Let $X$ and $Y$ be i.i.d. $operatorname{Geom}(p)$, and $N = X + Y$. Find the joint PMF of $X, Y, N$












1












$begingroup$



Let $X$ and $Y$ be i.i.d. $operatorname{Geom}(p)$, and $N = X + Y$. Find the joint PMF of $X, Y, N$.




I have generally difficulties with such problems, as I get easily confused. Below I detailed my (most probability incorrect) approach. Besides the correct approach, I also would like to have some more general advice on how to think about and tackle such problems.



The random variables $X$ and $Y$ have geometric distributions, so $P(X=x)=(1-p)^xp$ and $P(Y=y)=(1-p)^yp$. I also know that the sum of two i.i.d. geometric random variables has a negative binomial distribution, thus $P(N=n) = binom{n+r-1}{n} (1-p)^n p^r$, where $r$ is the number of success. In our case $r=2$, as we are dealing with the sum of two random variables. Hence, $P(N=n) = (n+1)(1-p)^n p^2$.



I'm asked to find $P(X=x, Y=y, N=n)$. I don't know how to write that down, so I condition on $N=n$ and write



begin{align}
P(X=x, Y=y, N=n) &= P(X=x, Y=y mid N=n)P(N=n) \
&= P(X=x, Y=y mid N=x+y)P(N=n),
end{align}
because I know that $n=x+y$. But conditional on $N=x+y$, $X=x$ and $Y=y$ are the same event. Thus,



begin{align}
P(X=x, Y=y, N=n) &= P(X=x mid N=x+y)P(N=n)
end{align}



I can assume that $X$ is the number of failures till the first of both successes, otherwise I could just consider $P(Y=y mid N=x+y)$. The probability that there are $X=x$ failures till the frist success, given that there are in total $N=x+y$ failures is the same is having the first success between the $x^{th}$ and $(x+1)^{th}$ failure. As both $X$ and $Y$ have the same probability of success, the probability of this to happen is $frac{1}{x+y+1}$. Hence,



begin{align}
P(X=x, Y=y, N=n) &= frac{1}{x+y+1} (n+1)(1-p)^n p^r \
&= (1-p)^n p^2.
end{align}



But this looks like nonsense.






probability solution-verification







share|cite|improve this question











$endgroup$

















    1












    $begingroup$



    Let $X$ and $Y$ be i.i.d. $operatorname{Geom}(p)$, and $N = X + Y$. Find the joint PMF of $X, Y, N$.




    I have generally difficulties with such problems, as I get easily confused. Below I detailed my (most probability incorrect) approach. Besides the correct approach, I also would like to have some more general advice on how to think about and tackle such problems.



    The random variables $X$ and $Y$ have geometric distributions, so $P(X=x)=(1-p)^xp$ and $P(Y=y)=(1-p)^yp$. I also know that the sum of two i.i.d. geometric random variables has a negative binomial distribution, thus $P(N=n) = binom{n+r-1}{n} (1-p)^n p^r$, where $r$ is the number of success. In our case $r=2$, as we are dealing with the sum of two random variables. Hence, $P(N=n) = (n+1)(1-p)^n p^2$.



    I'm asked to find $P(X=x, Y=y, N=n)$. I don't know how to write that down, so I condition on $N=n$ and write



    begin{align}
    P(X=x, Y=y, N=n) &= P(X=x, Y=y mid N=n)P(N=n) \
    &= P(X=x, Y=y mid N=x+y)P(N=n),
    end{align}
    because I know that $n=x+y$. But conditional on $N=x+y$, $X=x$ and $Y=y$ are the same event. Thus,



    begin{align}
    P(X=x, Y=y, N=n) &= P(X=x mid N=x+y)P(N=n)
    end{align}



    I can assume that $X$ is the number of failures till the first of both successes, otherwise I could just consider $P(Y=y mid N=x+y)$. The probability that there are $X=x$ failures till the frist success, given that there are in total $N=x+y$ failures is the same is having the first success between the $x^{th}$ and $(x+1)^{th}$ failure. As both $X$ and $Y$ have the same probability of success, the probability of this to happen is $frac{1}{x+y+1}$. Hence,



    begin{align}
    P(X=x, Y=y, N=n) &= frac{1}{x+y+1} (n+1)(1-p)^n p^r \
    &= (1-p)^n p^2.
    end{align}



    But this looks like nonsense.






    probability solution-verification







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      Let $X$ and $Y$ be i.i.d. $operatorname{Geom}(p)$, and $N = X + Y$. Find the joint PMF of $X, Y, N$.




      I have generally difficulties with such problems, as I get easily confused. Below I detailed my (most probability incorrect) approach. Besides the correct approach, I also would like to have some more general advice on how to think about and tackle such problems.



      The random variables $X$ and $Y$ have geometric distributions, so $P(X=x)=(1-p)^xp$ and $P(Y=y)=(1-p)^yp$. I also know that the sum of two i.i.d. geometric random variables has a negative binomial distribution, thus $P(N=n) = binom{n+r-1}{n} (1-p)^n p^r$, where $r$ is the number of success. In our case $r=2$, as we are dealing with the sum of two random variables. Hence, $P(N=n) = (n+1)(1-p)^n p^2$.



      I'm asked to find $P(X=x, Y=y, N=n)$. I don't know how to write that down, so I condition on $N=n$ and write



      begin{align}
      P(X=x, Y=y, N=n) &= P(X=x, Y=y mid N=n)P(N=n) \
      &= P(X=x, Y=y mid N=x+y)P(N=n),
      end{align}
      because I know that $n=x+y$. But conditional on $N=x+y$, $X=x$ and $Y=y$ are the same event. Thus,



      begin{align}
      P(X=x, Y=y, N=n) &= P(X=x mid N=x+y)P(N=n)
      end{align}



      I can assume that $X$ is the number of failures till the first of both successes, otherwise I could just consider $P(Y=y mid N=x+y)$. The probability that there are $X=x$ failures till the frist success, given that there are in total $N=x+y$ failures is the same is having the first success between the $x^{th}$ and $(x+1)^{th}$ failure. As both $X$ and $Y$ have the same probability of success, the probability of this to happen is $frac{1}{x+y+1}$. Hence,



      begin{align}
      P(X=x, Y=y, N=n) &= frac{1}{x+y+1} (n+1)(1-p)^n p^r \
      &= (1-p)^n p^2.
      end{align}



      But this looks like nonsense.






      probability solution-verification







      share|cite|improve this question











      $endgroup$





      Let $X$ and $Y$ be i.i.d. $operatorname{Geom}(p)$, and $N = X + Y$. Find the joint PMF of $X, Y, N$.




      I have generally difficulties with such problems, as I get easily confused. Below I detailed my (most probability incorrect) approach. Besides the correct approach, I also would like to have some more general advice on how to think about and tackle such problems.



      The random variables $X$ and $Y$ have geometric distributions, so $P(X=x)=(1-p)^xp$ and $P(Y=y)=(1-p)^yp$. I also know that the sum of two i.i.d. geometric random variables has a negative binomial distribution, thus $P(N=n) = binom{n+r-1}{n} (1-p)^n p^r$, where $r$ is the number of success. In our case $r=2$, as we are dealing with the sum of two random variables. Hence, $P(N=n) = (n+1)(1-p)^n p^2$.



      I'm asked to find $P(X=x, Y=y, N=n)$. I don't know how to write that down, so I condition on $N=n$ and write



      begin{align}
      P(X=x, Y=y, N=n) &= P(X=x, Y=y mid N=n)P(N=n) \
      &= P(X=x, Y=y mid N=x+y)P(N=n),
      end{align}
      because I know that $n=x+y$. But conditional on $N=x+y$, $X=x$ and $Y=y$ are the same event. Thus,



      begin{align}
      P(X=x, Y=y, N=n) &= P(X=x mid N=x+y)P(N=n)
      end{align}



      I can assume that $X$ is the number of failures till the first of both successes, otherwise I could just consider $P(Y=y mid N=x+y)$. The probability that there are $X=x$ failures till the frist success, given that there are in total $N=x+y$ failures is the same is having the first success between the $x^{th}$ and $(x+1)^{th}$ failure. As both $X$ and $Y$ have the same probability of success, the probability of this to happen is $frac{1}{x+y+1}$. Hence,



      begin{align}
      P(X=x, Y=y, N=n) &= frac{1}{x+y+1} (n+1)(1-p)^n p^r \
      &= (1-p)^n p^2.
      end{align}



      But this looks like nonsense.






      probability solution-verification




      probability solution-verification






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 30 '14 at 15:03







      NoBackingDown

















      asked Dec 30 '14 at 14:55









      NoBackingDownNoBackingDown

      8021233




      8021233






















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          $begingroup$

          Hint::
          $$P(X=x,Y=y,N=n) = begin{cases}P(X = x, Y = y) & text{if }x+y=n \ 0 & text{otherwise}end{cases}$$






          share|cite|improve this answer









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            $begingroup$

            Hint::
            $$P(X=x,Y=y,N=n) = begin{cases}P(X = x, Y = y) & text{if }x+y=n \ 0 & text{otherwise}end{cases}$$






            share|cite|improve this answer









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              0












              $begingroup$

              Hint::
              $$P(X=x,Y=y,N=n) = begin{cases}P(X = x, Y = y) & text{if }x+y=n \ 0 & text{otherwise}end{cases}$$






              share|cite|improve this answer









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                $begingroup$

                Hint::
                $$P(X=x,Y=y,N=n) = begin{cases}P(X = x, Y = y) & text{if }x+y=n \ 0 & text{otherwise}end{cases}$$






                share|cite|improve this answer









                $endgroup$



                Hint::
                $$P(X=x,Y=y,N=n) = begin{cases}P(X = x, Y = y) & text{if }x+y=n \ 0 & text{otherwise}end{cases}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 30 '14 at 15:07









                ronnoronno

                4,9051740




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