Let $X_n \sim \mathbf{B}(n,n)$ (Beta distribution), with pdf

$$ f_n(x) = \frac{1}{\text{B}(n,n)}x^{n-1}(1 - x)^{n-1},~~ x \in (0,1). $$

Knowing that $\text{E}(X_n) = 1/2$ and that $\text{Var}(X_n) = 1/[4(2n+1)]$, prove that

$$ 2\sqrt{2n + 1}(X_n - \small{\frac{1}{2}}) \stackrel{D}{\longrightarrow} N(0,1). $$


Definition. A sequence of random variables $X_1, X_2, ...$, converges in distribution to a random variable X if

$$ \text{lim}_{n \to \infty} F_{X_n}(x) = F_X(x) $$

So we have to prove that

$$ \text{lim}_{n \to \infty} F_{Y_n}(x) = \int_{-\infty}^{x} \frac{1}{ \sqrt{2\pi}} e^{-y^2/2}dy $$

Where $Y_n = 2\sqrt{2n + 1}(X_n - \small{\frac{1}{2}}) $.


$$ \begin{align} P(Y_n \leq x) & = P(2\sqrt{2n + 1}(X_n - \small{\frac{1}{2}}) \leq x) \\ & = P(X_n - 1/2 \leq \frac{x}{2\sqrt{2n+1}} \\ & = P(X_n \leq \frac{x}{2\sqrt{2n+1}} + 1/2) \\ & = F_{X_n} \Bigl( \frac{x}{2\sqrt{2n+1}} + \frac{1}{2} \Bigr) \\ & = \frac{1}{B(n,n)}\int_{0}^{ \frac{x}{2\sqrt{2n+1}} + 1/2 } t^{n-1}(1 - t)^{n-1}dt \end{align} $$

We use Stirling's approximation to $\text{B}(n,n)$:

$$ B(a, b) \approx \sqrt{2\pi} \frac{a^{a - 1/2}b^{b - 1/2}}{(a + b)^{a + b - 1/2}} $$

So $\text{B}(n, n) \approx \frac{\sqrt{\pi}}{2^{2n - 1}} \frac{1}{\sqrt{n}} $, after simplification.

Substituting the Stirling approximation (we do this because it converges asymptotically and we're taking the limit), we obtain

$$ \frac{1}{\frac{\sqrt{\pi}}{2^{2n - 1}} \frac{1}{\sqrt{n}}}\int_{0}^{ \frac{x}{2\sqrt{2n+1}} + 1/2 } t^{n-1}(1 - t)^{n-1}dt. $$

So what's left to do is prove that

$$ \text{lim}_{n \to \infty} \frac{1}{\frac{\sqrt{\pi}}{2^{2n - 1}} \frac{1}{\sqrt{n}}}\int_{0}^{ \frac{x}{2\sqrt{2n+1}} + 1/2 } t^{n-1}(1 - t)^{n-1}dt = \int_{-\infty}^{x} \frac{1}{ \sqrt{2\pi}} e^{-y^2/2}dy. $$

I don't know how to do this last step, finishing the proof. I asked my professor for guidance on how to finish the last step. All he said was "apply the limit theorem to solve directly".

  • $\begingroup$ Try a change of variable like $y=\log t/(1-t)$ and the dominated convergence theorem. $\endgroup$
    – Xi'an
    Commented Sep 30, 2020 at 15:26
  • $\begingroup$ I'm not familiar with the dominated convergence theorem. It's above the level of my mathematical statistics class. $\endgroup$
    – Sigma
    Commented Sep 30, 2020 at 16:14
  • 2
    $\begingroup$ Then check this previous answer. $\endgroup$
    – Xi'an
    Commented Sep 30, 2020 at 18:18
  • 1
    $\begingroup$ Hint: re-express the integrand in terms of $s=(t-1/2)\sqrt{8(n-1)}$ and remember that for large $n,$ $(1-s^2/(2n))^n \approx \exp(-s^2/2).$ $\endgroup$
    – whuber
    Commented Oct 1, 2020 at 22:18
  • $\begingroup$ Cross-post: math.stackexchange.com/q/3845562/321264. $\endgroup$ Commented Oct 3, 2020 at 5:34

1 Answer 1


I was pondering how to formulate the simplest possible elementary solution to this problem and it occurred to me we can avoid any consideration of Beta functions (no Stirling's approximation needed; indeed, even information about the moments of Beta distributions is unnecessary). The result is extremely general and, I hope, interesting.

Here, for the record, is what I will show:

Let $f$ be a positive multiple of any probability density function that is bounded, unimodal, and twice differentiable in a neighborhood of its mode. Let the second derivative at the mode equal $-a$. Then any sequence of random variables $X_n$ with distribution functions proportional to $$t\to f^n\left(\frac{t}{\sqrt{an}}\right)$$ converges in distribution to the Standard Normal distribution.

Notation, assumptions, and preliminary simplifications

Permit me to use $n+1$ rather than $n$ as the index, so that $$f_n(t)\ \propto\ t^n(1-t)^n = (t(1-t))^n = f(t)^n$$ (for $0\le t\le 1$), thereby avoiding writing "$n-1$" too often. In the question $f(t) = t(1-t)$ for $0\le t \le 1$ (and otherwise equals zero). However, this formula is a distracting, irrelevant detail.

Here's all we need to assume about $f:$

  1. There is a constant $c$ for which $cf$ is a probability density function. This means it is defined almost everywhere on all real numbers, integrable, with unit integral. Obviously $c^{-1}=\int f(t)\,\mathrm{d}t.$
  2. $f$ is bounded and unimodal. That is, $f$ has a unique finite maximum value.
  3. $f$ has a second derivative in a neighborhood of its mode.

These are clearly true of the $f$ in the question.

Letting $\mu$ be the mode, we may with no loss of generality analyze the function $t\to f(t-\mu),$ which has all the properties assumed of $f$ and whose mode is $0.$


$$f(t) = 1 - \frac{a}{2}\left(1 + g(t)\right)t^2,$$

the third assumption implies

$$\lim_{t\to 0} g(t) = 0$$

and there is some positive number $\epsilon$ for which whenever $|t|\le \epsilon,$ $g(t) \ge 0.$ Moreover, since $0$ is the unique mode, $a$ must be positive.

Without any loss of generality, replace $f$ by the function $t\to f(t)/f(0),$ making the largest value of $f$ exactly $1,$ attained at its mode $0.$

We are going to consider a sequence of probability density functions determined by powers of $f.$ First we need to normalize those powers, so let

$$c_n^{-1} = \int f^n (t)\,\mathrm{d}t.$$

This is always possible because

$$\int f^n(t)\,\mathrm{d}t \le \sup(f)\int f^{n-1}(t)\,\mathrm{d}t\ = \int f^{n-1}(t)\,\mathrm{d}t$$

shows recursively that the integrals of $f^n$ cannot increase and therefore are bounded.

A final preliminary manipulation is to standardize $f^n:$ we are going to analyze the sequence

$$f_n(t) = f\left(\frac{t}{\sqrt{an}}\right)^n.$$

The next few steps will show why this is effective at producing just the right cancellation of factors in the calculation. First, though, let's look at an example.

Figure: plots of f and f_n for n=1,2,3,4,5

As $n$ grows, $f$ spreads out from its mode, pushing all "satellites" out and dampening them, leaving a graph that rapidly approaches a multiple of a Normal pdf. (The plot of $f$ in the upper left corner has not yet been rescaled to a height of $1$ at its mode. The next plot of $f_1$ has been so scaled and is plotted on an $x$ axis expanded by a factor of $\sqrt{a}$ to show detail.)


Let $t$ be any real number. Once $n$ exceeds $N(t)=t^2 / (a\epsilon^2),$ $|t|/\sqrt{an}\le \epsilon$ puts this value into the neighborhood where $f$ behaves nicely. From now on take $n\gt N(t).$

We are going to estimate the value of $f^n(t)$ by using logarithms. This is the crux of the matter and it is where all the algebra is done. Fortunately, it's easy:

$$\begin{aligned} \log\left(f^n(t)\right) &= n \log(f(t)) \\ &= n \log f\left(\frac{t}{\sqrt{an}}\right) \\ &= n \log \left(1 - \frac{a}{2}\left(\frac{t}{\sqrt{an}}\right)^2\left(1 + g\left(\frac{t}{\sqrt{an}}\right) \right) \right) \\ &= n\log\left(1 - \frac{t^2}{2n}\left(1 + g\left(\frac{t}{\sqrt{an}}\right)\right)\right) \end{aligned}$$

Because $g$ shrinks to $0$ for small arguments, a sufficiently large value of $n$ assures that the argument of the logarithm in that last expression is of the form $1-u$ for an arbitrarily small value of $u.$ This permits us to approximate the logarithm using Taylor's Theorem (with remainder), giving

$$\begin{aligned} n\log\left(f^n(t)\right) &= -\frac{t^2}{2}\left(1 + g\left(\frac{t}{\sqrt{an}}\right)\right) + \frac{R}{n}\, \tilde{t}^4 \left(1 + g\left(\frac{\tilde t}{\sqrt{an}}\right)\right)^2 \end{aligned}$$

where $0\le |\tilde{t}| \le |t|$ and $R$ is some number (related to the remainder term in the Taylor expansion). Taking the limit as $n\to\infty$ makes the remainder and all the $g()$ terms disappear, leaving

$$\lim_{n\to\infty} \log\left(f(t)^n\right) = -\frac{t^2}{2},$$


$$\lim_{n\to\infty} f(t)^n = \exp\left(-\frac{t^2}{2}\right).$$

It follows (requiring only an intuitive, elementary proof) that the sequence of normalizing constants $c_n$ must approach the normalizing constant for the right hand side--which exists and, as is well known, equals $\sqrt{2\pi}.$ Consequently

$$\lim_{n\to\infty} f_n(t) = \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{t^2}{2}\right),$$

which is the standard Normal density $\phi.$


When $X_n$ is a sequence of random variables having densities $f_n,$ for every number $t$ the limit of their densities is $\phi(t).$ It follows easily that the limit of their distribution functions is $\Phi,$ the standard Normal distribution.

In the case of the Beta$(n,n)$ distributions, $f(t)=t(1-t)$ has a unique mode at $\mu=1/2,$ where it can be expressed (up to a constant multiple) as

$$4f(t) = 1 - \frac{8}{2}(t-1/2)^2.$$

From this we can read off the value $a=8.$ Following our preliminary simplifications, this says the distribution of $\sqrt{an}(X_n - \mu) = \sqrt{8n}(X_n-\mu)$ converges to the standard Normal distribution. Because asymptotically the ratio of $\sqrt{8n}$ and $2\sqrt{2n+1}$ becomes unity, the statement in the original question is proven.

  • $\begingroup$ Fantastic answer my friend, but it’s not elementary at all! My proof is simpler than this. $\endgroup$
    – Sigma
    Commented Oct 2, 2020 at 19:13
  • 5
    $\begingroup$ "Elementary," in mathematics, does not mean "simple." My approach is elementary in the sense of using basic definitions and a minimum amount of information. In that sense it is far more elementary than any approach requiring Stirling's approximation and properties of Beta distributions. Often, elementary approaches are insightful: they expose the crux of the matter. Regardless, the real advantage is the scope of the result. For instance, without any additional work you can immediately draw conclusions about many sequences of Beta variables and other distributions. $\endgroup$
    – whuber
    Commented Oct 2, 2020 at 19:30
  • 3
    $\begingroup$ I should also add that you haven't yet exhibited a proof ;-). $\endgroup$
    – whuber
    Commented Oct 2, 2020 at 19:30
  • $\begingroup$ My proof is simple: standardize the Beta(n, n) rv, then it is enough to proof that the limit of this standardized Beta(n, n) density = density of a standard normal. After you’ve done that, invoke Scheffé’s theorem to prove convergence in distribution. $\endgroup$
    – Sigma
    Commented Oct 2, 2020 at 19:35
  • $\begingroup$ Okay, that's good. But it should now be clear you don't need Scheffé’s Theorem :-). The truly interesting thing revealed by the standardization (in both our approaches) is that asymptotically the result depends only on the germ of $f$ at its mode--not on any other behavior. $\endgroup$
    – whuber
    Commented Oct 2, 2020 at 19:38

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