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In finding the asymptotic value of a certain quantity I ended up with the following:

$$f(n)=(n-1)B_{1/2}(n,n+2)$$ $$g(n)=(n+1)B_{1/2}(n+2,n)$$ $$ h(n)=4^n \left( f(n)-g(n)\right)$$ Numerical simulations lead me to believe that h(n) approaches 1 as n approaches infinity.However ,i am not able to prove it analytically Can somebody kindly help me with the evaluation of $ \lim_{n \to \infty} h(n)$?
If somebody could also plug it in Mathematica, I would be highly obliged. Thanks for any help in advance.
P.S.: In the above the notation $B_z(a,b)$ stands for the incomplete beta function defined by: $$B_z(a,b)=\int\limits_0^z u^{a-1}(1-u)^{b-1} \mathrm{d}u.$$

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Exploit the symmetry

$$B(1/2;n,n+2) + B(1/2;n+2,n) = B(1;n,n+2) = B(n,n+2) = \frac{\Gamma(n)\Gamma(n+2)}{\Gamma(2n+2)}$$

(the Beta function) to express this in terms of $B(1/2;n,n+2)$ and $B(n,n+2).$ Integrate by parts to express $B(1/2;n,n+2)$ in terms of $B(1/2;n+1,n+1) = B(n+1,n+1)/2$ (by the symmetry of its integrand) to obtain

$$f(n) - g(n) = 2^{-2n} + (n+1)(B(n+1,n+1) - B(n,n+2)).$$

Simple algebra then reduces this to

$$4^{n}(f(n) - g(n)) = 1 - \frac{n+1}{n(2n+1)}\,\frac{2^{2n}}{\binom{2n}{n}}.$$

It is well-known (and readily deducible from, say, Stirling's asymptotic approximation to the $\Gamma$ function) that $2^{-2n}\binom{2n}{n} = O(n^{-1/2})$ and the limit follows.

(Why did I express the result in this way? Because $\binom{2n}{n}/2^{2n}$ is a familiar quantity: it expresses the proportion of row $2n$ in Pascal's Triangle occupied by the middle value. To statisticians, this is the chance of obtaining exactly $n$ heads in $2n$ independent flips of a fair coin. The Normal approximation to the Binomial$(2n,1/2)$ distribution (which has mean $n$ and variance $n/2$) estimates this proportion as

$$2^{-2n}\binom{2n}{n} \approx \Phi\left(\frac{1/2}{\sqrt{n/2}}\right) - \Phi\left(-\frac{1/2}{\sqrt{n/2}}\right) \approx \phi(0)\left(\frac{\sqrt 2}{\sqrt{n}}\right)=\sqrt{\frac{1}{n\pi}}.$$

The approximation uses the first order expansion of the standard Normal cdf $\Phi$ at $0.$)

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  • $\begingroup$ The mean and variance should be $n$ and $\frac{n}2$. Then, the approximation in the second to last line will be $\sqrt{\frac{1}{n \pi}}$. $\endgroup$ – John L Mar 8 at 14:19
  • $\begingroup$ @John Right: I mistakenly replaced $2n$ with $n$ in my computations. Thank you for catching that. $\endgroup$ – whuber Mar 8 at 14:31
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Mathematica cannot evaluate the limit.

Use these equalities:
$$B_z(a,b)=B(a,b)-B_{1-z}(b,a)$$ $$B_z(a,b)=\frac{1}{a}((1-z)^b z^a+(a+b)B_z(a+1,b))$$ $$B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ $$B_{1/2}(a,a)=\frac{\Gamma(a)^2}{2\Gamma(2a)}$$

and you can show $$h(n)=4^n \left(2 \left(\left(\frac{1}2 \right)^{2n + 2} + \left(\left(\frac{1}2 \right)^{2n + 2} + \frac{(2n +3) \Gamma(n + 2)^2}{ \Gamma(2n + 2 )} \right) \right)\\-(n+1) \frac{\Gamma(n+2)\Gamma(n)}{\Gamma(2n+2)} \right)$$ $$=1-\frac{\sqrt{\pi}}{2n^{1/2}}-\frac{5 \sqrt{\pi}}{16 n^{3/2}}+...$$

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