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<title>The Convexity Property Related to Beta Densities - You don't need to prove this</title>
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<h1>The Convexity Property Related to Beta Densities</h1>
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<span> Posted on Sat 01 December 2018 in <a href="https://newptcai.github.io/category/math.html" style="font-style: italic">math</a>
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<p><img alt="Density function of Beta[3,7]" src="https://newptcai.github.io/images/2018-12-01-beta/beta-density.png"></p>
<p><em>(This is a question asked by <a href="https://katalog.uu.se/profile/?id=N15-1140">Tilo Wiklund</a> recently on our group meeting.)</em></p>
<h2>The problem</h2>
<p>A <a href="https://en.wikipedia.org/wiki/Beta_distribution">Beta distribution</a> with parameters (<span class="math">\(\alpha\)</span>,<span class="math">\(\beta\)</span>) has the density function
</p>
<div class="math">$$
f(x;\alpha,\beta) = \frac{x^{\alpha -1} (1-x)^{\beta -1}}{B(\alpha ,\beta )}, \qquad x \in
[0,1]
$$</div>
<p>
As shown in the above picture, the density of this function, is concave, not convex. So you may think there is a mistake in
the title.</p>
<p>No, actually we are going to talk a related function <span class="math">\(r(x)\)</span> constructed from <span class="math">\(f(x)\)</span>. Let <span class="math">\(m\)</span> be the
point where <span class="math">\(f(x)\)</span> reaches maximum. With a bit calculus, we have
</p>
<div class="math">$$
m=\frac{\alpha -1}{\alpha +\beta -2}.
$$</div>
<p>
(So in the above example, we have <span class="math">\(m=1/4\)</span>, which can also be seen from the picture.)
Let <span class="math">\(f^{-1}_{1}\)</span> and <span class="math">\(f^{-1}_{2}\)</span> be the inverse of <span class="math">\(f(x)\)</span> for <span class="math">\(x \in (0,m)\)</span> and <span class="math">\(x \in (m,1)\)</span>
respectively. They would look like this
<img alt="The inverse of the density function of Beta[3,7]" src="https://newptcai.github.io/images/2018-12-01-beta/beta-density-inverse.png">
Then we define this function
</p>
<div class="math">$$
r(x)
=
\begin{array}{cc}
\left\{
\begin{array}{cc}
f^{-1}_{2} (f1(x)) & x \in (0,m) \\
f^{-1}_{1} (f1(x)) & x \in [m,1) \\
\end{array}
\right.
\\
\end{array}
$$</div>
<p>
What does this weird function do? It takes one point <span class="math">\(x \in [0,1]\)</span>, and gives you the point <span class="math">\(r(x)
\in [0,1]\)</span>, such that <span class="math">\(f(x)=f(r(x))\)</span> and <span class="math">\(x \ne r(x)\)</span>. In other words, it reflects the point <span class="math">\(x\)</span>
through the curve of beta density.</p>
<p><img alt="r[x] reflects x through the density function of Beta[3,7]" src="https://newptcai.github.io/images/2018-12-01-beta/beta-density-reflect.png"></p>
<p>So what is the conjecture? Tilo suggests that if <span class="math">\(\beta>\alpha>1\)</span>, then <span class="math">\(r(x)\)</span> must be convex. This
looks reasonable if you look at the picture.</p>
<p><img alt="r[x] for the density function of Beta[3,7]" src="https://newptcai.github.io/images/2018-12-01-beta/beta-density-r.png"></p>
<h2>How to prove it?</h2>
<p>This is not an easy question as it seems. The main difficulty is that we do not how to inverse
<span class="math">\(x^{\alpha-1}(1-x)^{\beta-1}\)</span>. So if we take second derivative of <span class="math">\(r(x)\)</span>, using the formula
</p>
<div class="math">$$
\frac{
\mathrm d
}
{
\mathrm d x
}
f^{-1}(x)
=
\frac{1}
{
f'(f^{-1}(x))
}
,
$$</div>
<p>
we get a bit chunk of mess.</p>
<p>Fortunately we live in 2018 and we have computers to help us. Yes, <span class="math">\(r''(x)\)</span> is very ugly. But with
the help of computer, it is not to difficult to see that when <span class="math">\(\beta>\alpha>1\)</span>, <span class="math">\(r''(x)>0\)</span> is
equivalent to the following, very simple inequality
</p>
<div class="math">$$
x + r(x) \ge 2 m.
$$</div>
<p>Look at the picture again and it won't take you long to convince yourself that this is true.
<img alt="r[x] reflects x through the density function of Beta[3,7]" src="https://newptcai.github.io/images/2018-12-01-beta/beta-density-reflect.png">
In this case <span class="math">\(m\)</span> at most <span class="math">\(1/2\)</span>. So when <span class="math">\(x=0\)</span>, the left hand side is <span class="math">\(1> 2m\)</span>. And when <span class="math">\(x=r(x)=m\)</span>,
we have equality.</p>
<p>Now note that for <span class="math">\(x \leq m\)</span>,
</p>
<div class="math">$$
x+r(x)\geq 2 m\Leftrightarrow {f_2^{-1}}(f(x))+x\geq 2 m\Leftrightarrow {f_2^{-1}}(f(x))\geq 2 m-x
$$</div>
<p><img alt="The inverse of the density function of Beta[3,7]" src="https://newptcai.github.io/images/2018-12-01-beta/beta-density-inverse.png"></p>
<p>Looking at the picture of <span class="math">\(f^{-1}(x)\)</span> again, we see that <span class="math">\({f_2^{-1}}\)</span> is decreasing.
Therefore, we only need to prove that
</p>
<div class="math">$$
f(x) \le f(2m - x), \qquad x \in (0,m).
$$</div>
<p>
The nice thing about this inequality is that all the inverse functions are gone!</p>
<p>As mathematicians often like say when they run out of time, "the rest is left as an exercise"! (Or if
you have Mathematica install, you can have a look of my <a href="https://newptcai.github.io/doc/beta-2.nb">somewhat messy code</a> and check the details.)</p>
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