# Bounds on cumulants in terms of moments

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.

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 $$$$f_1(t)f_2(t)\le1, \tag{1}$$$$ where $$\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 $$$$f_1\le g_1 \tag{2},$$$$ $$$$f_2\le g_2 \tag{3},$$$$ $$$$g_1g_2\le1 \tag{4}$$$$ on $$(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 $$3$$). $$\Box$$

• Thanks. Very nice. Did you see this proof before? – Boby Aug 10 '20 at 12:58
• @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? – Iosif Pinelis Aug 10 '20 at 16:14
• 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) – Boby Aug 10 '20 at 16:20
• 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. – Iosif Pinelis Aug 10 '20 at 17:41
• In the paper, the author say this ..one can only find implicit inequalities (see, for example, [8]):" where refence [8] is the book. – Boby Aug 10 '20 at 18:12