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Homework Help: Exponential Distribution memory loss

  1. Apr 11, 2013 #1


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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]



    3. The attempt at a solution

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


    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


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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


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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


    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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