Poisson process: finding probability of 1 count in an interval given that 0 counts happen in a subinterval












1












$begingroup$


This was in my exam today and I'm not sure what's the correct answer.



Let's say that the number of people that enter into a store in the interval $(0,t]$ (in hours) is a Poisson process where $30$ people enter per hour. What is the probability of only $1$ person entering the store between the minute $0$ and the minute $6$, given that $0$ persons enter the store between the minute $0$ and $4$?



I thought of solving this using the variables of the inter-arrival times, which in a Poisson process $X(t) sim text{Po}(30t)$ are $T_i sim text{Exp}(30)$. Since $T_1$ is the time between $0$ and the first event, I formulated this as follows:



$Pr(frac6{60}geq T_1>0mid T_1 > frac4{60})$



This sounded correct to me, because we want to find the probability that the first event lands between the $0$th and $6$th minute, given that the first event did not happen before the $4$th minute.



The calculations I did go as such:



$Pr(frac6{60}geq T_1>0mid T_1 > frac4{60})=frac{Pr(frac6{60}geq T_1 > 0,T_1 > frac4{60})}{Pr(T_1 > frac4{60})}$



the intersection in the numerator is $Pr(frac6{60}geq T_1 > frac4{60})$. Knowing $T_1$ is exponential with parameter $30$, this gives a probability of $e^{-2}-e^{-3}$. The denominator is just $e^{-2}$, so the probability is approximately $0.63212$.



Is this correct? I know I made a stupid error calculating $Pr(T_1 > frac4{60})$ where I got $1-e^{-2}$ instead of $e^{-2}$, but I think my reasoning should be sound...






probability stochastic-processes poisson-process point-processes







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    1












    $begingroup$


    This was in my exam today and I'm not sure what's the correct answer.



    Let's say that the number of people that enter into a store in the interval $(0,t]$ (in hours) is a Poisson process where $30$ people enter per hour. What is the probability of only $1$ person entering the store between the minute $0$ and the minute $6$, given that $0$ persons enter the store between the minute $0$ and $4$?



    I thought of solving this using the variables of the inter-arrival times, which in a Poisson process $X(t) sim text{Po}(30t)$ are $T_i sim text{Exp}(30)$. Since $T_1$ is the time between $0$ and the first event, I formulated this as follows:



    $Pr(frac6{60}geq T_1>0mid T_1 > frac4{60})$



    This sounded correct to me, because we want to find the probability that the first event lands between the $0$th and $6$th minute, given that the first event did not happen before the $4$th minute.



    The calculations I did go as such:



    $Pr(frac6{60}geq T_1>0mid T_1 > frac4{60})=frac{Pr(frac6{60}geq T_1 > 0,T_1 > frac4{60})}{Pr(T_1 > frac4{60})}$



    the intersection in the numerator is $Pr(frac6{60}geq T_1 > frac4{60})$. Knowing $T_1$ is exponential with parameter $30$, this gives a probability of $e^{-2}-e^{-3}$. The denominator is just $e^{-2}$, so the probability is approximately $0.63212$.



    Is this correct? I know I made a stupid error calculating $Pr(T_1 > frac4{60})$ where I got $1-e^{-2}$ instead of $e^{-2}$, but I think my reasoning should be sound...






    probability stochastic-processes poisson-process point-processes







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      This was in my exam today and I'm not sure what's the correct answer.



      Let's say that the number of people that enter into a store in the interval $(0,t]$ (in hours) is a Poisson process where $30$ people enter per hour. What is the probability of only $1$ person entering the store between the minute $0$ and the minute $6$, given that $0$ persons enter the store between the minute $0$ and $4$?



      I thought of solving this using the variables of the inter-arrival times, which in a Poisson process $X(t) sim text{Po}(30t)$ are $T_i sim text{Exp}(30)$. Since $T_1$ is the time between $0$ and the first event, I formulated this as follows:



      $Pr(frac6{60}geq T_1>0mid T_1 > frac4{60})$



      This sounded correct to me, because we want to find the probability that the first event lands between the $0$th and $6$th minute, given that the first event did not happen before the $4$th minute.



      The calculations I did go as such:



      $Pr(frac6{60}geq T_1>0mid T_1 > frac4{60})=frac{Pr(frac6{60}geq T_1 > 0,T_1 > frac4{60})}{Pr(T_1 > frac4{60})}$



      the intersection in the numerator is $Pr(frac6{60}geq T_1 > frac4{60})$. Knowing $T_1$ is exponential with parameter $30$, this gives a probability of $e^{-2}-e^{-3}$. The denominator is just $e^{-2}$, so the probability is approximately $0.63212$.



      Is this correct? I know I made a stupid error calculating $Pr(T_1 > frac4{60})$ where I got $1-e^{-2}$ instead of $e^{-2}$, but I think my reasoning should be sound...






      probability stochastic-processes poisson-process point-processes







      share|cite|improve this question









      $endgroup$




      This was in my exam today and I'm not sure what's the correct answer.



      Let's say that the number of people that enter into a store in the interval $(0,t]$ (in hours) is a Poisson process where $30$ people enter per hour. What is the probability of only $1$ person entering the store between the minute $0$ and the minute $6$, given that $0$ persons enter the store between the minute $0$ and $4$?



      I thought of solving this using the variables of the inter-arrival times, which in a Poisson process $X(t) sim text{Po}(30t)$ are $T_i sim text{Exp}(30)$. Since $T_1$ is the time between $0$ and the first event, I formulated this as follows:



      $Pr(frac6{60}geq T_1>0mid T_1 > frac4{60})$



      This sounded correct to me, because we want to find the probability that the first event lands between the $0$th and $6$th minute, given that the first event did not happen before the $4$th minute.



      The calculations I did go as such:



      $Pr(frac6{60}geq T_1>0mid T_1 > frac4{60})=frac{Pr(frac6{60}geq T_1 > 0,T_1 > frac4{60})}{Pr(T_1 > frac4{60})}$



      the intersection in the numerator is $Pr(frac6{60}geq T_1 > frac4{60})$. Knowing $T_1$ is exponential with parameter $30$, this gives a probability of $e^{-2}-e^{-3}$. The denominator is just $e^{-2}$, so the probability is approximately $0.63212$.



      Is this correct? I know I made a stupid error calculating $Pr(T_1 > frac4{60})$ where I got $1-e^{-2}$ instead of $e^{-2}$, but I think my reasoning should be sound...






      probability stochastic-processes poisson-process point-processes




      probability stochastic-processes poisson-process point-processes






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 18 '18 at 12:02









      AstlyDichrarAstlyDichrar

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          $begingroup$

          It's simpler than that. The probability that one person enters between $0$ and $6$ given that no people enter between $0$ and $4$ is precisely the probability that one person enters between $4$ and $6$ given that no people enter between $0$ and $4$. But these two events are independent.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So the answer should just be $Pr(X(frac2{60})=1)=e^{-1}$? Where does my reasoning go wrong?
            $endgroup$
            – AstlyDichrar
            Dec 18 '18 at 12:14










          • $begingroup$
            Where you go wrong is that it's not sufficient for $T_1$ to be between $4$ and $6$, since you also need $T_2>6$. What you have calculated would be for "at least one" instead of "exactly one", and it does give the right value for that question.
            $endgroup$
            – Especially Lime
            Dec 18 '18 at 12:20










          • $begingroup$
            I didn't think of that... and that seems to explain why my answer is $1-e^{-1}$ and yours is $e^{-1}$ :( Thanks, that's totally correct. I thought of solving the problem the way you did it but I never trust the fact that the HPP has stationary and independent increments :(
            $endgroup$
            – AstlyDichrar
            Dec 18 '18 at 12:23













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          $begingroup$

          It's simpler than that. The probability that one person enters between $0$ and $6$ given that no people enter between $0$ and $4$ is precisely the probability that one person enters between $4$ and $6$ given that no people enter between $0$ and $4$. But these two events are independent.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So the answer should just be $Pr(X(frac2{60})=1)=e^{-1}$? Where does my reasoning go wrong?
            $endgroup$
            – AstlyDichrar
            Dec 18 '18 at 12:14










          • $begingroup$
            Where you go wrong is that it's not sufficient for $T_1$ to be between $4$ and $6$, since you also need $T_2>6$. What you have calculated would be for "at least one" instead of "exactly one", and it does give the right value for that question.
            $endgroup$
            – Especially Lime
            Dec 18 '18 at 12:20










          • $begingroup$
            I didn't think of that... and that seems to explain why my answer is $1-e^{-1}$ and yours is $e^{-1}$ :( Thanks, that's totally correct. I thought of solving the problem the way you did it but I never trust the fact that the HPP has stationary and independent increments :(
            $endgroup$
            – AstlyDichrar
            Dec 18 '18 at 12:23


















          0












          $begingroup$

          It's simpler than that. The probability that one person enters between $0$ and $6$ given that no people enter between $0$ and $4$ is precisely the probability that one person enters between $4$ and $6$ given that no people enter between $0$ and $4$. But these two events are independent.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So the answer should just be $Pr(X(frac2{60})=1)=e^{-1}$? Where does my reasoning go wrong?
            $endgroup$
            – AstlyDichrar
            Dec 18 '18 at 12:14










          • $begingroup$
            Where you go wrong is that it's not sufficient for $T_1$ to be between $4$ and $6$, since you also need $T_2>6$. What you have calculated would be for "at least one" instead of "exactly one", and it does give the right value for that question.
            $endgroup$
            – Especially Lime
            Dec 18 '18 at 12:20










          • $begingroup$
            I didn't think of that... and that seems to explain why my answer is $1-e^{-1}$ and yours is $e^{-1}$ :( Thanks, that's totally correct. I thought of solving the problem the way you did it but I never trust the fact that the HPP has stationary and independent increments :(
            $endgroup$
            – AstlyDichrar
            Dec 18 '18 at 12:23
















          0












          0








          0





          $begingroup$

          It's simpler than that. The probability that one person enters between $0$ and $6$ given that no people enter between $0$ and $4$ is precisely the probability that one person enters between $4$ and $6$ given that no people enter between $0$ and $4$. But these two events are independent.






          share|cite|improve this answer









          $endgroup$



          It's simpler than that. The probability that one person enters between $0$ and $6$ given that no people enter between $0$ and $4$ is precisely the probability that one person enters between $4$ and $6$ given that no people enter between $0$ and $4$. But these two events are independent.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 18 '18 at 12:06









          Especially LimeEspecially Lime

          22.6k23059




          22.6k23059












          • $begingroup$
            So the answer should just be $Pr(X(frac2{60})=1)=e^{-1}$? Where does my reasoning go wrong?
            $endgroup$
            – AstlyDichrar
            Dec 18 '18 at 12:14










          • $begingroup$
            Where you go wrong is that it's not sufficient for $T_1$ to be between $4$ and $6$, since you also need $T_2>6$. What you have calculated would be for "at least one" instead of "exactly one", and it does give the right value for that question.
            $endgroup$
            – Especially Lime
            Dec 18 '18 at 12:20










          • $begingroup$
            I didn't think of that... and that seems to explain why my answer is $1-e^{-1}$ and yours is $e^{-1}$ :( Thanks, that's totally correct. I thought of solving the problem the way you did it but I never trust the fact that the HPP has stationary and independent increments :(
            $endgroup$
            – AstlyDichrar
            Dec 18 '18 at 12:23




















          • $begingroup$
            So the answer should just be $Pr(X(frac2{60})=1)=e^{-1}$? Where does my reasoning go wrong?
            $endgroup$
            – AstlyDichrar
            Dec 18 '18 at 12:14










          • $begingroup$
            Where you go wrong is that it's not sufficient for $T_1$ to be between $4$ and $6$, since you also need $T_2>6$. What you have calculated would be for "at least one" instead of "exactly one", and it does give the right value for that question.
            $endgroup$
            – Especially Lime
            Dec 18 '18 at 12:20










          • $begingroup$
            I didn't think of that... and that seems to explain why my answer is $1-e^{-1}$ and yours is $e^{-1}$ :( Thanks, that's totally correct. I thought of solving the problem the way you did it but I never trust the fact that the HPP has stationary and independent increments :(
            $endgroup$
            – AstlyDichrar
            Dec 18 '18 at 12:23


















          $begingroup$
          So the answer should just be $Pr(X(frac2{60})=1)=e^{-1}$? Where does my reasoning go wrong?
          $endgroup$
          – AstlyDichrar
          Dec 18 '18 at 12:14




          $begingroup$
          So the answer should just be $Pr(X(frac2{60})=1)=e^{-1}$? Where does my reasoning go wrong?
          $endgroup$
          – AstlyDichrar
          Dec 18 '18 at 12:14












          $begingroup$
          Where you go wrong is that it's not sufficient for $T_1$ to be between $4$ and $6$, since you also need $T_2>6$. What you have calculated would be for "at least one" instead of "exactly one", and it does give the right value for that question.
          $endgroup$
          – Especially Lime
          Dec 18 '18 at 12:20




          $begingroup$
          Where you go wrong is that it's not sufficient for $T_1$ to be between $4$ and $6$, since you also need $T_2>6$. What you have calculated would be for "at least one" instead of "exactly one", and it does give the right value for that question.
          $endgroup$
          – Especially Lime
          Dec 18 '18 at 12:20












          $begingroup$
          I didn't think of that... and that seems to explain why my answer is $1-e^{-1}$ and yours is $e^{-1}$ :( Thanks, that's totally correct. I thought of solving the problem the way you did it but I never trust the fact that the HPP has stationary and independent increments :(
          $endgroup$
          – AstlyDichrar
          Dec 18 '18 at 12:23






          $begingroup$
          I didn't think of that... and that seems to explain why my answer is $1-e^{-1}$ and yours is $e^{-1}$ :( Thanks, that's totally correct. I thought of solving the problem the way you did it but I never trust the fact that the HPP has stationary and independent increments :(
          $endgroup$
          – AstlyDichrar
          Dec 18 '18 at 12:23




















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