Any help on proving a bound such that:
$\mathbb E(X^n)-\mathbb E(X)^n \le c; \quad n \in \mathbb Z_+.$
$X$ is the R.V. in the range of $[0,1]$
I am looking for something like Popoviciu's inequality but for n-th moments..
https://en.wikipedia.org/wiki/Popoviciu%27s_inequality_on_variances
Thanks!
On
Let $X: \Omega\rightarrow [0,1]$ be the random variable, and $P$ the probability measure on $\Omega$. Then the expected value is integration w.r.t. $P$.
So you are looking for a global upper bound for $\int\limits_{\Omega} X^n \,dP - (\int\limits_{\Omega} X \,dP)^n$. By approximating $X$ with step functions, we can discretize the problem.
So it is enough to answer the following question: Given $x_1, \ldots, x_r\in [0,1]$, and a probability distribution $p_1, \ldots, p_k$, what is the maximum of $(p_1x_1^n+\cdots p_rx_r^n) - (p_1x_1+\cdots p_rx_r)^n$?
This is a "Jensen-type" problem: the function is $f: [0,1]\rightarrow [0,1]$, $x\rightarrow x^n$. Given the values $x_1, \ldots, x_r\in [0,1]$, the difference is a convex combination of $f(x_1), \ldots, f(x_r)$ and the value of $f$ at the convex combination (with the same coefficients) of $x_1, \ldots, x_r$. Both things are points in the unit square, so the difference is at most $1$. It can be arbitrarily close to $1$: let $r=2$, $x_1=0, x_2=1$, $p_1=\varepsilon, p_2=1-\varepsilon$. Then $E(X^n)=1-\varepsilon$, and $(EX)^n= (1-\varepsilon)^n\rightarrow 0$ if $\varepsilon>0$ is fixed and $n\rightarrow \infty$.
Since $0\leq X\leq 1$, we have $$\mathbb{E}\big[X(1-X^{n-1})\big]\geq 0\,.$$ Thus, $$\mathbb{E}[X^n]\leq \mathbb{E}[X]=:\mu\,.$$ That is, $$\mathbb{E}[X^n]-\big(\mathbb{E}[X]\big)^n\leq \mu-\mu^n=\mu(1-\mu^{n-1})\,.$$ By the Weighted AM-GM Inequality, we see that $$\begin{align}\mu(1-\mu^{n-1}) &=\frac{1}{(n-1)^{\frac{1}{n-1}}}\left((n-1)\mu^{n-1}\right)^{\frac{1}{n-1}}\left(1-\mu^{n-1}\right) \\ &\leq \frac{1}{(n-1)^{\frac{1}{n-1}}}\left(\frac{\mu^{n-1}+(1-\mu^{n-1})}{\frac{1}{n-1}+1}\right)^{\frac{1}{n-1}+1} \\ &=\frac{1}{(n-1)^{\frac1{n-1}}}\left(\frac{n-1}{n}\right)^{\frac{n}{n-1}}=\frac{n-1}{n^{\frac{n}{n-1}}}\,,\end{align}$$ for every real number $n>1$. That is, $$\mathbb{E}[X^n]-\big(\mathbb{E}[X]\big)^n\leq \frac{n-1}{n^{\frac{n}{n-1}}}\,.$$ The equality holds if and only if $X\in\{0,1\}$ almost surely and the probability that $X=1$ is $\dfrac{1}{n^{\frac{1}{n-1}}}$. Thus, the optimal value of $c$ for $n=1$ is $c_1=0$, and the optimal value of $c$ for $n>1$ is $$c_n:=\frac{n-1}{n^{\frac{n}{n-1}}}\,.$$ It is not difficult to see that $c_n$ is an increasing function in $n\in\mathbb{R}_{\geq1}$.
Note that, as $n\to\infty$, $c_n\to 1$ from below. Therefore, the optimal global constant $c$ that works for every $n\geq1$ is $c_{\infty}:=\lim\limits_{n\to\infty}\,c_n=1$.
Here is a solution if $n$ is a positive integer and if $a\leq X\leq b$ for some real numbers $a$ and $b$ such that $0\leq a<b$. For $n=1$, the best constant $c$ is obviously $c_1:=0$. From now on, we suppose that $n>1$.
First note that, for $j=1,2,\ldots,n$, $$\mathbb{E}\big[(X-a)((b^{j-1}-X^{j-1})\big]\geq 0\,.$$ Thus, $$\mathbb{E}[X^j]-a\,\mathbb{E}[X^{j-1}]\leq b^{j-1}\,\mathbb{E}[X]-ab^{j-1}\,.$$ Consequently, with the convention that $0^0:=1$, we get $$\mathbb{E}[X^n]-a^n=\sum_{j=1}^n\,a^{n-j}\,\Big(\mathbb{E}[X^j]-a\,\mathbb{E}[X^{j-1}]\Big)\leq \sum_{j=1}^n\,a^{n-j}\,\Big( b^{j-1}\,\mu-ab^{j-1}\Big)\,,$$ where $\mu:=\mathbb{E}[X]$. That is, $$\mathbb{E}[X^n]\leq a^n+a^{n-1}\left(\frac{\left(\frac{b}{a}\right)^{n}-1}{\frac{b}{a}-1}\right)(\mu-a)=a\,\left(a^{n-1}-\frac{b^n-a^n}{b-a}\right)+\left(\frac{b^n-a^n}{b-a}\right)\mu\,.$$ Thus, $$f_n(X):=\mathbb{E}[X^n]-\big(\mathbb{E}[X]\big)^n\leq- ab\,\left(\frac{b^{n-1}-a^{n-1}}{b-a}\right)+\left(\frac{b^n-a^n}{b-a}\right)\,\mu-\mu^n\,.$$
We perform a similar trick. First, write $t_n:=\dfrac{b^n-a^n}{b-a}\geq b^{n-1}\geq \mu^{n-1}$. The inequality above can be written as $$f_n(X)\leq -s_n+\mu(t_n-\mu^{n-1})\,.$$ As before, from the Weighted AM-GM Inequality, we get $$\begin{align} f_n(X)&\leq -s_n+\frac{1}{(n-1)^{\frac{1}{n-1}}}\,\big((n-1)\mu^{n-1}\big)^{\frac{1}{n-1}}(t_n-\mu^{n-1}) \\ &\leq -s_n+\frac{1}{(n-1)^{\frac{1}{n-1}}}\,\left(\frac{\mu^{n-1}+(t_n-\mu^{n-1})}{\frac{1}{n-1}+1}\right)^{\frac{1}{n-1}+1} \\ &=-s_n+\left(\frac{n-1}{n^{\frac{n}{n-1}}}\right)\,t_n^{\frac{n}{n-1}}\,. \end{align}$$ Thus, we have $$\mathbb{E}[X^n]-\big(\mathbb{E}[X]\big)^n\leq \left(\frac{n-1}{n^{\frac{n}{n-1}}}\right)\left(\frac{b^n-a^n}{b-a}\right)^{\frac{n}{n-1}}-ab\left(\frac{b^{n-1}-a^{n-1}}{b-a}\right)\,.$$ The equality holds if and only if $X\in\{a,b\}$ almost surely and the probability that $X=b$ equals $$p_n:=\frac{1}{b-a}\,\left(\frac{1}{n^{\frac{1}{n-1}}}\,\left(\frac{b^{n}-a^{n}}{b-a}\right)^{\frac{1}{n-1}}-a\right)\,.$$ Observe that $p_n> 0$ always as $$(b-a)\,p_n=\left(\frac{b^{n-1}+b^{n-2}a+\ldots+ba^{n-2}+a^{n-1}}{n}\right)^{\frac{1}{n-1}}-a>\big(a^{n-1}\big)^{\frac{1}{n-1}}-a=0\,.$$ Thus, the best constant $c$ is $$c_n:=\left(\frac{n-1}{n^{\frac{n}{n-1}}}\right)\left(\frac{b^n-a^n}{b-a}\right)^{\frac{n}{n-1}}-ab\left(\frac{b^{n-1}-a^{n-1}}{b-a}\right)\,.$$ It would be interesting to find the best constant $c$ when $n>1$ is not an integer.