswritten- probably a typo. You mean
    [tex]P(X> x_1+ s_2|X> s_1)= \frac{P(X> s_1+ s_2)}{P(X> s_1)}[/tex]
    That is, the numerator on the right is NOT the condition probability (if it were you could cancel it with the left).



    That would be
    [tex]\frac{1- \beta\int_0^{s_1+ s_2}e^{-t/\beta} dt}{1- \beta\int_0^{s_1} e^{-t/\beta}dt}[/tex]
     
  4. Apr 11, 2013 #3

    Ray Vickson

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    This property is normally called memoryless, not memory loss. Anyway, if ##s_1, s_2 > 0## then ##P(X > s_1+s_2, X > s_1) = P(X > s_1 + s_2),## because if ##X > s_1 + s_2## then we automatically have ##X > s_1##.
     
  5. Apr 12, 2013 #4

    Mentallic

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    Actually I had it right the first time, but I didn't quite understand what it meant. I now realize that my notes' mention of

    [tex]P(X>s_1+s_2,X>s_1)[/tex]

    is equivalent to

    [tex]P(X>s_1+s_2 \cap X>s_1)[/tex]

    and Ray supported that claim:

    Thanks guys!

    Yeah, that's my fault :biggrin:
     
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