Results 1 to 4 of 4

Thread: Probabilities Dealing with Independent Exponential RVs

  1. #1
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    5

    Probabilities Dealing with Independent Exponential RVs

    This has been bothering me for a while now:

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

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

    Here's what I figured out so far:

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

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

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

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

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

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

    I would appreciate any help with this!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    May 2010
    Posts
    1,034
    Thanks
    28
    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

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

    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; Jun 29th 2010 at 02:47 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Chris L T521 View Post
    $\displaystyle \begin{aligned}\mathbb{P}(X_2<X_3\mid\min(X_2,X_3) \geq x) &= \mathbb{P}((X_2-x<X_3-x)\mid\min(X_2,X_3)\geq x)\\ &=\mathbb{P}(X_2<X_3)\\ &=\frac{\lambda_2}{\lambda_2+\lambda_3}\end{aligne d}$
    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 :

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

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

    So finally, $\displaystyle P(X_1<X_2<X_3)=P(X_2<X_3)P(X_1<X_2)$

    Did it go wrong somewhere ? oO


    Springfan25 : there's a problem with your boundaries, since for example $\displaystyle x_3$ appears after having integrated with respect to $\displaystyle x_3$, same for $\displaystyle x_2$.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    May 2010
    Posts
    1,034
    Thanks
    28
    Quote Originally Posted by Moo View Post

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

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


    it must also be true that

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

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

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



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

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

    or$\displaystyle \displaystyle P(X_1<X_2 | X_2 < X_3) = P(X_1 < X_2 | X_2 > X_3)$. 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 \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
    $

    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; Jun 30th 2010 at 02:46 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Exponent C When Dealing With Probabilities?
    Posted in the Statistics Forum
    Replies: 4
    Last Post: Jan 31st 2011, 12:03 AM
  2. Replies: 0
    Last Post: Apr 13th 2010, 08:37 PM
  3. Independent probabilities question
    Posted in the Statistics Forum
    Replies: 2
    Last Post: Feb 3rd 2010, 03:38 PM
  4. Joint PDF of two independent exponential RV
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: Dec 1st 2009, 07:38 AM
  5. Independent exponential R.V.
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: May 4th 2009, 05:14 AM

Search Tags


/mathhelpforum @mathhelpforum