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What is probabilistic or statistical interpretation of the identity given below? \begin{equation}\label{BIIntro1} \sum_{k=0}^{n}\binom{2k}{k}\binom{2n-2k}{n-k}=4^{n} \end{equation}

sudha pandey
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  • The combinatorial interpretation is given here: https://math.stackexchange.com/a/688370 and here https://math.stackexchange.com/a/360780. Following the last link, $\binom{2k}{k}\binom{2n-2k}{n-k}/4^{n}$ is the conditional probability, given that there are more heads than tails in $2n+1$ throws of fair coin, that the longest prefix which containes equal number of heads and tails, has length exactly $2k$. – NCh Jun 11 '17 at 09:01
  • @NCh Thank you very much for your response. Could you please explain conditional probability approach, this is not clear to me. – sudha pandey Jun 11 '17 at 09:23
  • In the linked answer, there proved that if we have word of length $2n+1$ from the alphabet ${A,B}$, then the total number of words which containes more letters $A$ than $B$, equals to $4^n=2^2n$. This is one half of total number of words $2^{2n+1}$. Let an event $X$ is that there are more $A$ than $B$ in a word of length $2n+1$ where each letter is chosen independently at random from the alphabet ${A,B}$. If we conditioned that this event happens, we consider only words from $X$ as the elementary events. Let $Y_k$ be an event that the longest prefix of a random ... – NCh Jun 11 '17 at 09:50
  • ...word which containes as many $A$ as $B$, has length $2k$. In the linked answer it is proved that the number of words in $X\cap Y_k$ equals to $\binom{2k}{k}\binom{2n-2k}{n-k}$. Then $$P(Y_k\mid X) =\frac{|X\cap Y_k|}{|X|}=\frac{\binom{2k}{k}\binom{2n-2k}{n-k}}{4^n}.$$ The coin looks more convinient to me. Simply replace $A$ by heads and $B$ by tails. – NCh Jun 11 '17 at 09:50
  • @NCh Thank you very much for your quick reply. Are these letters are only combinations of letter A and B? For example ABBBBABAA, AAAABAAAB etc. ? – sudha pandey Jun 11 '17 at 09:55
  • @NCh Is any statistical interpretation for this identity is possible? If yes, please share with me I shall be highly thankful for this help. – sudha pandey Jun 11 '17 at 09:58
  • Not sure that I understood yuor question correctly. Say, for $n=2$, $2n+1=5$, there are $32$ words total. Among them there are $16$ words where more $A$ than $B$: $AAAAA$, $AAABB$, $ABABA$, $BAAAA$ and so on. Then $|X|=4^n=16$. If $k=0$, the event $Y_0={AAAAA, AAABB, \ldots}$ has the probability $P(Y_0\mid X)=\binom00\binom42/16=6/16$. $Y_0$ means that no prefix containes equal number of $A$ and $B$. Next, $Y_1={ABAAB, BAAAB, \ldots}$ - prefix of max length $2$ contains equal number $A$ and $B$. $P(Y_1\mid X)=\binom21\binom{4-2}{2-1}=4/16$ too. And so on. – NCh Jun 11 '17 at 10:23
  • Unfortunately, I do not know what statistical interpretation is possible. – NCh Jun 11 '17 at 10:23
  • @NCh Thank you very much for your response. – sudha pandey Jun 11 '17 at 12:12

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