Using Binomial distribution to evaluate sums












0












$begingroup$


We know if $X$ is binomial random variable with parameter $n,p$, the mass function is given by $p_X(i) = {n choose i} p^i (1-p)^{n-i} $ and $sum_{i=0}^n p_X(i) = 1 $.



If we put $p=1/2$, we see that $sum {n choose i} = 2^n $. Is this a valid reasoning? or is it circular reasoning?



In general, one can write



$$ sum_{i=0}^n {n choose i} left( frac{p}{1-p} right)^i = frac{1}{(1-p)^n} $$



where $0 leq p leq 1$



For instance, one can put $p = sin^2 alpha $ and we obtain an identity



$$ sum_{i=0}^n {n choose i} tan^{2i} alpha = frac{1}{cos^{2n} alpha} $$



which is probably useless, but anyway, my question is, can we use the distribution as such or we are just in a circular reasoning?






calculus probability







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    0












    $begingroup$


    We know if $X$ is binomial random variable with parameter $n,p$, the mass function is given by $p_X(i) = {n choose i} p^i (1-p)^{n-i} $ and $sum_{i=0}^n p_X(i) = 1 $.



    If we put $p=1/2$, we see that $sum {n choose i} = 2^n $. Is this a valid reasoning? or is it circular reasoning?



    In general, one can write



    $$ sum_{i=0}^n {n choose i} left( frac{p}{1-p} right)^i = frac{1}{(1-p)^n} $$



    where $0 leq p leq 1$



    For instance, one can put $p = sin^2 alpha $ and we obtain an identity



    $$ sum_{i=0}^n {n choose i} tan^{2i} alpha = frac{1}{cos^{2n} alpha} $$



    which is probably useless, but anyway, my question is, can we use the distribution as such or we are just in a circular reasoning?






    calculus probability







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      We know if $X$ is binomial random variable with parameter $n,p$, the mass function is given by $p_X(i) = {n choose i} p^i (1-p)^{n-i} $ and $sum_{i=0}^n p_X(i) = 1 $.



      If we put $p=1/2$, we see that $sum {n choose i} = 2^n $. Is this a valid reasoning? or is it circular reasoning?



      In general, one can write



      $$ sum_{i=0}^n {n choose i} left( frac{p}{1-p} right)^i = frac{1}{(1-p)^n} $$



      where $0 leq p leq 1$



      For instance, one can put $p = sin^2 alpha $ and we obtain an identity



      $$ sum_{i=0}^n {n choose i} tan^{2i} alpha = frac{1}{cos^{2n} alpha} $$



      which is probably useless, but anyway, my question is, can we use the distribution as such or we are just in a circular reasoning?






      calculus probability







      share|cite|improve this question









      $endgroup$




      We know if $X$ is binomial random variable with parameter $n,p$, the mass function is given by $p_X(i) = {n choose i} p^i (1-p)^{n-i} $ and $sum_{i=0}^n p_X(i) = 1 $.



      If we put $p=1/2$, we see that $sum {n choose i} = 2^n $. Is this a valid reasoning? or is it circular reasoning?



      In general, one can write



      $$ sum_{i=0}^n {n choose i} left( frac{p}{1-p} right)^i = frac{1}{(1-p)^n} $$



      where $0 leq p leq 1$



      For instance, one can put $p = sin^2 alpha $ and we obtain an identity



      $$ sum_{i=0}^n {n choose i} tan^{2i} alpha = frac{1}{cos^{2n} alpha} $$



      which is probably useless, but anyway, my question is, can we use the distribution as such or we are just in a circular reasoning?






      calculus probability




      calculus probability






      share|cite|improve this question













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      asked Jan 26 at 10:09









      Jimmy SabaterJimmy Sabater

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          $begingroup$

          All comes down to the equality:$$sum_{i=0}^nbinom{n}ix^iy^{n-i}=(x+y)^ntag1$$where $n$ is a nonnegative integer and $x,y$ are real (or complex if you like) numbers.



          It can be shown that this is true if $x,ygeq0$ with $x+y=1$ (corresponding with binomial distribution) purely on base of probability theory.



          Then on base of that result it is also possible to deduce the more general case $(1)$.



          So the reasoning you mention is valid.



          I would not advice though to involve binomial distribution to evaluate sums.



          Why should we if we have $(1)$ on our plate already?






          share|cite|improve this answer









          $endgroup$













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            $begingroup$

            All comes down to the equality:$$sum_{i=0}^nbinom{n}ix^iy^{n-i}=(x+y)^ntag1$$where $n$ is a nonnegative integer and $x,y$ are real (or complex if you like) numbers.



            It can be shown that this is true if $x,ygeq0$ with $x+y=1$ (corresponding with binomial distribution) purely on base of probability theory.



            Then on base of that result it is also possible to deduce the more general case $(1)$.



            So the reasoning you mention is valid.



            I would not advice though to involve binomial distribution to evaluate sums.



            Why should we if we have $(1)$ on our plate already?






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              All comes down to the equality:$$sum_{i=0}^nbinom{n}ix^iy^{n-i}=(x+y)^ntag1$$where $n$ is a nonnegative integer and $x,y$ are real (or complex if you like) numbers.



              It can be shown that this is true if $x,ygeq0$ with $x+y=1$ (corresponding with binomial distribution) purely on base of probability theory.



              Then on base of that result it is also possible to deduce the more general case $(1)$.



              So the reasoning you mention is valid.



              I would not advice though to involve binomial distribution to evaluate sums.



              Why should we if we have $(1)$ on our plate already?






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                All comes down to the equality:$$sum_{i=0}^nbinom{n}ix^iy^{n-i}=(x+y)^ntag1$$where $n$ is a nonnegative integer and $x,y$ are real (or complex if you like) numbers.



                It can be shown that this is true if $x,ygeq0$ with $x+y=1$ (corresponding with binomial distribution) purely on base of probability theory.



                Then on base of that result it is also possible to deduce the more general case $(1)$.



                So the reasoning you mention is valid.



                I would not advice though to involve binomial distribution to evaluate sums.



                Why should we if we have $(1)$ on our plate already?






                share|cite|improve this answer









                $endgroup$



                All comes down to the equality:$$sum_{i=0}^nbinom{n}ix^iy^{n-i}=(x+y)^ntag1$$where $n$ is a nonnegative integer and $x,y$ are real (or complex if you like) numbers.



                It can be shown that this is true if $x,ygeq0$ with $x+y=1$ (corresponding with binomial distribution) purely on base of probability theory.



                Then on base of that result it is also possible to deduce the more general case $(1)$.



                So the reasoning you mention is valid.



                I would not advice though to involve binomial distribution to evaluate sums.



                Why should we if we have $(1)$ on our plate already?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 26 at 10:31









                drhabdrhab

                103k545136




                103k545136






























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