Solving $p_k=frac{1}{2}p_{k+1} + frac{1}{2}p_{k-1}$












0












$begingroup$


I am trying to solve the recurrence relation



$$p_k=frac{1}{2}p_{k+1} + frac{1}{2}p_{k-1}.$$



The context of this recurrence relation is as follows: if I start with $20, and I win $1 for every head and lose $1 for every tail, I obtain the above recurrence relation for $p$. I lose the overall game if I lose all my money, and win the overall game if I accumulate $100.



Now, I understand that substituting $p_k = p^k$ is the usual approach, which gives $p=1$. While clearly this is a solution, in the context of this above game something’s not right – where in my logic have I gone wrong? Thank you!






probability recurrence-relations problem-solving







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$endgroup$








  • 1




    $begingroup$
    What do you mean by p?
    $endgroup$
    – Martund
    Dec 20 '18 at 3:55












  • $begingroup$
    This may help.
    $endgroup$
    – Jam
    Dec 20 '18 at 4:04










  • $begingroup$
    The relationship between your recurrence relation and the stated problem is not clear. What do these $p_k$'s represent?
    $endgroup$
    – Michael Burr
    Dec 20 '18 at 4:23
















0












$begingroup$


I am trying to solve the recurrence relation



$$p_k=frac{1}{2}p_{k+1} + frac{1}{2}p_{k-1}.$$



The context of this recurrence relation is as follows: if I start with $20, and I win $1 for every head and lose $1 for every tail, I obtain the above recurrence relation for $p$. I lose the overall game if I lose all my money, and win the overall game if I accumulate $100.



Now, I understand that substituting $p_k = p^k$ is the usual approach, which gives $p=1$. While clearly this is a solution, in the context of this above game something’s not right – where in my logic have I gone wrong? Thank you!






probability recurrence-relations problem-solving







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What do you mean by p?
    $endgroup$
    – Martund
    Dec 20 '18 at 3:55












  • $begingroup$
    This may help.
    $endgroup$
    – Jam
    Dec 20 '18 at 4:04










  • $begingroup$
    The relationship between your recurrence relation and the stated problem is not clear. What do these $p_k$'s represent?
    $endgroup$
    – Michael Burr
    Dec 20 '18 at 4:23














0












0








0





$begingroup$


I am trying to solve the recurrence relation



$$p_k=frac{1}{2}p_{k+1} + frac{1}{2}p_{k-1}.$$



The context of this recurrence relation is as follows: if I start with $20, and I win $1 for every head and lose $1 for every tail, I obtain the above recurrence relation for $p$. I lose the overall game if I lose all my money, and win the overall game if I accumulate $100.



Now, I understand that substituting $p_k = p^k$ is the usual approach, which gives $p=1$. While clearly this is a solution, in the context of this above game something’s not right – where in my logic have I gone wrong? Thank you!






probability recurrence-relations problem-solving







share|cite|improve this question









$endgroup$




I am trying to solve the recurrence relation



$$p_k=frac{1}{2}p_{k+1} + frac{1}{2}p_{k-1}.$$



The context of this recurrence relation is as follows: if I start with $20, and I win $1 for every head and lose $1 for every tail, I obtain the above recurrence relation for $p$. I lose the overall game if I lose all my money, and win the overall game if I accumulate $100.



Now, I understand that substituting $p_k = p^k$ is the usual approach, which gives $p=1$. While clearly this is a solution, in the context of this above game something’s not right – where in my logic have I gone wrong? Thank you!






probability recurrence-relations problem-solving




probability recurrence-relations problem-solving






share|cite|improve this question













share|cite|improve this question











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asked Dec 20 '18 at 3:51









user107224user107224

463314




463314








  • 1




    $begingroup$
    What do you mean by p?
    $endgroup$
    – Martund
    Dec 20 '18 at 3:55












  • $begingroup$
    This may help.
    $endgroup$
    – Jam
    Dec 20 '18 at 4:04










  • $begingroup$
    The relationship between your recurrence relation and the stated problem is not clear. What do these $p_k$'s represent?
    $endgroup$
    – Michael Burr
    Dec 20 '18 at 4:23














  • 1




    $begingroup$
    What do you mean by p?
    $endgroup$
    – Martund
    Dec 20 '18 at 3:55












  • $begingroup$
    This may help.
    $endgroup$
    – Jam
    Dec 20 '18 at 4:04










  • $begingroup$
    The relationship between your recurrence relation and the stated problem is not clear. What do these $p_k$'s represent?
    $endgroup$
    – Michael Burr
    Dec 20 '18 at 4:23








1




1




$begingroup$
What do you mean by p?
$endgroup$
– Martund
Dec 20 '18 at 3:55






$begingroup$
What do you mean by p?
$endgroup$
– Martund
Dec 20 '18 at 3:55














$begingroup$
This may help.
$endgroup$
– Jam
Dec 20 '18 at 4:04




$begingroup$
This may help.
$endgroup$
– Jam
Dec 20 '18 at 4:04












$begingroup$
The relationship between your recurrence relation and the stated problem is not clear. What do these $p_k$'s represent?
$endgroup$
– Michael Burr
Dec 20 '18 at 4:23




$begingroup$
The relationship between your recurrence relation and the stated problem is not clear. What do these $p_k$'s represent?
$endgroup$
– Michael Burr
Dec 20 '18 at 4:23










3 Answers
3






active

oldest

votes


















2












$begingroup$

Write
$p_k=frac{1}{2}p_{k+1} + frac{1}{2}p_{k-1}.
$
in the form
$p_{k+1}
=2p_k-p_{k-1}
$.



This has characteristic polynomial
$d^2 = 2d-1$
or
$d^2-2d+1 = 0$
or
$(d-1)^2 = 0$.



The repeated root means that
the two solutions are
$r^k$
and $kr^k$.



The $r^k$ solution gives
$r^{k+1}
=2r^k-r^{k-1}
$
or
$r^2
=2r-1
$,
so $r = 1$
as you got.



The $kr^k$ solution gives,
since $r=1$,
just $k$.



To check, if
$p_k = k$
then
$2p_k-p_{k-1}
=2k-(k-1)
=k+1
=p_{k+1}
$.



Therefore the solutions are
$1$ and $k$,
so the general solution is
$a + bk$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Perfect answer. I'd like to add just one thing. $(d-a)^2$ itself gives $a^k$ and $ka^k$ ($a=1$ in this case). You need not put r and evaluate it separately.
    $endgroup$
    – Ankit Kumar
    Dec 20 '18 at 5:00












  • $begingroup$
    I forgot about the repeated root, thank you!
    $endgroup$
    – user107224
    Dec 20 '18 at 9:47



















0












$begingroup$

If $p_0$ and $p_1$ are given, then an easy proof by induction gives



$$p_n=n(p_1-p_0)+p_0.$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    A simple argument lets us avoid solving a recurrence relation.



    The game is fair, so the expected amount of money you end with is equal to the money you start with. You begin with $20; you end with $100 with some unknown probability $x$, and $0 otherwise. Therefore $20 = 100x$, which yields $x = frac15$.



    (Formally, we should check the hypotheses of Wald's identity to know that the expected amount of money you end with is equal to the money you start with. This means we want the number of steps in the game to be a stopping time (which it is: whether you have won or lost after $n$ games depends only on the outcomes of those $n$ games) and finite with probability $1$ (which is true in any finite Markov chain), while the bets should be bounded in addition to being fair.)






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Write
      $p_k=frac{1}{2}p_{k+1} + frac{1}{2}p_{k-1}.
      $
      in the form
      $p_{k+1}
      =2p_k-p_{k-1}
      $      .



      This has characteristic polynomial
      $d^2 = 2d-1$
      or
      $d^2-2d+1 = 0$
      or
      $(d-1)^2 = 0$.



      The repeated root means that
      the two solutions are
      $r^k$
      and $kr^k$.



      The $r^k$ solution gives
      $r^{k+1}
      =2r^k-r^{k-1}
      $
      or
      $r^2
      =2r-1
      $      ,
      so $r = 1$
      as you got.



      The $kr^k$ solution gives,
      since $r=1$,
      just $k$.



      To check, if
      $p_k = k$
      then
      $2p_k-p_{k-1}
      =2k-(k-1)
      =k+1
      =p_{k+1}
      $      .



      Therefore the solutions are
      $1$ and $k$,
      so the general solution is
      $a + bk$.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        Perfect answer. I'd like to add just one thing. $(d-a)^2$ itself gives $a^k$ and $ka^k$ ($a=1$ in this case). You need not put r and evaluate it separately.
        $endgroup$
        – Ankit Kumar
        Dec 20 '18 at 5:00












      • $begingroup$
        I forgot about the repeated root, thank you!
        $endgroup$
        – user107224
        Dec 20 '18 at 9:47
















      2












      $begingroup$

      Write
      $p_k=frac{1}{2}p_{k+1} + frac{1}{2}p_{k-1}.
      $
      in the form
      $p_{k+1}
      =2p_k-p_{k-1}
      $      .



      This has characteristic polynomial
      $d^2 = 2d-1$
      or
      $d^2-2d+1 = 0$
      or
      $(d-1)^2 = 0$.



      The repeated root means that
      the two solutions are
      $r^k$
      and $kr^k$.



      The $r^k$ solution gives
      $r^{k+1}
      =2r^k-r^{k-1}
      $
      or
      $r^2
      =2r-1
      $      ,
      so $r = 1$
      as you got.



      The $kr^k$ solution gives,
      since $r=1$,
      just $k$.



      To check, if
      $p_k = k$
      then
      $2p_k-p_{k-1}
      =2k-(k-1)
      =k+1
      =p_{k+1}
      $      .



      Therefore the solutions are
      $1$ and $k$,
      so the general solution is
      $a + bk$.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        Perfect answer. I'd like to add just one thing. $(d-a)^2$ itself gives $a^k$ and $ka^k$ ($a=1$ in this case). You need not put r and evaluate it separately.
        $endgroup$
        – Ankit Kumar
        Dec 20 '18 at 5:00












      • $begingroup$
        I forgot about the repeated root, thank you!
        $endgroup$
        – user107224
        Dec 20 '18 at 9:47














      2












      2








      2





      $begingroup$

      Write
      $p_k=frac{1}{2}p_{k+1} + frac{1}{2}p_{k-1}.
      $
      in the form
      $p_{k+1}
      =2p_k-p_{k-1}
      $      .



      This has characteristic polynomial
      $d^2 = 2d-1$
      or
      $d^2-2d+1 = 0$
      or
      $(d-1)^2 = 0$.



      The repeated root means that
      the two solutions are
      $r^k$
      and $kr^k$.



      The $r^k$ solution gives
      $r^{k+1}
      =2r^k-r^{k-1}
      $
      or
      $r^2
      =2r-1
      $      ,
      so $r = 1$
      as you got.



      The $kr^k$ solution gives,
      since $r=1$,
      just $k$.



      To check, if
      $p_k = k$
      then
      $2p_k-p_{k-1}
      =2k-(k-1)
      =k+1
      =p_{k+1}
      $      .



      Therefore the solutions are
      $1$ and $k$,
      so the general solution is
      $a + bk$.






      share|cite|improve this answer









      $endgroup$



      Write
      $p_k=frac{1}{2}p_{k+1} + frac{1}{2}p_{k-1}.
      $
      in the form
      $p_{k+1}
      =2p_k-p_{k-1}
      $      .



      This has characteristic polynomial
      $d^2 = 2d-1$
      or
      $d^2-2d+1 = 0$
      or
      $(d-1)^2 = 0$.



      The repeated root means that
      the two solutions are
      $r^k$
      and $kr^k$.



      The $r^k$ solution gives
      $r^{k+1}
      =2r^k-r^{k-1}
      $
      or
      $r^2
      =2r-1
      $      ,
      so $r = 1$
      as you got.



      The $kr^k$ solution gives,
      since $r=1$,
      just $k$.



      To check, if
      $p_k = k$
      then
      $2p_k-p_{k-1}
      =2k-(k-1)
      =k+1
      =p_{k+1}
      $      .



      Therefore the solutions are
      $1$ and $k$,
      so the general solution is
      $a + bk$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 20 '18 at 4:17









      marty cohenmarty cohen

      74.5k549129




      74.5k549129








      • 1




        $begingroup$
        Perfect answer. I'd like to add just one thing. $(d-a)^2$ itself gives $a^k$ and $ka^k$ ($a=1$ in this case). You need not put r and evaluate it separately.
        $endgroup$
        – Ankit Kumar
        Dec 20 '18 at 5:00












      • $begingroup$
        I forgot about the repeated root, thank you!
        $endgroup$
        – user107224
        Dec 20 '18 at 9:47














      • 1




        $begingroup$
        Perfect answer. I'd like to add just one thing. $(d-a)^2$ itself gives $a^k$ and $ka^k$ ($a=1$ in this case). You need not put r and evaluate it separately.
        $endgroup$
        – Ankit Kumar
        Dec 20 '18 at 5:00












      • $begingroup$
        I forgot about the repeated root, thank you!
        $endgroup$
        – user107224
        Dec 20 '18 at 9:47








      1




      1




      $begingroup$
      Perfect answer. I'd like to add just one thing. $(d-a)^2$ itself gives $a^k$ and $ka^k$ ($a=1$ in this case). You need not put r and evaluate it separately.
      $endgroup$
      – Ankit Kumar
      Dec 20 '18 at 5:00






      $begingroup$
      Perfect answer. I'd like to add just one thing. $(d-a)^2$ itself gives $a^k$ and $ka^k$ ($a=1$ in this case). You need not put r and evaluate it separately.
      $endgroup$
      – Ankit Kumar
      Dec 20 '18 at 5:00














      $begingroup$
      I forgot about the repeated root, thank you!
      $endgroup$
      – user107224
      Dec 20 '18 at 9:47




      $begingroup$
      I forgot about the repeated root, thank you!
      $endgroup$
      – user107224
      Dec 20 '18 at 9:47











      0












      $begingroup$

      If $p_0$ and $p_1$ are given, then an easy proof by induction gives



      $$p_n=n(p_1-p_0)+p_0.$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        If $p_0$ and $p_1$ are given, then an easy proof by induction gives



        $$p_n=n(p_1-p_0)+p_0.$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          If $p_0$ and $p_1$ are given, then an easy proof by induction gives



          $$p_n=n(p_1-p_0)+p_0.$$






          share|cite|improve this answer









          $endgroup$



          If $p_0$ and $p_1$ are given, then an easy proof by induction gives



          $$p_n=n(p_1-p_0)+p_0.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 20 '18 at 4:19









          FredFred

          48.6k11849




          48.6k11849























              0












              $begingroup$

              A simple argument lets us avoid solving a recurrence relation.



              The game is fair, so the expected amount of money you end with is equal to the money you start with. You begin with $20; you end with $100 with some unknown probability $x$, and $0 otherwise. Therefore $20 = 100x$, which yields $x = frac15$.



              (Formally, we should check the hypotheses of Wald's identity to know that the expected amount of money you end with is equal to the money you start with. This means we want the number of steps in the game to be a stopping time (which it is: whether you have won or lost after $n$ games depends only on the outcomes of those $n$ games) and finite with probability $1$ (which is true in any finite Markov chain), while the bets should be bounded in addition to being fair.)






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                A simple argument lets us avoid solving a recurrence relation.



                The game is fair, so the expected amount of money you end with is equal to the money you start with. You begin with $20; you end with $100 with some unknown probability $x$, and $0 otherwise. Therefore $20 = 100x$, which yields $x = frac15$.



                (Formally, we should check the hypotheses of Wald's identity to know that the expected amount of money you end with is equal to the money you start with. This means we want the number of steps in the game to be a stopping time (which it is: whether you have won or lost after $n$ games depends only on the outcomes of those $n$ games) and finite with probability $1$ (which is true in any finite Markov chain), while the bets should be bounded in addition to being fair.)






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  A simple argument lets us avoid solving a recurrence relation.



                  The game is fair, so the expected amount of money you end with is equal to the money you start with. You begin with $20; you end with $100 with some unknown probability $x$, and $0 otherwise. Therefore $20 = 100x$, which yields $x = frac15$.



                  (Formally, we should check the hypotheses of Wald's identity to know that the expected amount of money you end with is equal to the money you start with. This means we want the number of steps in the game to be a stopping time (which it is: whether you have won or lost after $n$ games depends only on the outcomes of those $n$ games) and finite with probability $1$ (which is true in any finite Markov chain), while the bets should be bounded in addition to being fair.)






                  share|cite|improve this answer









                  $endgroup$



                  A simple argument lets us avoid solving a recurrence relation.



                  The game is fair, so the expected amount of money you end with is equal to the money you start with. You begin with $20; you end with $100 with some unknown probability $x$, and $0 otherwise. Therefore $20 = 100x$, which yields $x = frac15$.



                  (Formally, we should check the hypotheses of Wald's identity to know that the expected amount of money you end with is equal to the money you start with. This means we want the number of steps in the game to be a stopping time (which it is: whether you have won or lost after $n$ games depends only on the outcomes of those $n$ games) and finite with probability $1$ (which is true in any finite Markov chain), while the bets should be bounded in addition to being fair.)







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 20 '18 at 4:45









                  Misha LavrovMisha Lavrov

                  47.8k657107




                  47.8k657107






























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                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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                      web3.py web3.isConnected() returns false always

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                      1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
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