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Given $x \in (0,\frac{1}{2})$ and $y \in (0,\frac{1}{2}]$, what is the value of the following limit:

$\lim_{n\rightarrow \infty}\sum_{k=0}^{n}{n \choose k}|x^{n-k}(1-x)^{k}-y^{n-k}(1-y)^{k}|?$

When $y = \frac{1}{2}$, we have:

$\lim_{n\rightarrow \infty}\sum_{k=0}^{n}{n \choose k}|x^{n-k}(1-x)^{k}-\frac{1}{2^{n}}|$

Any idea can help me... Or a book which may help, it's fine too...

thanks for your attention

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if x = y the value is 0 but if $x \ne y$ it is 2. By a large deviation estimate the sum involving the x terms will concentrate on an interval $k \in ((x-\epsilon) n, (x + \epsilon n))$ and similarly for the y, and no other interval will contribute much. Again by large deviation and local limit theorems, the y terms are much smaller than the x terms on x's big interval and vice versa.

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  • $\begingroup$ I didn`t expect for this answer, Professor @Michael. But, I couldn't understand your answer. What do you mean by sum involving the x terms" and y terms? Because we have a modulus there... it's very strange to me this answer... could you say the theorems you use to obtain it? Because, 2 is the value of the limit when you change the minus signal for plus signal... Thanks again for your attention $\endgroup$
    – Bruno Brogni Uggioni
    Commented Oct 8, 2015 at 14:38
  • $\begingroup$ I mean that according to the Law of Large Numbers $\sum_{k \notin ((x-\epsilon) , ((x + \epsilon) n}n \choose k}|x^{n-k}(1-x)^{k}|$ goes to 0, and similarly for the y's. Also $\sum_{k \in ((x-\epsilon) , ((x + \epsilon) n}n \choose k}|y^{n-k}(1-y)^{k}|$ and similarly for the x's. I mentioned large deviation results, but I think this is all you need to get the limit of the sums. $\endgroup$
    – Michael
    Commented Oct 8, 2015 at 15:20
  • $\begingroup$ Sorry, I tried to reply by my tex style is not up to it. $\endgroup$
    – Michael
    Commented Oct 8, 2015 at 15:29
  • $\begingroup$ Please add $ where apppropriate so we can read formulas more easily, thanks. $\endgroup$
    – Jean Duchon
    Commented Oct 8, 2015 at 15:31
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    $\begingroup$ I will try and explain @Michael's answer some more: $\binom nk x^k(1-x)^{n-k}$ is the probability distribution of Binomial$(n,x)$. As Michael says, almost all the mass of this is close to $nx$. On the other hand, $y^k(1-y)^{n-k}$ is the probability distribution of Binomial$(n,y)$. This one is concentrated near $ny$ (within $\pm n^{\frac 12+\epsilon}$). Hence for large $n$, the two distributions are essentially disjointly supported. Now when you subtract them, there is basically no cancellation. When you take the abs. value, it's essentially the same as adding the two distributions, giving 2. $\endgroup$
    – Anthony Quas
    Commented Oct 8, 2015 at 18:00

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