$X$ is a Poisson process and $T$ is exponentially distributed?











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$X = {X(t): t geq 0}$ is a Poisson process (with intensity $lambda$). Independent of $X$, $T$ is a random variable that is exponentially distributed with intensity $theta$.



I need to find the PMF for $Y=X(T)$ and also Var$[X(T)]$.




There is an example in my book that talks about finding the PMF recursively. I'll denote $p_n=P[X(T)]$. If I let $p_0=e^{-lambda}$, then supposedly $$p_n=frac{lambda}{n}p_{n-1}$$
So $$p_1=frac{lambda}{n}e^{-lambda}$$
$$p_2=frac{lambda^2}{n^2}e^{-lambda}$$
Then, this just becomes
$$p_n=frac{lambda^n}{n^n}e^{-lambda}$$



Is that right? I don't understand the first recursive part; I just copied that from a similar example from the book. I don't actually know where that comes from.



There's another example that looks exactly the same but has the answer $$frac{(lambda t)^n e^{(-lambda t)}}{n!}$$



Can someone explain how I should be doing this? Would really appreciate it.






probability stochastic-processes







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    up vote
    0
    down vote

    favorite












    $X = {X(t): t geq 0}$ is a Poisson process (with intensity $lambda$). Independent of $X$, $T$ is a random variable that is exponentially distributed with intensity $theta$.



    I need to find the PMF for $Y=X(T)$ and also Var$[X(T)]$.




    There is an example in my book that talks about finding the PMF recursively. I'll denote $p_n=P[X(T)]$. If I let $p_0=e^{-lambda}$, then supposedly $$p_n=frac{lambda}{n}p_{n-1}$$
    So $$p_1=frac{lambda}{n}e^{-lambda}$$
    $$p_2=frac{lambda^2}{n^2}e^{-lambda}$$
    Then, this just becomes
    $$p_n=frac{lambda^n}{n^n}e^{-lambda}$$



    Is that right? I don't understand the first recursive part; I just copied that from a similar example from the book. I don't actually know where that comes from.



    There's another example that looks exactly the same but has the answer $$frac{(lambda t)^n e^{(-lambda t)}}{n!}$$



    Can someone explain how I should be doing this? Would really appreciate it.






    probability stochastic-processes







    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      $X = {X(t): t geq 0}$ is a Poisson process (with intensity $lambda$). Independent of $X$, $T$ is a random variable that is exponentially distributed with intensity $theta$.



      I need to find the PMF for $Y=X(T)$ and also Var$[X(T)]$.




      There is an example in my book that talks about finding the PMF recursively. I'll denote $p_n=P[X(T)]$. If I let $p_0=e^{-lambda}$, then supposedly $$p_n=frac{lambda}{n}p_{n-1}$$
      So $$p_1=frac{lambda}{n}e^{-lambda}$$
      $$p_2=frac{lambda^2}{n^2}e^{-lambda}$$
      Then, this just becomes
      $$p_n=frac{lambda^n}{n^n}e^{-lambda}$$



      Is that right? I don't understand the first recursive part; I just copied that from a similar example from the book. I don't actually know where that comes from.



      There's another example that looks exactly the same but has the answer $$frac{(lambda t)^n e^{(-lambda t)}}{n!}$$



      Can someone explain how I should be doing this? Would really appreciate it.






      probability stochastic-processes







      share|cite|improve this question















      $X = {X(t): t geq 0}$ is a Poisson process (with intensity $lambda$). Independent of $X$, $T$ is a random variable that is exponentially distributed with intensity $theta$.



      I need to find the PMF for $Y=X(T)$ and also Var$[X(T)]$.




      There is an example in my book that talks about finding the PMF recursively. I'll denote $p_n=P[X(T)]$. If I let $p_0=e^{-lambda}$, then supposedly $$p_n=frac{lambda}{n}p_{n-1}$$
      So $$p_1=frac{lambda}{n}e^{-lambda}$$
      $$p_2=frac{lambda^2}{n^2}e^{-lambda}$$
      Then, this just becomes
      $$p_n=frac{lambda^n}{n^n}e^{-lambda}$$



      Is that right? I don't understand the first recursive part; I just copied that from a similar example from the book. I don't actually know where that comes from.



      There's another example that looks exactly the same but has the answer $$frac{(lambda t)^n e^{(-lambda t)}}{n!}$$



      Can someone explain how I should be doing this? Would really appreciate it.






      probability stochastic-processes




      probability stochastic-processes






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 24 at 2:43









      Rócherz

      2,7262721




      2,7262721










      asked Mar 21 '16 at 21:15









      Taylor

      325111




      325111






















          1 Answer
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          Since $X(t)$ is a Poisson process and $T$ is independent of ${X(t):tgeq 0}$, it follows that
          $$ mathbb{P}(X(T)=n|T=t)=mathbb{P}(X(t)=n)=frac{(lambda t)^ne^{-lambda t}}{n!} $$
          for $n=0,1,2,dots$.



          Therefore if $f_T(t)=theta e^{-theta t}1_{t>0}$ is the pdf of $T$, then
          $$ mathbb{P}(X(T)=n)=int_{mathbb{R}}mathbb{P}(X(T)=n|T=t)f_T(t);dt=int_{0}^{infty}frac{(lambda t)^ne^{-lambda t}}{n!}theta e^{-theta t};dt $$
          $$= frac{thetalambda^n}{n!}int_0^{infty}t^ne^{-(theta+lambda)t};dt=frac{thetalambda^n}{(lambda+theta)^{n+1}n!}int_0^{infty}u^ne^{-u};du=frac{thetalambda^n}{(lambda+theta)^{n+1}}$$
          since the last integral is $Gamma(n+1)=n!$



          The expected value and variance of $X(T)$ can be computed in a similar way.



          In particular, since $mathbb{E}[X(t)]=lambda t$, it follows that
          $$ mathbb{E}[X(T)]=int_0^{infty}mathbb{E}[X(T)|T=t]f_T(t);dt=int_0^{infty}mathbb{E}[X(t)]f_T(t);dt=int_0^{infty}lambda tcdottheta e^{-theta t};dt=frac{lambda}{theta} $$






          share|cite|improve this answer























          • Ah I sort of see where this all comes from. Would you mind showing the integral I need to evaluate for $E[X]$?
            – Taylor
            Mar 21 '16 at 21:39











          Your Answer





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          active

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          up vote
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          down vote



          accepted










          Since $X(t)$ is a Poisson process and $T$ is independent of ${X(t):tgeq 0}$, it follows that
          $$ mathbb{P}(X(T)=n|T=t)=mathbb{P}(X(t)=n)=frac{(lambda t)^ne^{-lambda t}}{n!} $$
          for $n=0,1,2,dots$.



          Therefore if $f_T(t)=theta e^{-theta t}1_{t>0}$ is the pdf of $T$, then
          $$ mathbb{P}(X(T)=n)=int_{mathbb{R}}mathbb{P}(X(T)=n|T=t)f_T(t);dt=int_{0}^{infty}frac{(lambda t)^ne^{-lambda t}}{n!}theta e^{-theta t};dt $$
          $$= frac{thetalambda^n}{n!}int_0^{infty}t^ne^{-(theta+lambda)t};dt=frac{thetalambda^n}{(lambda+theta)^{n+1}n!}int_0^{infty}u^ne^{-u};du=frac{thetalambda^n}{(lambda+theta)^{n+1}}$$
          since the last integral is $Gamma(n+1)=n!$



          The expected value and variance of $X(T)$ can be computed in a similar way.



          In particular, since $mathbb{E}[X(t)]=lambda t$, it follows that
          $$ mathbb{E}[X(T)]=int_0^{infty}mathbb{E}[X(T)|T=t]f_T(t);dt=int_0^{infty}mathbb{E}[X(t)]f_T(t);dt=int_0^{infty}lambda tcdottheta e^{-theta t};dt=frac{lambda}{theta} $$






          share|cite|improve this answer























          • Ah I sort of see where this all comes from. Would you mind showing the integral I need to evaluate for $E[X]$?
            – Taylor
            Mar 21 '16 at 21:39















          up vote
          1
          down vote



          accepted










          Since $X(t)$ is a Poisson process and $T$ is independent of ${X(t):tgeq 0}$, it follows that
          $$ mathbb{P}(X(T)=n|T=t)=mathbb{P}(X(t)=n)=frac{(lambda t)^ne^{-lambda t}}{n!} $$
          for $n=0,1,2,dots$.



          Therefore if $f_T(t)=theta e^{-theta t}1_{t>0}$ is the pdf of $T$, then
          $$ mathbb{P}(X(T)=n)=int_{mathbb{R}}mathbb{P}(X(T)=n|T=t)f_T(t);dt=int_{0}^{infty}frac{(lambda t)^ne^{-lambda t}}{n!}theta e^{-theta t};dt $$
          $$= frac{thetalambda^n}{n!}int_0^{infty}t^ne^{-(theta+lambda)t};dt=frac{thetalambda^n}{(lambda+theta)^{n+1}n!}int_0^{infty}u^ne^{-u};du=frac{thetalambda^n}{(lambda+theta)^{n+1}}$$
          since the last integral is $Gamma(n+1)=n!$



          The expected value and variance of $X(T)$ can be computed in a similar way.



          In particular, since $mathbb{E}[X(t)]=lambda t$, it follows that
          $$ mathbb{E}[X(T)]=int_0^{infty}mathbb{E}[X(T)|T=t]f_T(t);dt=int_0^{infty}mathbb{E}[X(t)]f_T(t);dt=int_0^{infty}lambda tcdottheta e^{-theta t};dt=frac{lambda}{theta} $$






          share|cite|improve this answer























          • Ah I sort of see where this all comes from. Would you mind showing the integral I need to evaluate for $E[X]$?
            – Taylor
            Mar 21 '16 at 21:39













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Since $X(t)$ is a Poisson process and $T$ is independent of ${X(t):tgeq 0}$, it follows that
          $$ mathbb{P}(X(T)=n|T=t)=mathbb{P}(X(t)=n)=frac{(lambda t)^ne^{-lambda t}}{n!} $$
          for $n=0,1,2,dots$.



          Therefore if $f_T(t)=theta e^{-theta t}1_{t>0}$ is the pdf of $T$, then
          $$ mathbb{P}(X(T)=n)=int_{mathbb{R}}mathbb{P}(X(T)=n|T=t)f_T(t);dt=int_{0}^{infty}frac{(lambda t)^ne^{-lambda t}}{n!}theta e^{-theta t};dt $$
          $$= frac{thetalambda^n}{n!}int_0^{infty}t^ne^{-(theta+lambda)t};dt=frac{thetalambda^n}{(lambda+theta)^{n+1}n!}int_0^{infty}u^ne^{-u};du=frac{thetalambda^n}{(lambda+theta)^{n+1}}$$
          since the last integral is $Gamma(n+1)=n!$



          The expected value and variance of $X(T)$ can be computed in a similar way.



          In particular, since $mathbb{E}[X(t)]=lambda t$, it follows that
          $$ mathbb{E}[X(T)]=int_0^{infty}mathbb{E}[X(T)|T=t]f_T(t);dt=int_0^{infty}mathbb{E}[X(t)]f_T(t);dt=int_0^{infty}lambda tcdottheta e^{-theta t};dt=frac{lambda}{theta} $$






          share|cite|improve this answer














          Since $X(t)$ is a Poisson process and $T$ is independent of ${X(t):tgeq 0}$, it follows that
          $$ mathbb{P}(X(T)=n|T=t)=mathbb{P}(X(t)=n)=frac{(lambda t)^ne^{-lambda t}}{n!} $$
          for $n=0,1,2,dots$.



          Therefore if $f_T(t)=theta e^{-theta t}1_{t>0}$ is the pdf of $T$, then
          $$ mathbb{P}(X(T)=n)=int_{mathbb{R}}mathbb{P}(X(T)=n|T=t)f_T(t);dt=int_{0}^{infty}frac{(lambda t)^ne^{-lambda t}}{n!}theta e^{-theta t};dt $$
          $$= frac{thetalambda^n}{n!}int_0^{infty}t^ne^{-(theta+lambda)t};dt=frac{thetalambda^n}{(lambda+theta)^{n+1}n!}int_0^{infty}u^ne^{-u};du=frac{thetalambda^n}{(lambda+theta)^{n+1}}$$
          since the last integral is $Gamma(n+1)=n!$



          The expected value and variance of $X(T)$ can be computed in a similar way.



          In particular, since $mathbb{E}[X(t)]=lambda t$, it follows that
          $$ mathbb{E}[X(T)]=int_0^{infty}mathbb{E}[X(T)|T=t]f_T(t);dt=int_0^{infty}mathbb{E}[X(t)]f_T(t);dt=int_0^{infty}lambda tcdottheta e^{-theta t};dt=frac{lambda}{theta} $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 21 '16 at 22:08

























          answered Mar 21 '16 at 21:31









          carmichael561

          46.9k54382




          46.9k54382












          • Ah I sort of see where this all comes from. Would you mind showing the integral I need to evaluate for $E[X]$?
            – Taylor
            Mar 21 '16 at 21:39


















          • Ah I sort of see where this all comes from. Would you mind showing the integral I need to evaluate for $E[X]$?
            – Taylor
            Mar 21 '16 at 21:39
















          Ah I sort of see where this all comes from. Would you mind showing the integral I need to evaluate for $E[X]$?
          – Taylor
          Mar 21 '16 at 21:39




          Ah I sort of see where this all comes from. Would you mind showing the integral I need to evaluate for $E[X]$?
          – Taylor
          Mar 21 '16 at 21:39


















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