This is a research question. Consider an univariate non-negative random variable $q$. I intend to have a desired tail probability, say $\mathcal{A}$. If I don't know the distribution of $q$, but I know only the mean and covariance estimated from collected data, I could use the chebyshev bound to get the required threshold, say $\alpha_{2}$ (because we used two moments) for a desired tail probability $\mathcal{A}$. Here, $\alpha_{2}$ will be very conservative as it has to agree for worst case distribution of $q$ with the given two moments. That is, \begin{align*} \sup_{P_q} P_q(q > \alpha_{2}) \leq \mathcal{A} \end{align*} Figure with thresholds with same tail Probability Here is my question - Higher order moments like skewness, kurtosis, etc (estimated from data again) reveal more information about the true unknown distribution of $q$. In that case, I can find another threshold called $\alpha_{k}$ using first $k$ moments such that the $\sup_{P_q} P_q(q > \alpha_{k}) \leq \mathcal{A}$. My intuition is that for a same tail probability $\mathcal{A}$, the threshold should get tightened using higher order moments. That is, \begin{equation} \forall k > 2, \alpha_{k} \leq \alpha_{2} \qquad \qquad (1) \end{equation} However, I am not sure how to prove this. I tried using Chebyshev bound on higher order moments to prove (1) namely \begin{align*} P_q \left[|q - \mathbb{E}[q]| \geq \alpha_k \right] \leq \frac{\mathbb{E}[q - \mathbb{E}[q]]^{k}}{\alpha^k_k} \end{align*} but it trivially fails if $q$ gets realized between (0,1) (like an exponential distribution with parameter >3) because in that case $\mathbb{E}[q]^{k} \leq \mathbb{E}[q]^{2}$. Is my intuition wrong ? Please help me to prove (1).
$\begingroup$
$\endgroup$
4
-
$\begingroup$ I'm having a hard time understanding what's at issue here. Any distribution agreeing with the first $k$ moments for $k>2$ automatically agrees with the first $2$ moments. So the "worst-case" distribution agreeing with the first $k$ moments can't be worse than the "worst-case" agreeing with the first $2$ only. As $k$ increases, you're maximising over a smaller set, so the obtained maximum decreases. $\endgroup$– James MartinCommented Dec 11, 2020 at 17:26
-
$\begingroup$ I understand your point with the argument using sup definition on a smaller subset of distributions. Also, by "obtained maximum decreases", do you refer to the threshold $\alpha_{k}$ decreasing for a given same tail probability? Does that mean that my intuition is correct ? Sorry for troubling you. I am a newbie to this forum. $\endgroup$– Venkatraman RenganathanCommented Dec 11, 2020 at 18:18
-
$\begingroup$ To be precise I should have said "does not increase" rather than "decreases". And then indeed you get that $\alpha_k$ is non-decreasing in $k$. To answer your specific question, let $\mathcal{M}_k$ be the set of distributions matching the first $k$ moments. Then $\mathcal{M}_k \subseteq \mathcal{M}_2$, and so $\sup_{P\in\mathcal{M}_k} P(q>\alpha_2) \leq \sup_{P\in\mathcal{M}_2} P(q>\alpha_2) \leq \mathcal{A}$. That gives you that the threshold $\alpha_k$ is no higher than $\alpha_2$. $\endgroup$– James MartinCommented Dec 11, 2020 at 19:39
-
$\begingroup$ Thank you so much. That explanation was crystal clear. $\endgroup$– Venkatraman RenganathanCommented Dec 11, 2020 at 19:43
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Very general results here are due to Mallows, who obtained exact upper and lower bounds on the cumulative distribution function $F$ of a random variable $X$ given a finite number of moments of $X$ and also, optionally, information about the monotonicity pattern(s) of $F$ and some of its derivatives.
answered Dec 11, 2020 at 17:24