This is a research question. Consider an univariate nonnegative 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).

$\begingroup$ 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 "worstcase" distribution agreeing with the first $k$ moments can't be worse than the "worstcase" agreeing with the first $2$ only. As $k$ increases, you're maximising over a smaller set, so the obtained maximum decreases. $\endgroup$– James MartinDec 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 RenganathanDec 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 nondecreasing 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 MartinDec 11, 2020 at 19:39

$\begingroup$ Thank you so much. That explanation was crystal clear. $\endgroup$– Venkatraman RenganathanDec 11, 2020 at 19:43
1 Answer
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.