# Rotating a PDF that is log-concave and symmetric

Would you mind checking out this proof?

Consider a symmetric, log-concave pdf $f(\cdot)$ with unbounded support and mean $\mu$, and define an associate pdf: $$g(w;\gamma)=\frac{1}{\gamma}f\left( \mu+\frac{w-\mu}{\gamma }\right)$$

I want to show that there exists an $\epsilon>0$ such that $\partial g(w)/ \partial \gamma<0$ for all $w\in(\mu-\epsilon,\mu+\epsilon)$, and $\partial g(w)/ \partial \gamma>0$ outside this interval.

To prove this I have done the following:

$$g'(w)=\frac{-1}{\gamma^2}\left[f\left( \mu+\frac{w-\mu}{\gamma }\right)+\frac{w-\mu}{\gamma }{f'\left( \mu+\frac{w-\mu}{\gamma }\right)}\right]$$

First, I have to prove that $g(\mu)<0$. Which amounts to proving that: $$f(\mu)>-wf'(\mu)/\gamma$$ Here I thought that since $f$ is symmetric then, $f'(\mu)=0$. However, I am not sure if this is alright. After knowing that, I would proceed by arguing that it must me the case that $g'(\mu+\epsilon)$ changes sign for some $\epsilon$. To see why argue by contradiction: if $g'(\mu+\epsilon)<0$ for all $\epsilon>0$ then:

$$f(\mu+\epsilon/\gamma)>-wf'(\mu+\epsilon/\gamma)/\gamma$$

which in turn implies that:

Edit: analysis after here is wrong, since it relied on a misscalculation. Thaks to the commentors for pointing it out.

$$f'(\mu+\epsilon/\gamma)/f''(\mu+\epsilon/\gamma)>f(\mu+\epsilon/\gamma)/f'(\mu+\epsilon/\gamma)>1/\gamma>0$$

where the first inequality is just a consequence of the log-concavity ($log f$ is concave). Note that since $f$ is a pdf, then it must be positive, which then implies that $f'$ and $f''$ must be positive too for all $\epsilon>0$. But then, this would imply that $f$ should be convex, which is a contradiction with the fact that it is log-concave (is this ok?). And therefore ot must be the case that there exists an $\epsilon_0>0$ such that $g'(\mu+\epsilon)$ for any $\epsilon>\epsilon_0$. By symmetry of $f$, these arguments should also work with $-\epsilon$.

What to you think about this proof? I am not sure about how I used log-concavity here. Also I am aware that I haven't proven that this treshold $\epsilon_0$ is unique. What would you suggest?

• 1. "Which amounts to proving that..." - no, you forgot to change the sign. 2. Further, you divide by $f'$, which is negative. 3. Hint Wlog you can assume that $\mu = 0$. Then, for fixed $x\neq 0$ (wlog positive), you are interested in the sign of $a(\omega) = 1 + x f'(\omega x)/f(\omega x)$. By log-concavity, this expression is non-increasing. However, it cannot be always negative or always positive - prove this, using that the total mass of $g$ is 1. (Btw $f$ should be strictly log-concave, otherwise it is possible that $g'_\gamma(\cdot,\gamma)$ is zero on some interval.) – zhoraster Nov 12 '17 at 9:07
• Thanks! I corrected the typo in (1); for your hint in (3), how do you argue that $1+f'(\omega)/(\omega)=1+xf'(x\omega)/f(x \omega)$? – Weierstraß Ramirez Nov 13 '17 at 15:33
• 1. You corrected only the first instance, further discussion is still incorrect. 2. $x = 1/\gamma$. – zhoraster Nov 13 '17 at 16:17
• Yeah, I realised that there was also a mistake in the derivative. It turns out that you can actually just study the behaviour of $a(x)=1+xf'(\mu+x)/f(\mu+x)$. I was thinking about arguing by contradiction: take $x>0$ and argue that $a(x')>0$ for all $x'>x$, then $f'(\mu+x')/f(\mu+x')<-1/x'$. But the we can consider $f'(\mu-x')/f(\mu-x')<f'(\mu+x')/f(\mu+x')$. If $a(-x)>0$ then $-1/x'<f'(\mu+x')/f(\mu+x')<f'(\mu-x')/f(\mu-x')<1/x'$. Note that this cannot be the case for $x'\to \infty$ (unbounded support). Then $a(-x)<0$, and I should also obtain some inconsistency here. What do you think? – Weierstraß Ramirez Nov 15 '17 at 9:46

1. Note that $g'(\mu)<0$. Since $g'(\mu)=-f(\mu)/\gamma^2$ and $f(\cdot)>0$ since it is a density;
2.Note that it must the case that $g(\mu+\epsilon)>0$ for and $\epsilon$ that is big enough. To see this, consider the function $\alpha(x)=f(\mu+x)+xf'(\mu+x)$ where $x$ can be defined as $x=(w-\mu)/\gamma$. Note that, from the last step, $\alpha(0)=f(\mu)>0$. Without loss of generality, take $x>0$ and argue by contradiction that $\alpha(x)>0$ for all $x\in \mathbb{R}^+$. Then it should be the case that: $$\frac{1}{x}>-\frac{f'(\mu+x)}{f(\mu+x)}, \quad \text{for all } x>0$$ now consider $\alpha(-x)$. If $\alpha(-x)>0$, then we would have that: $$-\frac{1}{x}<-\frac{f'(\mu-x)}{f(\mu-x)}, \quad \text{for all } x>0$$ and, using the fact that $f$ is log-concave and hence $-f'/f$ is increasing, we have that: \begin{align*} \frac{1}{x}>-\frac{f'(\mu+x)}{f(\mu+x)}>-\frac{f'(\mu-x)}{f(\mu-x)}>-\frac{1}{x} \end{align*} it is clear that this inequality cannot be met for $x\to \infty$ ($f$ has unbounded support). Therefore it must be the case that $\alpha(-x)<0$. Using the fact that the symmetry of $f$ implies$^*$ that $f'(\mu+x)/f(\mu+x)=-f'(\mu-x)/f(\mu-x)$, his implies that: \begin{align*} -\frac{1}{x}>-\frac{f'(\mu-x)}{f(\mu-x)}=\frac{f'(\mu+x)}{f(\mu+x)}{>}-\frac{1}{x} \end{align*} which is of course a contradiction. Therefore, it must be the case that there exists an $\epsilon>0$ so that $\alpha(\mu+x)<0$ for any $x>\epsilon$.
$^*$We have that $f(\mu+x)=f(\mu-x)$, which implies that $f'(\mu+x)=-f'(\mu-x)$, which yields the desired condition.
• Too many words, but overall seems correct. I thought of another conclusion of the argument: if $dg(\omega,\gamma)/d\gamma$ had the same sign for all $\omega$, then $\int_{\mathbb R} g(\omega,\gamma) d\omega =1$ would increase or decrease, which is absurd. – zhoraster Nov 15 '17 at 16:36