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Math Help - Probabilities Dealing with Independent Exponential RVs

  1. #1
    Rhymes with Orange Chris L T521's Avatar
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    Probabilities Dealing with Independent Exponential RVs

    This has been bothering me for a while now:

    Q: If X_i,\,i=1,2,3 are independent exponential random variables with rates \lambda_i,\,i=1,2,3, find

    (a) \mathbb{P}(X_1<X_2<X_3)

    Here's what I figured out so far:

    Let Y=\min(X_2,X_3) Then Y is \mathcal{E}(\lambda_2+\lambda_3) and is independent of X_1. Now observe that \mathbb{P}(X_1<Y)=\frac{\lambda_1}{\lambda_1+\lamb  da_2+\lambda_3}. We now see that

    \begin{aligned}\mathbb{P}(X_1<X_2<X_3) &= \mathbb{P}(X_1=\min(X_1,X_2,X_3))\mathbb{P}(X_2<X_  3\mid X_1=\min(X_1,X_2,X_3))\\&=\frac{\lambda_1}{\lambda  _1+\lambda_2+\lambda_3}\mathbb{P}(X_2<X_3\mid X_1=\min(X_1,X_2,X_3))\end{aligned}

    Now, take note that the distribution of X_2-x and X_3-x given that they are larger than X_1 = x is the same as that of X_2 and X_3, so

    \begin{aligned}\mathbb{P}(X_2<X_3\mid\min(X_2,X_3)  \geq x) &= \mathbb{P}((X_2-x<X_3-x)\mid\min(X_2,X_3)\geq x)\\ &=\mathbb{P}(X_2<X_3)\\ &=\frac{\lambda_2}{\lambda_2+\lambda_3}\end{aligne  d}

    ----------------------------------------

    At this stage, I'm not sure how to tie it all together to get \mathbb{P}(X_1<X_2<X_3). Is it valid for me to state that we should simply get \mathbb{P}(X_1<X_2<X_3)=\left(\frac{\lambda_1}{\la  mbda_1+\lambda_2+\lambda_3}\right)\left(\frac{\lam  bda_2}{\lambda_2+\lambda_3}\right)?

    I would appreciate any help with this!
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  2. #2
    MHF Contributor
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    I cant comment on your solution but i think there might be an easier way
    Spoiler:

    you can get the joint pdf easily enough, the probability you want is then

    \displaystyle \intop_{x_1=0}^{x_1=x_2} ~ \intop_{x_2=x_1}^{x_2=x_3} ~ \intop_{x_3=x_2}^{x_3=\infty}f(x1,x2,x3) dx_3 dx_2 dx_1

    I haven't done the integration but i would have thought that it was tractable, given that f() is going to be very simple? If nothing else it could verify your answer
    Last edited by SpringFan25; June 29th 2010 at 02:47 AM.
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  3. #3
    Moo
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    Quote Originally Posted by Chris L T521 View Post
    \begin{aligned}\mathbb{P}(X_2<X_3\mid\min(X_2,X_3)  \geq x) &= \mathbb{P}((X_2-x<X_3-x)\mid\min(X_2,X_3)\geq x)\\ &=\mathbb{P}(X_2<X_3)\\ &=\frac{\lambda_2}{\lambda_2+\lambda_3}\end{aligne  d}
    Are you sure of the second equality ?

    Anyway, if you really want to do something along the lines of what you've already done, can't you write it this way :

    P(X_1<X_2<X_3)=P(X_2<X_3)P(X_1<\min(X_2,X_3)\mid X_2<X_3)

    But P(X_1<\min(X_2,X_3)\mid X_2<X_3)=P(X_1<X_2)

    So finally, P(X_1<X_2<X_3)=P(X_2<X_3)P(X_1<X_2)

    Did it go wrong somewhere ? oO


    Springfan25 : there's a problem with your boundaries, since for example x_3 appears after having integrated with respect to x_3, same for x_2.
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  4. #4
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    Quote Originally Posted by Moo View Post

    But P(X_1<\min(X_2,X_3)\mid X_2<X_3)=P(X_1<X_2)
    Are you sure about that one?

    it must be true that
     \displaystyle (1)~~~~~~P(X_1< X_2) =P(X_1<X_2 | X_2 < X_3)P(X_2 < X_3) + P(X_1 < X_2 | X_2 > X_3)P(X_2 > X_3)
    \displaystyle (1.1)~~~P(X_1< X_2) =P(X_1<X_2 | X_2 < X_3)P(X_2 < X_3) + P(X_1 < X_2 | X_2 > X_3)(1-P(X_2 < X_3))


    it must also be true that

    \displaystyle (2) ~~~P(X_1<\min(X_2,X_3)\mid X_2<X_3) = P(X_1 < X_2 | X_2 < X_3)

    You propose
    \displaystyle (3)~~~ P(X_1<X_2)=P(X_1<\min(X_2,X_3)\mid X_2<X_3)

    (2) and (3) Together imply:
    \displaystyle (4)~~~P(X_1<X_2) = P(X_1 < X_2 | X_2 < X_3)



    But i think (4) contradicts (1.1)
    \displaystyle (1.1) ~~~P(X_1< X_2) =P(X_1<X_2 | X_2 < X_3)P(X_2 < X_3) + P(X_1 < X_2 | X_2 > X_3)(1-P(X_2 < X_3))
    \displaystyle (4)~~~~~P(X_1<X_2) = P(X_1 < X_2 | X_2 < X_3)

    I can only see 2 ways that (4) could be consistent with (1.1)
    either \displaystyle P(X_2 < X_3) =1 (not true in general)

    or \displaystyle P(X_1<X_2 | X_2 < X_3) = P(X_1 < X_2 | X_2 > X_3). This is clearly not the case for any given value of x3 (eg, consider x3=0), and therefore (i think?!) not true....but im not sure about that reasoning.









    In the unlikely event that i am right about the above, i think the integration approach could still work. I may have the limits wrong but there is a constructable region R (albeit unbounded) on which the integral needs to be done. Here's my second go:

    Spoiler:

    \displaystyle \intop_{x_1=0}^{x_1=\infty} \intop_{x_2=x_1}^{x_2=\infty} \intop_{x_3=x_2}^{x_3=\infty}f_{x_1,x_2,x_3} dx_3 dx_2 dx_1<br />

    Im not sure about those upper limits though. I think it would work for a finite upper limit that was the same for x1, x2, x3; but i dont know if it works with infinity.

    Last edited by SpringFan25; June 30th 2010 at 02:46 PM.
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