combinatorics floor summation












2












$begingroup$


While solving a problem I came across a rather interesting identity, and I do not see how I could prove it.



$sum_{i=0}^{floor(frac{n}{2})} binom{n}{2i}p^{2i}(1-p)^{n-2i} = frac{1}{2}((2p-1)^{n}+1)$



Any ideas?






probability combinatorics summation binomial-coefficients binomial-distribution







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  • 1




    $begingroup$
    Flip $n$ coins each having probability $p$ of landing heads. The lefthand side is the probability of flipping an even number of heads. I don't immediately see how the righthand side calculates the same probability.
    $endgroup$
    – Austin Mohr
    2 hours ago






  • 1




    $begingroup$
    I believe you have a mistake: the RHS should be $frac12((1-2p)^n+1)$.
    $endgroup$
    – Mike Earnest
    2 hours ago
















2












$begingroup$


While solving a problem I came across a rather interesting identity, and I do not see how I could prove it.



$sum_{i=0}^{floor(frac{n}{2})} binom{n}{2i}p^{2i}(1-p)^{n-2i} = frac{1}{2}((2p-1)^{n}+1)$



Any ideas?






probability combinatorics summation binomial-coefficients binomial-distribution







share|cite|improve this question









New contributor




SzymonSzymon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Flip $n$ coins each having probability $p$ of landing heads. The lefthand side is the probability of flipping an even number of heads. I don't immediately see how the righthand side calculates the same probability.
    $endgroup$
    – Austin Mohr
    2 hours ago






  • 1




    $begingroup$
    I believe you have a mistake: the RHS should be $frac12((1-2p)^n+1)$.
    $endgroup$
    – Mike Earnest
    2 hours ago














2












2








2





$begingroup$


While solving a problem I came across a rather interesting identity, and I do not see how I could prove it.



$sum_{i=0}^{floor(frac{n}{2})} binom{n}{2i}p^{2i}(1-p)^{n-2i} = frac{1}{2}((2p-1)^{n}+1)$



Any ideas?






probability combinatorics summation binomial-coefficients binomial-distribution







share|cite|improve this question









New contributor




SzymonSzymon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




While solving a problem I came across a rather interesting identity, and I do not see how I could prove it.



$sum_{i=0}^{floor(frac{n}{2})} binom{n}{2i}p^{2i}(1-p)^{n-2i} = frac{1}{2}((2p-1)^{n}+1)$



Any ideas?






probability combinatorics summation binomial-coefficients binomial-distribution




probability combinatorics summation binomial-coefficients binomial-distribution






share|cite|improve this question









New contributor




SzymonSzymon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




SzymonSzymon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Austin Mohr

20.5k35098




20.5k35098






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asked 2 hours ago









SzymonSzymonSzymonSzymon

111




111




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SzymonSzymon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





SzymonSzymon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






SzymonSzymon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    Flip $n$ coins each having probability $p$ of landing heads. The lefthand side is the probability of flipping an even number of heads. I don't immediately see how the righthand side calculates the same probability.
    $endgroup$
    – Austin Mohr
    2 hours ago






  • 1




    $begingroup$
    I believe you have a mistake: the RHS should be $frac12((1-2p)^n+1)$.
    $endgroup$
    – Mike Earnest
    2 hours ago














  • 1




    $begingroup$
    Flip $n$ coins each having probability $p$ of landing heads. The lefthand side is the probability of flipping an even number of heads. I don't immediately see how the righthand side calculates the same probability.
    $endgroup$
    – Austin Mohr
    2 hours ago






  • 1




    $begingroup$
    I believe you have a mistake: the RHS should be $frac12((1-2p)^n+1)$.
    $endgroup$
    – Mike Earnest
    2 hours ago








1




1




$begingroup$
Flip $n$ coins each having probability $p$ of landing heads. The lefthand side is the probability of flipping an even number of heads. I don't immediately see how the righthand side calculates the same probability.
$endgroup$
– Austin Mohr
2 hours ago




$begingroup$
Flip $n$ coins each having probability $p$ of landing heads. The lefthand side is the probability of flipping an even number of heads. I don't immediately see how the righthand side calculates the same probability.
$endgroup$
– Austin Mohr
2 hours ago




1




1




$begingroup$
I believe you have a mistake: the RHS should be $frac12((1-2p)^n+1)$.
$endgroup$
– Mike Earnest
2 hours ago




$begingroup$
I believe you have a mistake: the RHS should be $frac12((1-2p)^n+1)$.
$endgroup$
– Mike Earnest
2 hours ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

Let $q=1-p$. Apply the binomial theorem twice, then add:



begin{array}{rrl}
&(p+q)^n&=;;sum_{i=0}^n binom{n}ip^iq^{n-i}\
+&(-p+q)^n&=;;sum_{i=0}^n binom{n}i(-p)^iq^{n-i}\hline
&(p+q)^n+(-p+q)^n &=2sum_{i=0}^{lfloor n/2rfloor}binom{n}{2i}p^{2i}q^{n-2i}
end{array}
To understand the RHS of the sum, note when $i$ is odd we have $p^i+(-p)^i=0$, while when $i$ is even, $p^i+(-p)^i=2p^i$. Replacing $q$ with $1-p$ makes the LHS $1+(1-2p)^n$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, that's a very neat proof!
    $endgroup$
    – SzymonSzymon
    2 hours ago



















1












$begingroup$

With the corrected RHS of $frac12((1-2p)^n+1)$ and Austin Mohr's interpretation of the LHS as the probability of an even number of heads occurring in $n$ tosses of coin which comes up heads with probability $p$, the result is easily established by induction.



Case $n = 0$: $frac12((1-2p)^0+1) = 1$, which is correct since zero tosses always produce zero heads.



Case $n > 0$: An even number of heads after $n$ tosses occurs either by having an even number of heads in $n-1$ tosses followed by tails, or an odd number in $n-1$ tosses followed by heads. The probability of this is
$$
begin{align}
&frac12((1-2p)^{n-1}+1)(1-p) + (1 - frac12((1-2p)^{n-1}+1))cdot p\
&= frac12(1-2p)^{n-1}(1-p-p) +frac12(1-p-p) + p\
&= frac12((1-2p)^n + 1),
end{align}
$$
as required.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Let $q=1-p$. Apply the binomial theorem twice, then add:



    begin{array}{rrl}
    &(p+q)^n&=;;sum_{i=0}^n binom{n}ip^iq^{n-i}\
    +&(-p+q)^n&=;;sum_{i=0}^n binom{n}i(-p)^iq^{n-i}\hline
    &(p+q)^n+(-p+q)^n &=2sum_{i=0}^{lfloor n/2rfloor}binom{n}{2i}p^{2i}q^{n-2i}
    end{array}
    To understand the RHS of the sum, note when $i$ is odd we have $p^i+(-p)^i=0$, while when $i$ is even, $p^i+(-p)^i=2p^i$. Replacing $q$ with $1-p$ makes the LHS $1+(1-2p)^n$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you, that's a very neat proof!
      $endgroup$
      – SzymonSzymon
      2 hours ago
















    2












    $begingroup$

    Let $q=1-p$. Apply the binomial theorem twice, then add:



    begin{array}{rrl}
    &(p+q)^n&=;;sum_{i=0}^n binom{n}ip^iq^{n-i}\
    +&(-p+q)^n&=;;sum_{i=0}^n binom{n}i(-p)^iq^{n-i}\hline
    &(p+q)^n+(-p+q)^n &=2sum_{i=0}^{lfloor n/2rfloor}binom{n}{2i}p^{2i}q^{n-2i}
    end{array}
    To understand the RHS of the sum, note when $i$ is odd we have $p^i+(-p)^i=0$, while when $i$ is even, $p^i+(-p)^i=2p^i$. Replacing $q$ with $1-p$ makes the LHS $1+(1-2p)^n$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you, that's a very neat proof!
      $endgroup$
      – SzymonSzymon
      2 hours ago














    2












    2








    2





    $begingroup$

    Let $q=1-p$. Apply the binomial theorem twice, then add:



    begin{array}{rrl}
    &(p+q)^n&=;;sum_{i=0}^n binom{n}ip^iq^{n-i}\
    +&(-p+q)^n&=;;sum_{i=0}^n binom{n}i(-p)^iq^{n-i}\hline
    &(p+q)^n+(-p+q)^n &=2sum_{i=0}^{lfloor n/2rfloor}binom{n}{2i}p^{2i}q^{n-2i}
    end{array}
    To understand the RHS of the sum, note when $i$ is odd we have $p^i+(-p)^i=0$, while when $i$ is even, $p^i+(-p)^i=2p^i$. Replacing $q$ with $1-p$ makes the LHS $1+(1-2p)^n$.






    share|cite|improve this answer











    $endgroup$



    Let $q=1-p$. Apply the binomial theorem twice, then add:



    begin{array}{rrl}
    &(p+q)^n&=;;sum_{i=0}^n binom{n}ip^iq^{n-i}\
    +&(-p+q)^n&=;;sum_{i=0}^n binom{n}i(-p)^iq^{n-i}\hline
    &(p+q)^n+(-p+q)^n &=2sum_{i=0}^{lfloor n/2rfloor}binom{n}{2i}p^{2i}q^{n-2i}
    end{array}
    To understand the RHS of the sum, note when $i$ is odd we have $p^i+(-p)^i=0$, while when $i$ is even, $p^i+(-p)^i=2p^i$. Replacing $q$ with $1-p$ makes the LHS $1+(1-2p)^n$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 2 hours ago

























    answered 2 hours ago









    Mike EarnestMike Earnest

    24.9k22151




    24.9k22151












    • $begingroup$
      Thank you, that's a very neat proof!
      $endgroup$
      – SzymonSzymon
      2 hours ago


















    • $begingroup$
      Thank you, that's a very neat proof!
      $endgroup$
      – SzymonSzymon
      2 hours ago
















    $begingroup$
    Thank you, that's a very neat proof!
    $endgroup$
    – SzymonSzymon
    2 hours ago




    $begingroup$
    Thank you, that's a very neat proof!
    $endgroup$
    – SzymonSzymon
    2 hours ago











    1












    $begingroup$

    With the corrected RHS of $frac12((1-2p)^n+1)$ and Austin Mohr's interpretation of the LHS as the probability of an even number of heads occurring in $n$ tosses of coin which comes up heads with probability $p$, the result is easily established by induction.



    Case $n = 0$: $frac12((1-2p)^0+1) = 1$, which is correct since zero tosses always produce zero heads.



    Case $n > 0$: An even number of heads after $n$ tosses occurs either by having an even number of heads in $n-1$ tosses followed by tails, or an odd number in $n-1$ tosses followed by heads. The probability of this is
    $$
    begin{align}
    &frac12((1-2p)^{n-1}+1)(1-p) + (1 - frac12((1-2p)^{n-1}+1))cdot p\
    &= frac12(1-2p)^{n-1}(1-p-p) +frac12(1-p-p) + p\
    &= frac12((1-2p)^n + 1),
    end{align}
    $$
    as required.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      With the corrected RHS of $frac12((1-2p)^n+1)$ and Austin Mohr's interpretation of the LHS as the probability of an even number of heads occurring in $n$ tosses of coin which comes up heads with probability $p$, the result is easily established by induction.



      Case $n = 0$: $frac12((1-2p)^0+1) = 1$, which is correct since zero tosses always produce zero heads.



      Case $n > 0$: An even number of heads after $n$ tosses occurs either by having an even number of heads in $n-1$ tosses followed by tails, or an odd number in $n-1$ tosses followed by heads. The probability of this is
      $$
      begin{align}
      &frac12((1-2p)^{n-1}+1)(1-p) + (1 - frac12((1-2p)^{n-1}+1))cdot p\
      &= frac12(1-2p)^{n-1}(1-p-p) +frac12(1-p-p) + p\
      &= frac12((1-2p)^n + 1),
      end{align}
      $$
      as required.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        With the corrected RHS of $frac12((1-2p)^n+1)$ and Austin Mohr's interpretation of the LHS as the probability of an even number of heads occurring in $n$ tosses of coin which comes up heads with probability $p$, the result is easily established by induction.



        Case $n = 0$: $frac12((1-2p)^0+1) = 1$, which is correct since zero tosses always produce zero heads.



        Case $n > 0$: An even number of heads after $n$ tosses occurs either by having an even number of heads in $n-1$ tosses followed by tails, or an odd number in $n-1$ tosses followed by heads. The probability of this is
        $$
        begin{align}
        &frac12((1-2p)^{n-1}+1)(1-p) + (1 - frac12((1-2p)^{n-1}+1))cdot p\
        &= frac12(1-2p)^{n-1}(1-p-p) +frac12(1-p-p) + p\
        &= frac12((1-2p)^n + 1),
        end{align}
        $$
        as required.






        share|cite|improve this answer









        $endgroup$



        With the corrected RHS of $frac12((1-2p)^n+1)$ and Austin Mohr's interpretation of the LHS as the probability of an even number of heads occurring in $n$ tosses of coin which comes up heads with probability $p$, the result is easily established by induction.



        Case $n = 0$: $frac12((1-2p)^0+1) = 1$, which is correct since zero tosses always produce zero heads.



        Case $n > 0$: An even number of heads after $n$ tosses occurs either by having an even number of heads in $n-1$ tosses followed by tails, or an odd number in $n-1$ tosses followed by heads. The probability of this is
        $$
        begin{align}
        &frac12((1-2p)^{n-1}+1)(1-p) + (1 - frac12((1-2p)^{n-1}+1))cdot p\
        &= frac12(1-2p)^{n-1}(1-p-p) +frac12(1-p-p) + p\
        &= frac12((1-2p)^n + 1),
        end{align}
        $$
        as required.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        FredHFredH

        2,1641016




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