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I am interested in finding bounds on cumulants in terms of moments. For example, this paper alludes to the bound \begin{align} |\kappa_n|\le n^n E[|X-E[X]|^n] \end{align} where $\kappa_n$ is the $n$-th cumulant. However, due to the language barrier, tracking down the proof of this is difficult.

I would like to see the proof of this bound or something similar to this.

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Let $k_n:=\kappa_n$, $a_n:=E(X-EX)^n$, $b_n:=E|X-EX|^n$, so that $|a_n|\le b_n$. We have to show that $$|k_n|\le n^n b_n$$ for natural $n$. For $n=1,2$ this is obvious. The key is the recursion $$k_n=a_n-\sum_{m=1}^{n-1}\binom{n-1}{m-1}k_m a_{n-m}$$ at the end of this section , which implies $$|k_n|\le b_n+\sum_{m=1}^{n-1}\binom{n-1}{m-1}|k_m| b_{n-m}.$$ This allows us to use the induction on $n$, which implies $$|k_n|\le l_n b_n,$$ where $$l_n:=1+\sum_{m=1}^{n-1}\binom{n-1}{m-1}m^m;$$ here we used the log-convexity of $b_j$ in $j$ together with the fact $b_0=1$ (or, equivalently, Hölder's inequality), which implies $b_m b_{n-m}\le b_n$.

It remains to show that $l_n\le n^n$ for $n\ge2$. This is trivial for $n=2$. For $n\ge3$, proceed again by induction: \begin{align*} l_n&=1+\sum_{m=1}^{n-1}\binom{n-2}{m-1}m^m+\sum_{m=1}^{n-1}\binom{n-2}{m-2}m^m \\ &=1+\sum_{m=1}^{n-2}\binom{n-2}{m-1}m^m+(n-1)^{n-1}+\sum_{m=2}^{n-1}\binom{n-2}{m-2}m^m \\ &=l_{n-1}+(n-1)^{n-1}+\sum_{j=1}^{n-2}\binom{n-2}{j-1}(j+1)^{j+1} \\ &\le l_{n-1}+(n-1)^{n-1}+\sum_{j=1}^{n-2}\binom{n-2}{j-1}j^j c_n \\ &=l_{n-1}+(n-1)^{n-1}+(l_{n-1}-1) c_n \\ &\le (n-1)^{n-1}+(n-1)^{n-1}+(n-1)^{n-1}c_n \\ &=(n-1)^{n-1}\Big(2+\frac{(n-1)^{n-1}}{(n-2)^{n-2}}\Big)=:r_n \end{align*} where $$c_n:=\max_{1\le j\le n-2}\frac{(j+1)^{j+1}}{j^j}=\frac{(n-1)^{n-1}}{(n-2)^{n-2}},$$ because $$\frac{(j+1)^{j+1}}{j^j}=\Big(1+\frac1j\Big)^j(j+1)$$ is increasing in $j\ge1$.

It remains to check that $r_n\le n^n$ for all real $n>2$, which, by substitution $n=1+\frac1t$, can be rewritten as \begin{equation} f_1(t)f_2(t)\le1, \tag{1} \end{equation} where \begin{equation*} f_1(t):=e \left(\frac{1}{t+1}\right)^{\frac{1}{t}+1},\quad f_2(t):=\frac{\left(\left(\frac{1}{t}\right)^{\frac{1}{t}} \left(\frac{1}{t}-1\right)^{\frac{t-1}{t}}+2\right) t}{e}; \end{equation*} everywhere here, $t\in(0,1]$.

To prove (1), it is enough to prove \begin{equation} f_1\le g_1 \tag{2}, \end{equation} \begin{equation} f_2\le g_2 \tag{3}, \end{equation} \begin{equation} g_1g_2\le1 \tag{4} \end{equation} on $(0,1]$, where \begin{equation*} g_1(t):=\frac{7 t^2}{24}-\frac{t}{2}+1,\quad g_2(t):=\left(\frac{2}{e}-\frac{1}{2}\right) t+1. \end{equation*}

To prove (2), consider \begin{equation*} d_1:=\ln f_1-\ln g_1 \end{equation*} and then $d_{11}(t):=t^2d_1'(t)$ and $d_{11}'$, which latter is a rational function, which is easily seen to be $<0$. So, $d_{11}$ decreases. Also, $d_{11}(0+)=0$. So, $d_{11}<0$ (on $(0,1]$) and hence $d_1$ decreases. Also, $d_1(0+)=0$. So, $d_1<0$, which yields (2).

To prove (3), rewrite it as
\begin{equation*} d_2(t):=\ln \left(\left(\frac{1}{t}-1\right)^{\frac{t-1}{t}} \left(\frac{1}{t}\right)^{\frac{1}{t}}\right)-\ln \left(\frac{e (2-t)}{2 t}\right)\le0 \end{equation*} (for $t\in(0,1]$). Consider then $d_{21}(t):=t^2d_2'(t)$ and $d_{21}'$, which latter is a rational function, which is easily seen to be $<0$ on $(0,1)$. So, $d_{21}$ decreases. Also, $d_{21}(0+)=0$. So, $d_{21}<0$ (on $(0,1]$) and hence $d_2$ decreases. Also, $d_2(0+)=0$. So, $d_2<0$, which yields (3).

Finally, (4) is elementary, since $g_1g_2$ is a polynomial (of degree $3$). $\Box$

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  • $\begingroup$ Thanks. Very nice. Did you see this proof before? $\endgroup$ – Boby Aug 10 '20 at 12:58
  • $\begingroup$ @Boby : No, I have not seen a proof of this. I tried to find one, but was unable to. Do you know where the original proof of this can be found, in whatever language? $\endgroup$ – Iosif Pinelis Aug 10 '20 at 16:14
  • $\begingroup$ I found a remark about this in the paper the I linked. They refer to the paper Yu. V. Prokhorov and Yu. A. Rozanov, Probability Theory [in Russian], Nauka, Moscow (1973) $\endgroup$ – Boby Aug 10 '20 at 16:20
  • $\begingroup$ Yu. V. Prokhorov and Yu. A. Rozanov, Probability Theory [in Russian], Nauka, Moscow (1973) is a book. There is no proof of this bound or reference to a proof there. $\endgroup$ – Iosif Pinelis Aug 10 '20 at 17:41
  • $\begingroup$ In the paper, the author say this ``..one can only find implicit inequalities (see, for example, [8]):" where refence [8] is the book. $\endgroup$ – Boby Aug 10 '20 at 18:12

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