Probability of Binomial twice of GeometricI've come up with an interesting result:Let X be the amount of failures of Bernoulli(p) until we get (p). X = G e o ( p ) B = B i n ( 2 X , 1 2 ) P ( B = X ) = ?Turns out: P ( B = X ) = p I found it using the Taylor expansion of 1 1 − p , where the coefficient of p i turns out to be P ( B i n ( 2 i , 1 2 ) = i ).I would like to see a probabilistic proof of this result.Explanation of the process in words:Roll a die with probability p of getting "X". Each time that we don't get "X", toss 2 balanced coins and accumulate the number of heads and tails. When you get "X", check if you got the same amount of heads and tails.Programmatic explanation: i = 0 w h i l e ( ! b e r n u l i ( p ) ) i++; B i n ( 2 i , 1 2 ) = ? iExample:If we succeed immediately (with probability p), X = 0, and P ( B i n ( 2 ⋅ 0 , 1 2 ) = 0 ) = 1, thus P ( B = 0 ) = 1, thus contributing p to the conditional sum, p &lt; p , and everything is alright.

Question

Probability of Binomial twice of GeometricI&#039;ve come up with an interesting result:Let X be the amount of failures of Bernoulli(p) until we get (p).  X  =  G  e  o  (  p  )  B  =  B  i  n      (    2    X    ,          1      2        )    P  (  B  =  X  )  =     ?Turns out:   P  (  B  =  X  )  =      p  I found it using the Taylor expansion of       1          1      −      p      , where the coefficient of       p    i   turns out to be   P  (  B  i  n  (  2  i  ,      1    2    )  =  i  ).I would like to see a probabilistic proof of this result.Explanation of the process in words:Roll a die with probability p of getting &quot;X&quot;. Each time that we don&#039;t get &quot;X&quot;, toss 2 balanced coins and accumulate the number of heads and tails. When you get &quot;X&quot;, check if you got the same amount of heads and tails.Programmatic explanation:  i  =  0  w  h  i  l  e  (  !  b  e  r  n  u  l  i  (  p  )  )     i++;  B  i  n      (    2    i    ,          1      2        )                      =                    ?              iExample:If we succeed immediately (with probability p),   X  =  0, and   P  (  B  i  n  (  2  ⋅  0  ,      1    2    )  =  0  )  =  1, thus   P  (  B  =  0  )  =  1, thus contributing p to the conditional sum,   p  &amp;lt;      p  , and everything is alright.

Jaydan Gilbert · Accepted Answer

Explanation:                    P        (        B        =        X        )                            =                              ∑                      x            =            0                    ∞                P        (        B        =        X                  |                X        =        x        )        P        (        X        =        x        )                                          =                              ∑                      x            =            0                    ∞                                                    (                                                          2              x                        x                                              )                                                            (                          1              2                        )                                2            x                          (        1        −        p                  )          x                p                                          =                              p

Roderick Bradley · Accepted Answer

Step 1Let       S    n   be simple random walk. For integer a, let       p    a    =  P  (  2  a  +      S          2      X        =  0  ) - that is starting the random walk at 2a rather then zero. The quantity of interest is       p    0  . Consider the single double-step of the random walk and recall memorylessness of X to get       p    a    =  p  [  a  =  0  ]  +  (  1  −  p  )  (      1    4        p          a      −      1        +      1    2        p    a    +      1    4        p          a      +      1        ).Solution to this recurrence is       p    i    =      p    0        r                  |            i              |             where   r  =            1      −              p                    1      +              p              ∼  1  −  2      p   and we obtain       p    0    =      p   through normalization.Step 2To get some further intuition as to how a square root appears, consider standard Brownian motion       B    X   with exponential clock X of rate   2  λ. Then density of       B    X   at 0 is       f                  B        X              (  0  )  =      ∫    0    ∞        1          2      π      x        2  λ      e          −      2      λ      x          d  x  =      λ   and more generally       f          a      +              B        X              (  0  )  =      λ        e          −      2              λ                    |            a              |            .

Turns out: P(B=X)=sqrt{p}. I found it using the Taylor expansion of 1/sqrt{1-p}, where the coefficient of pi turns out to be P(Bin(2i,1/2)=i). I would like to see a probabilistic proof of this result.

Answered question

Answer & Explanation

New Questions in High school geometry