I would really appreciate some help with showing that an inequality holds for a particular class of probability distributions. In what follows, I will describe you the problem and my solution so far:

There is a continuous random variable $\theta$, which is distributed according to some CDF $F(\cdot)$ with strictly positive density $f(\cdot)$ on a compact intervall $[\underline{\theta},\overline{\theta}]$, where $\underline{\theta} \ge 0 $.

Furthermore, consider the following assumptions:

1.) $f(\cdot)$ is log concave (Therefore $F(\cdot)$ is log concave)

2.) $\underline \theta \cdot f(\underline \theta) \le 1$

Now to the problem:

Let $\tilde\theta \in (\underline{\theta},\overline{\theta})$ be the solution to $(1 - F(\tilde\theta))^2 = F(\tilde\theta)$ such that $\tilde\theta = F^{-1}(\frac{1}{2}\cdot(3-\sqrt{5}))$. Therefore $\tilde\theta$ is the $\frac{1}{2}\cdot(3-\sqrt{5})$-quantil of the distribution.

I want to show that the following inequality holds: \begin{equation} \tilde\theta - \frac{1-2F(\tilde\theta)}{f(\tilde\theta)} \ge0 \end{equation} which we can simply rewrite as

\begin{equation} \tilde\theta \cdot f(\tilde\theta) + F(\tilde\theta) \ge 1 - F(\tilde\theta) \end{equation}

Let us first consider all distributions such that $\tilde\theta \le \theta^m$, where $\theta^m$ is the mode of the distribution. Due to the log concavity of the distribution and since $\tilde\theta \le \theta^m$, it follows that $f'(\theta) \ge 0 $ for all $\theta \le \tilde\theta$ and hence $F(\theta)$ is convex on $[\underline\theta, \tilde \theta]$. Due to the convexity it follows that $\tilde\theta \cdot f(\tilde\theta) \ge F(\tilde\theta) $ and hence our inequality is fulfilled with strict inequality since \begin{equation} \tilde\theta \cdot f(\tilde\theta) + F(\tilde\theta) \ge 2\cdot F(\tilde\theta) > 1 - F(\tilde\theta) \end{equation} $\square$

The problem arises, once I consider the remaining distributions where $\tilde \theta > \theta^m$. Those are right skewed distributions with a probability mass concentration on the lower boundary of the support. So far I was not able to prove that the inequality holds for this case. However, I also did not find any counter example when testing the inequality with particular distributions. For instance, I considered a Weibull distribution with parameter restrictions such that the density remains log concave. The inequality holds for all possible parameter values. So even when in some sense considering extremely right skewed distributions such that the density is still log concave, this inequality holds. Once I use a Pareto Distribution, which is not log concave, I find parameter values such that the inequality is not fullfiled.

My intuition tells me, that I somehow need to use the definition of log concavity to show that the inequality holds for the case when $\tilde \theta > \theta^m$. But so far I failed, miserably...

I really hope that someone can help me with this, at least with some ideas on how to approach the proof, since it seems that I am too dumb to do it by myself :(

Thanks! Nima

Edit: Still haven't solved it :( anybody got an idea?

Second Edit: Im still searching for any suggestions :)


Am I wrong or does it require only the transformation of inequality? The only assumption that we have to ensure is that $f(\bar{\theta}) > 0$. \begin{equation} \bar{\theta} \cdot f(\bar{\theta}) + F(\bar{\theta}) \geq 1- F(\bar{\theta}) \leftrightarrow \bar{\theta} \cdot f(\bar{\theta}) \geq 1- 2F(\bar{\theta}) \leftrightarrow \bar{\theta} \geq \frac{1- 2F(\bar{\theta}) }{f(\bar{\theta}) } \leftrightarrow \bar{\theta} - \frac{1- 2F(\bar{\theta}) }{f(\bar{\theta}) } \geq 0 \end{equation}

  • $\begingroup$ Hi, thanks for your comment! What you wrote will only ensure that by Bolzanos Theorem there actually exists a zero of the function $h(\theta) = \theta - \frac{1-2\cdot F(\theta)}{f(\theta)}$ in $[\underline{\theta}, \overline{\theta}]$, since the function is continuous, strictly increasing due to assumption 1 and negative for $\underline{\theta}$ due to assumption 2! What I want to show is that the function is positive at a particular point, namely $h(\tilde{\theta}) \ge 0$, where the point $\tilde{\theta}$ is defined above. $\endgroup$
    – P3rs3rk3r
    Mar 18 '19 at 9:20

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