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
    $\begingroup$ 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.) $\endgroup$ – zhoraster Nov 12 '17 at 9:07
  • $\begingroup$ Thanks! I corrected the typo in (1); for your hint in (3), how do you argue that $1+f'(\omega)/(\omega)=1+xf'(x\omega)/f(x \omega)$? $\endgroup$ – Weierstraß Ramirez Nov 13 '17 at 15:33
  • $\begingroup$ 1. You corrected only the first instance, further discussion is still incorrect. 2. $x = 1/\gamma$. $\endgroup$ – zhoraster Nov 13 '17 at 16:17
  • $\begingroup$ 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? $\endgroup$ – Weierstraß Ramirez Nov 15 '17 at 9:46

Thanks to @zhoraster's comments, I arrived to the following proof.

I proceed in two steps:

  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.

  • 1
    $\begingroup$ 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. $\endgroup$ – zhoraster Nov 15 '17 at 16:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.