Poisson Process Probability of a machine failing before other












2












$begingroup$



Machine $1$ is working now. Machine $2$ will be switched on at time
$t$. Suppose that machine $1$ fails at rate $λ_1$ and $2$ at rate
$λ_2$with an exponential waiting time. What is the probability that
machine $2$ fails first?




I thought:



It should be $P(X_2+t<X_1)$ or $P(X_2<X_1|X_2=t)$ but I don't know how to go about calculating it.



Can someone give me a hint?






probability poisson-distribution poisson-process







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Calculate the probability M1 fails before time $t$. Once both are working, probability M1<M2 is $frac {lambda_1}{lambda_1+lambda_2}$
    $endgroup$
    – A.S.
    Mar 4 '16 at 10:17










  • $begingroup$
    This is definitely $P(X_2+t<X_1)$ (I cannot even fathom what $P(X_2<X_1|X_2=t)$ stands for...) and one knows that, for every nonnegative $x$, $P(X_1>x)=e^{-lambda_1x}$ hence, by independence, $P(X_1>X_2+t)=E(e^{-lambda_1(X_2+t)})=e^{-lambda_1t}E(e^{-lambda_1X_2})$. Now, can you compute $E(e^{-sX_2})$ for every $s$?
    $endgroup$
    – Did
    Mar 4 '16 at 10:17












  • $begingroup$
    @A.S. would this be simply $P(X_1>t)P(X_1>X_2)$?
    $endgroup$
    – GRS
    Mar 4 '16 at 10:25










  • $begingroup$
    Yes. I was computing the complementary probability instead. Your answer is coincides with Did's term-for-term.
    $endgroup$
    – A.S.
    Mar 4 '16 at 10:27












  • $begingroup$
    @Did I had a think about it, but I don't understand why we are taking expectation? I know how to calculate $E(X_2)=1/lambda_2$
    $endgroup$
    – GRS
    Mar 4 '16 at 10:27
















2












$begingroup$



Machine $1$ is working now. Machine $2$ will be switched on at time
$t$. Suppose that machine $1$ fails at rate $λ_1$ and $2$ at rate
$λ_2$with an exponential waiting time. What is the probability that
machine $2$ fails first?




I thought:



It should be $P(X_2+t<X_1)$ or $P(X_2<X_1|X_2=t)$ but I don't know how to go about calculating it.



Can someone give me a hint?






probability poisson-distribution poisson-process







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Calculate the probability M1 fails before time $t$. Once both are working, probability M1<M2 is $frac {lambda_1}{lambda_1+lambda_2}$
    $endgroup$
    – A.S.
    Mar 4 '16 at 10:17










  • $begingroup$
    This is definitely $P(X_2+t<X_1)$ (I cannot even fathom what $P(X_2<X_1|X_2=t)$ stands for...) and one knows that, for every nonnegative $x$, $P(X_1>x)=e^{-lambda_1x}$ hence, by independence, $P(X_1>X_2+t)=E(e^{-lambda_1(X_2+t)})=e^{-lambda_1t}E(e^{-lambda_1X_2})$. Now, can you compute $E(e^{-sX_2})$ for every $s$?
    $endgroup$
    – Did
    Mar 4 '16 at 10:17












  • $begingroup$
    @A.S. would this be simply $P(X_1>t)P(X_1>X_2)$?
    $endgroup$
    – GRS
    Mar 4 '16 at 10:25










  • $begingroup$
    Yes. I was computing the complementary probability instead. Your answer is coincides with Did's term-for-term.
    $endgroup$
    – A.S.
    Mar 4 '16 at 10:27












  • $begingroup$
    @Did I had a think about it, but I don't understand why we are taking expectation? I know how to calculate $E(X_2)=1/lambda_2$
    $endgroup$
    – GRS
    Mar 4 '16 at 10:27














2












2








2





$begingroup$



Machine $1$ is working now. Machine $2$ will be switched on at time
$t$. Suppose that machine $1$ fails at rate $λ_1$ and $2$ at rate
$λ_2$with an exponential waiting time. What is the probability that
machine $2$ fails first?




I thought:



It should be $P(X_2+t<X_1)$ or $P(X_2<X_1|X_2=t)$ but I don't know how to go about calculating it.



Can someone give me a hint?






probability poisson-distribution poisson-process







share|cite|improve this question









$endgroup$





Machine $1$ is working now. Machine $2$ will be switched on at time
$t$. Suppose that machine $1$ fails at rate $λ_1$ and $2$ at rate
$λ_2$with an exponential waiting time. What is the probability that
machine $2$ fails first?




I thought:



It should be $P(X_2+t<X_1)$ or $P(X_2<X_1|X_2=t)$ but I don't know how to go about calculating it.



Can someone give me a hint?






probability poisson-distribution poisson-process




probability poisson-distribution poisson-process






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 4 '16 at 10:12









GRSGRS

1,140927




1,140927








  • 1




    $begingroup$
    Calculate the probability M1 fails before time $t$. Once both are working, probability M1<M2 is $frac {lambda_1}{lambda_1+lambda_2}$
    $endgroup$
    – A.S.
    Mar 4 '16 at 10:17










  • $begingroup$
    This is definitely $P(X_2+t<X_1)$ (I cannot even fathom what $P(X_2<X_1|X_2=t)$ stands for...) and one knows that, for every nonnegative $x$, $P(X_1>x)=e^{-lambda_1x}$ hence, by independence, $P(X_1>X_2+t)=E(e^{-lambda_1(X_2+t)})=e^{-lambda_1t}E(e^{-lambda_1X_2})$. Now, can you compute $E(e^{-sX_2})$ for every $s$?
    $endgroup$
    – Did
    Mar 4 '16 at 10:17












  • $begingroup$
    @A.S. would this be simply $P(X_1>t)P(X_1>X_2)$?
    $endgroup$
    – GRS
    Mar 4 '16 at 10:25










  • $begingroup$
    Yes. I was computing the complementary probability instead. Your answer is coincides with Did's term-for-term.
    $endgroup$
    – A.S.
    Mar 4 '16 at 10:27












  • $begingroup$
    @Did I had a think about it, but I don't understand why we are taking expectation? I know how to calculate $E(X_2)=1/lambda_2$
    $endgroup$
    – GRS
    Mar 4 '16 at 10:27














  • 1




    $begingroup$
    Calculate the probability M1 fails before time $t$. Once both are working, probability M1<M2 is $frac {lambda_1}{lambda_1+lambda_2}$
    $endgroup$
    – A.S.
    Mar 4 '16 at 10:17










  • $begingroup$
    This is definitely $P(X_2+t<X_1)$ (I cannot even fathom what $P(X_2<X_1|X_2=t)$ stands for...) and one knows that, for every nonnegative $x$, $P(X_1>x)=e^{-lambda_1x}$ hence, by independence, $P(X_1>X_2+t)=E(e^{-lambda_1(X_2+t)})=e^{-lambda_1t}E(e^{-lambda_1X_2})$. Now, can you compute $E(e^{-sX_2})$ for every $s$?
    $endgroup$
    – Did
    Mar 4 '16 at 10:17












  • $begingroup$
    @A.S. would this be simply $P(X_1>t)P(X_1>X_2)$?
    $endgroup$
    – GRS
    Mar 4 '16 at 10:25










  • $begingroup$
    Yes. I was computing the complementary probability instead. Your answer is coincides with Did's term-for-term.
    $endgroup$
    – A.S.
    Mar 4 '16 at 10:27












  • $begingroup$
    @Did I had a think about it, but I don't understand why we are taking expectation? I know how to calculate $E(X_2)=1/lambda_2$
    $endgroup$
    – GRS
    Mar 4 '16 at 10:27








1




1




$begingroup$
Calculate the probability M1 fails before time $t$. Once both are working, probability M1<M2 is $frac {lambda_1}{lambda_1+lambda_2}$
$endgroup$
– A.S.
Mar 4 '16 at 10:17




$begingroup$
Calculate the probability M1 fails before time $t$. Once both are working, probability M1<M2 is $frac {lambda_1}{lambda_1+lambda_2}$
$endgroup$
– A.S.
Mar 4 '16 at 10:17












$begingroup$
This is definitely $P(X_2+t<X_1)$ (I cannot even fathom what $P(X_2<X_1|X_2=t)$ stands for...) and one knows that, for every nonnegative $x$, $P(X_1>x)=e^{-lambda_1x}$ hence, by independence, $P(X_1>X_2+t)=E(e^{-lambda_1(X_2+t)})=e^{-lambda_1t}E(e^{-lambda_1X_2})$. Now, can you compute $E(e^{-sX_2})$ for every $s$?
$endgroup$
– Did
Mar 4 '16 at 10:17






$begingroup$
This is definitely $P(X_2+t<X_1)$ (I cannot even fathom what $P(X_2<X_1|X_2=t)$ stands for...) and one knows that, for every nonnegative $x$, $P(X_1>x)=e^{-lambda_1x}$ hence, by independence, $P(X_1>X_2+t)=E(e^{-lambda_1(X_2+t)})=e^{-lambda_1t}E(e^{-lambda_1X_2})$. Now, can you compute $E(e^{-sX_2})$ for every $s$?
$endgroup$
– Did
Mar 4 '16 at 10:17














$begingroup$
@A.S. would this be simply $P(X_1>t)P(X_1>X_2)$?
$endgroup$
– GRS
Mar 4 '16 at 10:25




$begingroup$
@A.S. would this be simply $P(X_1>t)P(X_1>X_2)$?
$endgroup$
– GRS
Mar 4 '16 at 10:25












$begingroup$
Yes. I was computing the complementary probability instead. Your answer is coincides with Did's term-for-term.
$endgroup$
– A.S.
Mar 4 '16 at 10:27






$begingroup$
Yes. I was computing the complementary probability instead. Your answer is coincides with Did's term-for-term.
$endgroup$
– A.S.
Mar 4 '16 at 10:27














$begingroup$
@Did I had a think about it, but I don't understand why we are taking expectation? I know how to calculate $E(X_2)=1/lambda_2$
$endgroup$
– GRS
Mar 4 '16 at 10:27




$begingroup$
@Did I had a think about it, but I don't understand why we are taking expectation? I know how to calculate $E(X_2)=1/lambda_2$
$endgroup$
– GRS
Mar 4 '16 at 10:27










1 Answer
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oldest

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$begingroup$

Let $X_i$ be the failure time of machine $i$. By lack of memory we have
begin{align}
mathbb P(X_2+t<X_1) &= mathbb P(X_1>X_2+tmid X_1>t)mathbb P(X_1>t)\
&= mathbb P(X_1>X_2)mathbb P(X_1>t)\
&= left(frac{lambda_2}{lambda_1+lambda_2}right) e^{-lambda_1 t}.
end{align}






share|cite|improve this answer











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    1












    $begingroup$

    Let $X_i$ be the failure time of machine $i$. By lack of memory we have
    begin{align}
    mathbb P(X_2+t<X_1) &= mathbb P(X_1>X_2+tmid X_1>t)mathbb P(X_1>t)\
    &= mathbb P(X_1>X_2)mathbb P(X_1>t)\
    &= left(frac{lambda_2}{lambda_1+lambda_2}right) e^{-lambda_1 t}.
    end{align}






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Let $X_i$ be the failure time of machine $i$. By lack of memory we have
      begin{align}
      mathbb P(X_2+t<X_1) &= mathbb P(X_1>X_2+tmid X_1>t)mathbb P(X_1>t)\
      &= mathbb P(X_1>X_2)mathbb P(X_1>t)\
      &= left(frac{lambda_2}{lambda_1+lambda_2}right) e^{-lambda_1 t}.
      end{align}






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Let $X_i$ be the failure time of machine $i$. By lack of memory we have
        begin{align}
        mathbb P(X_2+t<X_1) &= mathbb P(X_1>X_2+tmid X_1>t)mathbb P(X_1>t)\
        &= mathbb P(X_1>X_2)mathbb P(X_1>t)\
        &= left(frac{lambda_2}{lambda_1+lambda_2}right) e^{-lambda_1 t}.
        end{align}






        share|cite|improve this answer











        $endgroup$



        Let $X_i$ be the failure time of machine $i$. By lack of memory we have
        begin{align}
        mathbb P(X_2+t<X_1) &= mathbb P(X_1>X_2+tmid X_1>t)mathbb P(X_1>t)\
        &= mathbb P(X_1>X_2)mathbb P(X_1>t)\
        &= left(frac{lambda_2}{lambda_1+lambda_2}right) e^{-lambda_1 t}.
        end{align}







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 24 '16 at 22:57

























        answered Apr 1 '16 at 11:35









        Math1000Math1000

        19.2k31745




        19.2k31745






























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