Find the CDF of a given pmf (discrete case)












0












$begingroup$


I am trying to simulate n random discrete variable which has the following
pmf



$P(X = k) = (1-p)^2kp^{k-1}$



I am thinking about using the inverse transform sampling method and I am trying to find the cdf.



$P(X le k) = 1 - P(X > k) = 1- P(X ge k+1) = sum_{x=k+1}^{infty} (1-p)^2kp^{k-1} = 1-[ (1-p)^2sum_{x=k+1}^{infty}kp^{k-1}]$



= $1-(1-p)^2[(k+1)p^k + (k+2)p^{k+1}+ (k+3)p^{k+4} + ...] $



I can't seem to calculate this sum ($sum_{x=k+1}^{infty}kp^{k-1}$).



Any help or hint will be appreciated !






calculus statistics







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    0












    $begingroup$


    I am trying to simulate n random discrete variable which has the following
    pmf



    $P(X = k) = (1-p)^2kp^{k-1}$



    I am thinking about using the inverse transform sampling method and I am trying to find the cdf.



    $P(X le k) = 1 - P(X > k) = 1- P(X ge k+1) = sum_{x=k+1}^{infty} (1-p)^2kp^{k-1} = 1-[ (1-p)^2sum_{x=k+1}^{infty}kp^{k-1}]$



    = $1-(1-p)^2[(k+1)p^k + (k+2)p^{k+1}+ (k+3)p^{k+4} + ...] $



    I can't seem to calculate this sum ($sum_{x=k+1}^{infty}kp^{k-1}$).



    Any help or hint will be appreciated !






    calculus statistics







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I am trying to simulate n random discrete variable which has the following
      pmf



      $P(X = k) = (1-p)^2kp^{k-1}$



      I am thinking about using the inverse transform sampling method and I am trying to find the cdf.



      $P(X le k) = 1 - P(X > k) = 1- P(X ge k+1) = sum_{x=k+1}^{infty} (1-p)^2kp^{k-1} = 1-[ (1-p)^2sum_{x=k+1}^{infty}kp^{k-1}]$



      = $1-(1-p)^2[(k+1)p^k + (k+2)p^{k+1}+ (k+3)p^{k+4} + ...] $



      I can't seem to calculate this sum ($sum_{x=k+1}^{infty}kp^{k-1}$).



      Any help or hint will be appreciated !






      calculus statistics







      share|cite|improve this question









      $endgroup$




      I am trying to simulate n random discrete variable which has the following
      pmf



      $P(X = k) = (1-p)^2kp^{k-1}$



      I am thinking about using the inverse transform sampling method and I am trying to find the cdf.



      $P(X le k) = 1 - P(X > k) = 1- P(X ge k+1) = sum_{x=k+1}^{infty} (1-p)^2kp^{k-1} = 1-[ (1-p)^2sum_{x=k+1}^{infty}kp^{k-1}]$



      = $1-(1-p)^2[(k+1)p^k + (k+2)p^{k+1}+ (k+3)p^{k+4} + ...] $



      I can't seem to calculate this sum ($sum_{x=k+1}^{infty}kp^{k-1}$).



      Any help or hint will be appreciated !






      calculus statistics




      calculus statistics






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      asked Dec 3 '18 at 5:39









      TatariaTataria

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          $begingroup$

          begin{align}S &= (k+1) p^k + (k+2) p^{k+1} + (k+3) p^{k+2} + cdots
          \
          pS &= phantom{(k+1) p^k +{}} (k+1) p^{k+1} + (k+2) p^{k+2} + (k+3) p^{k+2} + cdots
          end{align}
          Subtracting the two equations yields
          $$(1-p) S = p^k + p^{k+1} + p^{k+2} + cdots.$$
          The right-hand side is a geometric series I assume you can compute. Then divide both sides by $1-p$.






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            $begingroup$

            begin{align}S &= (k+1) p^k + (k+2) p^{k+1} + (k+3) p^{k+2} + cdots
            \
            pS &= phantom{(k+1) p^k +{}} (k+1) p^{k+1} + (k+2) p^{k+2} + (k+3) p^{k+2} + cdots
            end{align}
            Subtracting the two equations yields
            $$(1-p) S = p^k + p^{k+1} + p^{k+2} + cdots.$$
            The right-hand side is a geometric series I assume you can compute. Then divide both sides by $1-p$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              begin{align}S &= (k+1) p^k + (k+2) p^{k+1} + (k+3) p^{k+2} + cdots
              \
              pS &= phantom{(k+1) p^k +{}} (k+1) p^{k+1} + (k+2) p^{k+2} + (k+3) p^{k+2} + cdots
              end{align}
              Subtracting the two equations yields
              $$(1-p) S = p^k + p^{k+1} + p^{k+2} + cdots.$$
              The right-hand side is a geometric series I assume you can compute. Then divide both sides by $1-p$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                begin{align}S &= (k+1) p^k + (k+2) p^{k+1} + (k+3) p^{k+2} + cdots
                \
                pS &= phantom{(k+1) p^k +{}} (k+1) p^{k+1} + (k+2) p^{k+2} + (k+3) p^{k+2} + cdots
                end{align}
                Subtracting the two equations yields
                $$(1-p) S = p^k + p^{k+1} + p^{k+2} + cdots.$$
                The right-hand side is a geometric series I assume you can compute. Then divide both sides by $1-p$.






                share|cite|improve this answer









                $endgroup$



                begin{align}S &= (k+1) p^k + (k+2) p^{k+1} + (k+3) p^{k+2} + cdots
                \
                pS &= phantom{(k+1) p^k +{}} (k+1) p^{k+1} + (k+2) p^{k+2} + (k+3) p^{k+2} + cdots
                end{align}
                Subtracting the two equations yields
                $$(1-p) S = p^k + p^{k+1} + p^{k+2} + cdots.$$
                The right-hand side is a geometric series I assume you can compute. Then divide both sides by $1-p$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 3 '18 at 5:47









                angryavianangryavian

                40.6k23280




                40.6k23280






























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