On the limiting distribution of sample central moments

Abstract

We investigate the limiting behavior of sample central moments, examining the special cases where the limiting (as the sample size tends to infinity) distribution is degenerate. Parent (non-degenerate) distributions with this property are called singular, and we show in this article that the singular distributions contain at most three supporting points. Moreover, using the delta-method, we show that the (second-order) limiting distribution of sample central moments from a singular distribution is either a multiple, or a difference of two multiples of independent Chi-square random variables with one degree of freedom. Finally, we present a new characterization of normality through the asymptotic independence of the sample mean and all sample central moments.

Appendix A Proofs

We shall make use of the following Lemmas. For the proof of Lemma 6 see, e.g., Gut (1988, p. 18); for more general results, see Afendras and Markatou (2016).

Lemma 6

If $X,X_1,\ldots ,X_n$ are independent and identically distributed with ${\mathbb {E}}(X)=\mu$, ${\textsf {Var}}(X)=\sigma ^2$ and ${\mathbb {E}}|X|^\delta <\infty$ for some $\delta \ge 2$, then, for any $\alpha \in (0,\delta ]$,

$$\begin{aligned} {\mathbb {E}}\left| \surd {n}\left( \bar{X}_n-\mu \right) \right| ^\alpha \rightarrow \sigma ^\alpha {\mathbb {E}}|Z|^\alpha , \end{aligned}$$

where $Z\sim N(0,1)$ and $\bar{X}_n=(X_1+\cdots +X_n)/n$.

Lemma 7

If $X,X_1,\ldots ,X_n$ are independent and identically distributed with ${\mathbb {E}}(X)=\mu$ and ${\mathbb {E}}|X|^\nu <\infty$ for some $\nu \in \{2,3,\ldots \}$, then, for any $j\in \{2,\ldots ,\nu \}$,

$$\begin{aligned} {\mathbb {E}}|m_{j,n}|^{{\nu }/{j}} \le {\mathbb {E}}|X-\mu |^{\nu }, \end{aligned}$$

(31)

where $m_{j,n}=n^{-1}\sum _{i=1}^n(X_i-\mu )^j$.

Proof

If $j=\nu$, then (31) follows by taking expectations to the obvious inequality $|m_{j,n}|\le \frac{1}{n}\sum _{i=1}^n|X_i-\mu |^j=\frac{1}{n} \sum _{i=1}^n|X_i-\mu |^{\nu }$. If $j<\nu$ (and thus, $\nu \ge 3$), we apply the inequality

$$\begin{aligned} \left| \sum _{i=1}^{n} x_i\right| ^p \le \left( \sum _{i=1}^{n} |x_i|\right) ^p \le n^{p-1} \sum _{i=1}^n |x_i|^p, \quad p>1, \end{aligned}$$

(the last inequality is a by-product of Hölder’s inequality) for $p=\nu /j$ and $x_i=(X_i-\mu )^j$. Then, we have

$$\begin{aligned} {\mathbb {E}}|m_{j,n}|^{{\nu }/{j}}&= \frac{1}{n^{{\nu }/{j}}} {\mathbb {E}}\left| \sum _{i=1}^n (X_i-\mu )^j\right| ^{{\nu }/{j}} \le \frac{1}{n^{{\nu }/{j}}} {\mathbb {E}}\left( \sum _{i=1}^n |X_i-\mu |^j\right) ^{{\nu }/{j}} \\&\le \frac{1}{n^{{\nu }/{j}}} {\mathbb {E}}\left( n^{{\nu }/{j}-1} \sum _{i=1}^n |X_i-\mu |^\nu \right) =\frac{1}{n} {\mathbb {E}}\left( \sum _{i=1}^n |X_i-\mu |^\nu \right) \\ {}&={\mathbb {E}}|X-\mu |^{\nu }. \end{aligned}$$

$\square$

Proof of Proposition 2

(a) Observe that the statement in Proposition 2(a) is equivalent to

$$\begin{aligned} {\mathbb {E}}[\surd {n}(M_{k,n}-\mu _k)]\rightarrow 0. \end{aligned}$$

(32)

Writing

$$\begin{aligned} M_{k,n}-\mu _k= (m_{k,n}-\mu _k)+(-\,1)^{k-1}(k-1) m_{1,n}^k +\sum _{j=2}^{k-1} (-\,1)^{k-j} {k\atopwithdelims ()j} m_{1,n}^{k-j} m_{j,n},\qquad \end{aligned}$$

(33)

it suffices to verify that

• (i) $\surd {n}{\mathbb {E}}(m_{k,n}-\mu _k)=0$,

• (ii) $\surd {n}{\mathbb {E}}(m_{1,n}^k)\rightarrow 0$,

• (iii) $\surd {n}{\mathbb {E}}(m_{1,n}^{k-j} m_{j,n}) \rightarrow 0$, $j=2,\ldots ,k-2$ (provided $k\ge 4$), and

• (iv) $\surd {n}{\mathbb {E}}(m_{1,n} m_{k-1,n}) \rightarrow 0$ (provided $k\ge 3$).

Now, (i) is obvious (since ${\mathbb {E}}(m_{k,n})=\mu _k$), (iv) follows from ${\mathbb {E}}(m_{1,n} m_{k-1,n})=\mu _k/n$ and (ii) can be seen by using Lemma 6 with $\alpha =\delta =k$, which shows that

$$\begin{aligned} \left| n^{k/2}{\mathbb {E}}\left( m_{1,n}^k\right) \right| \le n^{k/2} {\mathbb {E}}|m_{1,n}|^k={\mathbb {E}}\left| \surd {n}\left( \bar{X}_n-\mu \right) \right| ^k\rightarrow \sigma ^{k}{\mathbb {E}}|Z|^k<\infty , \end{aligned}$$

and thus, $|\surd {n}{\mathbb {E}}(m_{1,n}^k)|\le n^{-(k-1)/2}{\mathbb {E}}|\surd {n}(\bar{X}_n-\mu )|^k\rightarrow 0$. To show (iii), we assume that $k\ge 4$ and $2\le j\le k-2$, and we use Hölder’s inequality with $p=k/(k-j)>1$, Lemma 7 with $\nu =k$ and Lemma 6 with $\alpha =\delta =k$ to obtain

$$\begin{aligned}&\left| \surd {n}{\mathbb {E}}\left( m_{1,n}^{k-j}m_{j,n}\right) \right| \\&\quad \le {\surd {n}{\mathbb {E}}\left( |m_{1,n}|^{k-j} |m_{j,n}|\right) \le \surd {n}\left( {\mathbb {E}}|m_{1,n}|^k\right) ^{{(k-j)}/{k}} \left( {\mathbb {E}}|m_{j,n}|^{{k}/{j}}\right) ^{{j}/{k}} } \\&\quad \le \surd {n} \left[ n^{-k/2}{\mathbb {E}}\left| \surd {n}\left( \bar{X}_n-\mu \right) \right| ^k\right] ^{{(k-j)}/{k}} \left( {\mathbb {E}}|X-\mu |^k\right) ^{{j}/{k}}\\&\quad ={n^{-(k-1-j)/2} \left[ {\mathbb {E}}\left| \surd {n}\left( \bar{X}_n-\mu \right) \right| ^k\right] ^{{(k-j)}/{k}} \left( {\mathbb {E}}|X-\mu |^k\right) ^{{j}/{k}} } \\&\quad = {n^{-(k-1-j)/2} O(1)\rightarrow 0,} \end{aligned}$$

because ${\mathbb {E}}|\surd {n}(\bar{X}_n-\mu )|^k\rightarrow \sigma ^{k}{\mathbb {E}}|Z|^k<\infty$.

(b) Observe that the statement in Proposition 2(b) is equivalent to

$$\begin{aligned} \mathsf {Cov}\left[ \surd {n}\left( \bar{X}_n-\mu \right) ,\surd {n}(M_{k,n}-\mu _k)\right] \rightarrow \mu _{k+1}-k\sigma ^2\mu _{k-1}, \end{aligned}$$

(34)

and since ${\mathbb {E}}(\bar{X}_n-\mu )=0$, it suffices to verify that

$$\begin{aligned} n{\mathbb {E}}\left[ \left( \bar{X}_n-\mu \right) (M_{k,n}-\mu _k)\right] =n{\mathbb {E}}[m_{1,n}(M_{k,n}-\mu _k)]\rightarrow \mu _{k+1}-k\sigma ^2\mu _{k-1}. \end{aligned}$$

(35)

If $k=2$, then $n{\mathbb {E}}[m_{1,n}(M_{2,n}-\mu _2)]=n{\mathbb {E}}[(\bar{X}_n-\mu )(m_{2,n}-\mu _2)] -n{\mathbb {E}}(\bar{X}_n-\mu )^3=\mu _3-n{\mathbb {E}}(\bar{X}_n-\mu )^3$, and it easily seen, by Lemma 6 with $\alpha =\delta =3$, that $|n{\mathbb {E}}(\bar{X}_n-\mu )^3| \le n^{-1/2} {\mathbb {E}}|\surd {n}(\bar{X}_n-\mu )|^3\rightarrow 0$; thus, $n{\mathbb {E}}[m_{1,n}(M_{2,n}-\mu _2)]\rightarrow \mu _3$. Since $\mu _1=0$, (35) is satisfied for $k=2$.

If $k=3$, $n{\mathbb {E}}[m_{1,n}(M_{3,n}-\mu _3)]=n{\mathbb {E}}[(\bar{X}_n-\mu )(m_{3,n}-\mu _3)] +2n{\mathbb {E}}(\bar{X}_n-\mu )^4-3n {\mathbb {E}}[m_{2,n}(\bar{X}_n-\mu )^2]$, and it is easy to see that $n{\mathbb {E}}[(\bar{X}_n-\mu )(m_{3,n}-\mu _3)]=\mu _4$. Also, by Lemma 6 with $\alpha =\delta =4$, $2n{\mathbb {E}}(\bar{X}_n-\mu )^4\rightarrow 0$. Finally, $-\,3n{\mathbb {E}}[m_{2,n}(\bar{X}_n-\mu )^2]=-\,3[\mu _4+(n-1)\mu _2^2]/n\rightarrow -3 \mu _2^2=-\,3\sigma ^4$, which verifies (35) for $k=3$.

In the general case when $k\ge 4$, we write $M_{k,n}-\mu _k$ as in (33) and we observe that for (35) to hold it suffices to verify that

• (i) $n {\mathbb {E}}[m_{1,n}(m_{k,n}-\mu _k)]=\mu _{k+1}$,

• (ii) $n{\mathbb {E}}(m_{1,n}^{k+1}) \rightarrow 0$,

• (iii) $n {\mathbb {E}}(m_{1,n}^{k+1-j} m_{j,n}) \rightarrow 0$, $j=2,\ldots ,k-2$, and

• (iv) $n{\mathbb {E}}(m_{1,n}^2 m_{k-1,n}) \rightarrow \sigma ^2 \mu _{k-1}$.

Calculating ${\mathbb {E}}[m_{1,n}(m_{k,n}-\mu _k)]={\mathbb {E}}[(\bar{X}_n-\mu )(m_{k,n}-\mu _k)] ={\mathbb {E}}[(\bar{X}_n-\mu )m_{k,n}] = n^{-2}\sum _{i_1=1}^n\sum _{i_2=1}^n {\mathbb {E}}[(X_{i_1}-\mu )(X_{i_2}-\mu )^k] ={\mu _{k+1}}/{n}$, we conclude (i), while (ii) follows by using Lemma 6 with $\alpha =\delta =k+1$. Also,

$$\begin{aligned} n{\mathbb {E}}\left( m_{1,n}^2 m_{k-1,n}\right)= & {} \frac{1}{n^2}\sum _{i_1=1}^n\sum _{i_2=1}^n\sum _{i_3=1}^n {\mathbb {E}}\left[ \left( X_{i_1}-\mu \right) \left( X_{i_2}-\mu \right) \left( X_{i_3}-\mu \right) ^{k-1}\right] \\= & {} { \frac{1}{n^2}\left[ n \mu _{k+1}+n(n-1)\sigma ^2\mu _{k-1}\right] \rightarrow \sigma ^2\mu _{k-1}, } \end{aligned}$$

which shows that (iv) is satisfied, and it remains to verify (iii). To this end, we use Hölder’s inequality with $p=(k+1)/(k+1-j)>1$ and Lemma 7 with $\nu =k+1$ to obtain

$$\begin{aligned}&\left| n {\mathbb {E}}\left( m_{1,n}^{k+1-j}m_{j,n}\right) \right| \\&\quad \le { n {\mathbb {E}}|m_{1,n}|^{k+1-j} |m_{j,n}| \le n \left( {\mathbb {E}}|m_{1,n}|^{k+1}\right) ^{{(k+1-j)}/{(k+1)}} \left( {\mathbb {E}}|m_{j,n}|^{(k+1)/{j}}\right) ^{{j}/{(k+1)}} }\\&\quad \le { n \left[ n^{-(k+1)/2}{\mathbb {E}}\left| \surd {n}\left( \bar{X}_n-\mu \right) \right| ^{k+1}\right] ^{{(k+1-j)}/{(k+1)}} \left( {\mathbb {E}}|X-\mu |^{k+1}\right) ^{{j}/{(k+1)}} }\\&\quad = { n^{-(k-1-j)/2} \left[ {\mathbb {E}}\left| \surd {n}\left( \bar{X}_n-\mu \right) \right| ^{k+1}\right] ^{{(k+1-j)}/{(k+1)}} \left( {\mathbb {E}}|X-\mu |^{k+1}\right) ^{{j}/{(k+1)}} \rightarrow 0, } \end{aligned}$$

because $n^{-(k-1-j)/2}\rightarrow 0$; and, by Lemma 6 with $\alpha =\delta =k+1$, ${\mathbb {E}}|\surd {n}(\bar{X}_n-\mu )|^{k+1}\rightarrow \sigma ^{k+1}{\mathbb {E}}|Z|^{k+1} <\infty$.

(c) Without loss of generality assume that $2\le r\le k$ and observe that the first statement of Proposition 2(c) is equivalent to

$$\begin{aligned} \mathsf {Cov}[\surd {n}(M_{r,n}-\mu _r),\surd {n}(M_{k,n}-\mu _k)]\rightarrow v_{rk}. \end{aligned}$$

(36)

Since ${\mathbb {E}}|X|^{r+k}<\infty$, (32) shows that ${\mathbb {E}}[\surd {n}(M_{k,n}-\mu _k)]\rightarrow 0$ and ${\mathbb {E}}[\surd {n}(\bar{X}_n-\mu )]\rightarrow 0$, and it suffices to verify that

$$\begin{aligned} n{\mathbb {E}}[(M_{r,n}-\mu _r)(M_{k,n}-\mu _k)] \rightarrow v_{rk}= & {} \mu _{r+k}-\mu _r\mu _k-r\mu _{r-1}\mu _{k+1}\nonumber \\&-\,k\mu _{r+1}\mu _{k-1}+rk\sigma ^2\mu _{r-1}\mu _{k-1}.\qquad \end{aligned}$$

(37)

The proof can be deduced by showing that (37) holds for each one of the cases $r=k=2$; $r=2$, $k=3$; $r=k=3$; $r=2$, $k\ge 4$; $r=3$, $k\ge 4$; $4\le r\le k$. In the following we shall present the details only for the case where $4\le r\le k$; the other cases can be treated using similar (and simpler) arguments.

Assume now that $4\le r\le k$. From (33), we have

$$\begin{aligned} M_{r,n}-\mu _r= & {} (m_{r,n}-\mu _r)-rm_{1,n}m_{r-1,n}\nonumber \\&+\sum _{j_1=2}^{r-2} (-\,1)^{r-j_1} {r\atopwithdelims ()j_1} m_{1,n}^{r-j_1} m_{j_1,n} +(-\,1)^{r-1}(r-1) m_{1,n}^r, \end{aligned}$$

(38)

$$\begin{aligned} M_{k,n}-\mu _k= & {} (m_{k,n}-\mu _k)-km_{1,n}m_{k-1,n}\nonumber \\&+\sum _{j_2=2}^{k-2} (-\,1)^{k-j_2} {k\atopwithdelims ()j_2} m_{1,n}^{k-j_2} m_{j_2,n} +(-\,1)^{k-1}(k-1) m_{1,n}^k. \end{aligned}$$

(39)

We shall show that the asymptotic covariance in (36) can be determined by using only the first two terms in (38) and (39). Indeed, it is easily seen that (37) holds true if it can be shown that

• (i) $n {\mathbb {E}}[(m_{r,n}-\mu _r)(m_{k,n}-\mu _k)]=\mu _{r+k}-\mu _r\mu _k$,

• (ii) $n{\mathbb {E}}[m_{1,n}m_{k-1,n}(m_{r,n}-\mu _r)] \rightarrow \mu _{r+1}\mu _{k-1}$,

• (iii) $n{\mathbb {E}}[m_{1,n}m_{r-1,n}(m_{k,n}-\mu _k)] \rightarrow \mu _{r-1}\mu _{k+1}$,

• (iv) $n{\mathbb {E}}(m_{1,n}^2m_{r-1,n}m_{k-1,n}) \rightarrow \sigma ^2 \mu _{r-1}\mu _{k-1}$,

• (v) $n {\mathbb {E}}[m_{1,n}^{k-j_2} m_{j_2,n}(m_{r,n}-\mu _r)] \rightarrow 0$, $j_2=2,\ldots ,k-2$,

• (vi) $n{\mathbb {E}}[m_{1,n}^k (m_{r,n}-\mu _r)] \rightarrow 0$,

• (vii) $n {\mathbb {E}}(m_{1,n}^{k+1-j_2} m_{j_2,n}m_{r-1,n}) \rightarrow 0$, $j_2=2,\ldots ,k-2$,

• (viii) $n{\mathbb {E}}(m_{1,n}^{k+1} m_{r-1,n}) \rightarrow 0$,

• (ix) $n {\mathbb {E}}[m_{1,n}^{r-j_1} m_{j_1,n}(m_{k,n}-\mu _k)] \rightarrow 0$, $j_1=2,\ldots ,r-2$,

• (x) $n{\mathbb {E}}(m_{1,n}^{r+1-j_1} m_{j_1,n} m_{k-1,n}) \rightarrow 0$, $j_1=2,\ldots ,r-2$,

• (xi) $n {\mathbb {E}}(m_{1,n}^{r+k-j_1-j_2} m_{j_1,n}m_{j_2,n}) \rightarrow 0$, $j_1=2,\ldots ,r-2$, $j_2=2,\ldots ,k-2$,

• (xii) $n{\mathbb {E}}(m_{1,n}^{r+k-j_1} m_{j_1,n}) \rightarrow 0$, $j_1=2,\ldots ,r-2$,

• (xiii) $n {\mathbb {E}}[m_{1,n}^{r} (m_{k,n}-\mu _k)] \rightarrow 0$,

• (xiv) $n{\mathbb {E}}(m_{1,n}^{r+1} m_{k-1,n}) \rightarrow 0$,

• (xv) $n {\mathbb {E}}(m_{1,n}^{r+k-j_2} m_{j_2,n}) \rightarrow 0$, $j_2=2,\ldots ,k-2$, and

• (xvi) $n{\mathbb {E}}(m_{1,n}^{r+k}) \rightarrow 0$.

We now proceed to verify (i)–(xvi). Since ${\mathbb {E}}(m_{r,n})=\mu _r$ and ${\mathbb {E}}(m_{k,n})=\mu _k$, we have

$$\begin{aligned}&n{\mathbb {E}}[(m_{r,n}-\mu _r)(m_{k,n}-\mu _k)] \\&\quad = { n[{\mathbb {E}}(m_{r,n}m_{k,n})-\mu _r\mu _k] = n\left\{ \frac{1}{n^2}\sum _{i_1=1}^n\sum _{i_2=1}^n {\mathbb {E}}\left[ \left( X_{i_1}-\mu \right) ^r\left( X_{i_2}-\mu \right) ^k\right] -\mu _r\mu _k\right\} } \\&\quad = { n\left\{ \frac{1}{n^2}[n\mu _{r+k}+n(n-1)\mu _r\mu _k]-\mu _r\mu _k\right\} =\mu _{r+k}-\mu _r\mu _k, } \end{aligned}$$

which shows (i). Also, (ii), (iii) and (iv) follow by straightforward computations; e.g., for (ii) we have

$$\begin{aligned}&n{\mathbb {E}}[m_{1,n}m_{k-1,n}(m_{r,n}-\mu _r)] \\&\quad =-\,\mu _r\mu _k +\frac{\mu _{r+k}+(n-1)(\mu _{r+1}\mu _{k-1}+ \mu _{r}\mu _{k})}{n}\rightarrow \mu _{r+1}\mu _{k-1}, \end{aligned}$$

while (iii) is similar to (ii), and (iv) can be deduced from

$$\begin{aligned}&n{\mathbb {E}}\left( m_{1,n}^2m_{r-1,n}m_{k-1,n}\right) \\&\quad =\frac{1}{n^3}\left[ n(n-1)(n-2)\sigma ^2\mu _{r-1}\mu _{k-1} +o\left( n^3\right) \right] \rightarrow \sigma ^2\mu _{r-1}\mu _{k-1}. \end{aligned}$$

The vanishing limits (vi)–(viii) and (x)–(xvi) are by-products of Lemmas 6 and 7 with $\alpha =\delta =\nu =r+k$, since ${\mathbb {E}}|X|^{r+k}<\infty$. Indeed, we have $|n{\mathbb {E}}(m_{1,n}^{r+k})|\le n{\mathbb {E}}|m_{1,n}|^{r+k} =n^{-(r+k-2)/2}{\mathbb {E}}|\surd {n}(\bar{X}_n-\mu )|^{r+k}\rightarrow 0$, which verifies (xvi). Also, using Hölder’s inequality with $p=(r+k)/(r+k-j_2)>1$, we obtain (xv) as follows:

$$\begin{aligned}&\left| n {\mathbb {E}}\left( m_{1,n}^{r+k-j_2}m_{j_2,n}\right) \right| \\&\quad \le {n {\mathbb {E}}\left( |m_{1,n}|^{r+k-j_2} |m_{j_2,n}|\right) \le n \left( {\mathbb {E}}|m_{1,n}|^{r+k}\right) ^{\frac{r+k-j_2}{r+k}} \left( {\mathbb {E}}|m_{j_2,n}|^{\frac{r+k}{j_2}}\right) ^{\frac{j_2}{r+k}}} \\&\quad \le n\left[ n^{-(r+k)/2}{\mathbb {E}}\left| \surd {n}\left( \bar{X}_n-\mu \right) \right| ^{r+k}\right] ^{\frac{r+k-j_2}{r+k}} \left( {\mathbb {E}}|X-\mu |^{r+k}\right) ^{\frac{j_2}{r+k}} \\&\quad ={n^{-(r+k-j_2-2)/2} \left[ {\mathbb {E}}\left| \surd {n}\left( \bar{X}_n-\mu \right) \right| ^{r+k}\right] ^{\frac{r+k-j_2}{r+k}} \left( {\mathbb {E}}|X-\mu |^{r+k}\right) ^{\frac{j_2}{r+k}} \rightarrow 0,} \end{aligned}$$

because $n^{-(r+k-j_2-2)/2}\rightarrow 0$ and ${\mathbb {E}}|\surd {n}(\bar{X}_n-\mu )|^{r+k}\rightarrow \sigma ^{r+k}{\mathbb {E}}|Z|^{r+k} <\infty$; (xii) is similar to (xv). For the limit (xiv) we have

$$\begin{aligned}&\left| n{\mathbb {E}}\left[ m_{1,n}^{r+1}m_{k-1,n}\right] \right| \\&\quad \le { n{\mathbb {E}}\left( |m_{1,n}|^{r+1} |m_{k-1,n}|\right) \le n \left( {\mathbb {E}}|m_{1,n}|^{r+k}\right) ^{\frac{r+1}{r+k}} \left( {\mathbb {E}}|m_{k-1,n}|^{\frac{r+k}{k-1}}\right) ^{\frac{k-1}{r+k}} }\\&\quad \le { n^{-(r-1)/2}\left[ {\mathbb {E}}\left| \surd {n}\left( \bar{X}_n-\mu \right) \right| ^{r+k}\right] ^{\frac{r+1}{r+k}} \left( {\mathbb {E}}|X-\mu |^{r+k}\right) ^{\frac{k-1}{r+k}} \rightarrow 0, } \end{aligned}$$

and similarly for (viii). In order to prove (xiii), it is sufficient to show that $n{\mathbb {E}}[m_{1,n}^r m_{k,n}]\rightarrow 0$ and $n{\mathbb {E}}(m_{1,n}^r)\rightarrow 0$. The second limit is obvious since, as for (xvi), one can easily verify that $|n{\mathbb {E}}(m_{1,n}^r)|\le n^{-(r-2)/2}{\mathbb {E}}|\surd {n}(\bar{X}_n-\mu )|^r =n^{-(r-2)/2}O(1)\rightarrow 0$. For the first limit, we have

$$\begin{aligned} \left| n{\mathbb {E}}\left( m_{1,n}^{r}m_{k,n}\right) \right|\le & {} { n{\mathbb {E}}\left( |m_{1,n}|^{r} |m_{k,n}|\right) \le n \left( {\mathbb {E}}|m_{1,n}|^{r+k}\right) ^{\frac{r}{r+k}} \left( {\mathbb {E}}|m_{k,n}|^{\frac{r+k}{k}}\right) ^{\frac{k}{r+k}} }\\\le & {} { n^{-(r-2)/2}\left[ {\mathbb {E}}\left| \surd {n}\left( \bar{X}_n-\mu \right) \right| ^{r+k}\right] ^{\frac{r}{r+k}} \left( {\mathbb {E}}|X-\mu |^{r+k}\right) ^{\frac{k}{r+k}} \rightarrow 0. } \end{aligned}$$

Limit (vi) is similar to (xiii), and its proof is omitted. Regarding (xi), we have

$$\begin{aligned}&\left| n{\mathbb {E}}\left( m_{1,n}^{r+k-j_1-j_2}m_{j_1,n}m_{j_2,n}\right) \right| \\&\quad \le n{\mathbb {E}}\left( |m_{1,n}|^{r+k-j_1-j_2} |m_{j_1,n}m_{j_2,n}|\right) \\&\quad \le n \left( {\mathbb {E}}|m_{1,n}|^{r+k}\right) ^{\frac{r+k-j_1-j_2}{r+k}} \left( {\mathbb {E}}|m_{j_1,n}m_{j_2,n}|^{\frac{r+k}{j_1+j_2}}\right) ^{\frac{j_1+j_2}{r+k}}\\&\quad \le n \left( {\mathbb {E}}|m_{1,n}|^{r+k}\right) ^{\frac{r+k-j_1-j_2}{r+k}} \left[ \left( {\mathbb {E}}|m_{j_1,n}|^{\frac{r+k}{j_1}}\right) ^{\frac{j_1}{j_1+j_2}} \left( {\mathbb {E}}|m_{j_2,n}|^{\frac{r+k}{j_2}}\right) ^{\frac{j_2}{j_1+j_2}}\right] ^{\frac{j_1+j_2}{r+k}}\\&\quad \le n \left( {\mathbb {E}}|m_{1,n}|^{r+k}\right) ^{\frac{r+k-j_1-j_2}{r+k}} \left( {\mathbb {E}}|X-\mu |^{r+k}\right) ^{\frac{j_1+j_2}{r+k}} \\&\quad = n^{-(r+k-j_1-j_2-2)/2} \left[ {\mathbb {E}}\left| \surd {n}\left( \bar{X}_n-\mu \right) \right| ^{r+k}\right] ^{\frac{r+k-j_1-j_2}{r+k}} \left( {\mathbb {E}}|X-\mu |^{r+k}\right) ^{\frac{j_1+j_2}{r+k}} \rightarrow 0. \end{aligned}$$

Similarly, for (x) we have

$$\begin{aligned}&\left| n{\mathbb {E}}\left( m_{1,n}^{r+1-j_1}m_{j_1,n}m_{k-1,n}\right) \right| \\&\quad \le n\left( {\mathbb {E}}|m_{1,n}|^{r+1-j_1} |m_{j_1,n}m_{k-1,n}|\right) \\&\quad \le n \left( {\mathbb {E}}|m_{1,n}|^{r+k}\right) ^{\frac{r+1-j_1}{r+k}} \left( {\mathbb {E}}|m_{j_1,n}m_{k-1,n}|^{\frac{r+k}{j_1+k-1}}\right) ^{\frac{j_1+k-1}{r+k}}\\&\quad \le n \left( {\mathbb {E}}|m_{1,n}|^{r+k}\right) ^{\frac{r+1-j_1}{r+k}} \left[ \left( {\mathbb {E}}|m_{j_1,n}|^{\frac{r+k}{j_1}}\right) ^{\frac{j_1}{j_1+k-1}} \left( {\mathbb {E}}|m_{k-1,n}|^{\frac{r+k}{k-1}}\right) ^{\frac{k-1}{j_1+k-1}}\right] ^{\frac{j_1+k-1}{r+k}}\\&\quad \le n \left( {\mathbb {E}}|m_{1,n}|^{r+k}\right) ^{\frac{r+1-j_1}{r+k}} \left( {\mathbb {E}}|X-\mu |^{r+k}\right) ^{\frac{j_1+k-1}{r+k}}\\&\quad = n^{-(r-j_1-1)/2} \left[ {\mathbb {E}}\left| \surd {n}\left( \bar{X}_n-\mu \right) \right| ^{r+k}\right] ^{\frac{r+1-j_1}{r+k}} \left( {\mathbb {E}}|X-\mu |^{r+k}\right) ^{\frac{j_1+k-1}{r+k}} \rightarrow 0, \end{aligned}$$

while (vii) is similar to (x).

It remains to verify (v) and (ix); but, since they are similar, it suffices to prove (v). If $j_2\in \{2,\ldots ,k-3\}$ (and hence, $k\ge 5$ and $j_2<k-2$), we have

$$\begin{aligned}&\left| n{\mathbb {E}}\left[ m_{1,n}^{k-j_2}m_{j_2,n}(m_{r,n}-\mu _r)\right] \right| \\&\quad \le n{\mathbb {E}}\left( |m_{1,n}|^{k-j_2}|m_{j_2,n}m_{r,n}|\right) + n |\mu _r| {\mathbb {E}}\left( |m_{1,n}|^{k-j_2}|m_{j_2,n}|\right) , \end{aligned}$$

and it suffices to prove that $n{\mathbb {E}}(|m_{1,n}|^{k-j_2}|m_{j_2,n}m_{r,n}|)\rightarrow 0$ and $n {\mathbb {E}}(|m_{1,n}|^{k-j_2}|m_{j_2,n}|)\rightarrow 0$. For the first quantity, we have

$$\begin{aligned}&n{\mathbb {E}}\left( \left| m_{1,n}\right| ^{k-j_2}\left| m_{j_2,n}m_{r,n}\right| \right) \\&\quad \le n \left( {\mathbb {E}}|m_{1,n}|^{r+k}\right) ^{\frac{k-j_2}{r+k}} \left( {\mathbb {E}}|m_{j_2,n}m_{r,n}|^{\frac{r+k}{r+j_2}}\right) ^{\frac{r+j_2}{r+k}}\\&\quad \le n \left( {\mathbb {E}}|m_{1,n}|^{r+k}\right) ^{\frac{k-j_2}{r+k}} \left[ \left( {\mathbb {E}}|m_{j_2,n}|^{\frac{r+k}{j_2}}\right) ^{\frac{j_2}{r+j_2}} \left( {\mathbb {E}}|m_{r,n}|^{\frac{r+k}{r}}\right) ^{\frac{r}{r+j_2}}\right] ^{\frac{r+j_2}{r+k}}\\&\quad \le n^{-(k-j_2-2)/2} \left[ {\mathbb {E}}\left| \surd {n}\left( \bar{X}_n-\mu \right) \right| ^{r+k}\right] ^{\frac{k-j_2}{r+k}} \left( {\mathbb {E}}|X-\mu |^{r+k}\right) ^{\frac{r+j_2}{r+k}} \rightarrow 0, \end{aligned}$$

because $k-j_2-2>0$. Similarly, for the second quantity we have

$$\begin{aligned}&n{\mathbb {E}}\left( \left| m_{1,n}\right| ^{k-j_2}\left| m_{j_2,n}\right| \right) \\&\quad \le n \left( {\mathbb {E}}|m_{1,n}|^{r+k}\right) ^{\frac{k-j_2}{r+k}} \left( {\mathbb {E}}|m_{j_2,n}|^{\frac{r+k}{r+j_2}}\right) ^{\frac{r+j_2}{r+k}}\\&\quad \le n \left( {\mathbb {E}}|m_{1,n}|^{r+k}\right) ^{\frac{k-j_2}{r+k}} \left( {\mathbb {E}}|m_{j_2,n}|^{\frac{r+k}{j_2}}\right) ^{\frac{j_2}{r+k}}\\&\quad \le n^{-(k-j_2-2)/2} \left[ {\mathbb {E}}\left| \surd {n}\left( \bar{X}_n-\mu \right) \right| ^{r+k}\right] ^{\frac{k-j_2}{r+k}} \left( {\mathbb {E}}|X-\mu |^{r+k}\right) ^{\frac{j_2}{r+k}} \rightarrow 0, \end{aligned}$$

because $k-j_2-2>0$. Finally, it remains to study the limit (v) when $j_2=k-2$; in this case the above limits do not necessarily vanish. However, since $j_2=k-2$ we have

$$\begin{aligned} n{\mathbb {E}}\left[ m_{1,n}^{k-j_2}m_{j_2,n}(m_{r,n}-\mu _r)\right] =n{\mathbb {E}}\left( m_{1,n}^2 m_{r,n} m_{k-2,n}\right) -n\mu _r {\mathbb {E}}\left( m_{1,n}^2m_{k-2,n}\right) , \end{aligned}$$

and direct computations show that

$$\begin{aligned} n{\mathbb {E}}\left( m_{1,n}^2 m_{r,n} m_{k-2,n}\right) =\frac{1}{n^3}\left[ n(n-1)(n-2) \sigma ^2\mu _r\mu _{k-2}+o\left( n^3\right) \right] \rightarrow \sigma ^2\mu _r\mu _{k-2} \end{aligned}$$

and

$$\begin{aligned} n{\mathbb {E}}\left( m_{1,n}^2 m_{k-2,n}\right) =\frac{1}{n^2}\left[ n(n-1) \sigma ^2\mu _{k-2}+o\left( n^2\right) \right] \rightarrow \sigma ^2\mu _{k-2}. \end{aligned}$$

Hence, when $j_2=k-2$ we have

$$\begin{aligned} n{\mathbb {E}}\left[ m_{1,n}^{k-j_2}m_{j_2,n}(m_{r,n}-\mu _r)\right]= & {} n{\mathbb {E}}\left( m_{1,n}^2 m_{r,n} m_{k-2,n}\right) \\&-\,n\mu _r {\mathbb {E}}\left( m_{1,n}^2 m_{k-2,n}\right) \rightarrow \sigma ^2\mu _r\mu _{k-2} \\&- \,\mu _r \sigma ^2\mu _{k-2}=0, \end{aligned}$$

and the proof is complete. $\square$

Proof of Theorem 3

Observe that $M_{k,n}-\mu _k=g_{k,k}({\varvec{m}}_{k,n})-g_{k,k}(\pmb {\mu }_{k})$; see in Sect. 2. Also, , where , see (1). Hence, Lemma 5 applies to ${\varvec{X}}_n={\varvec{m}}_{k,n}$, provided (21) is fulfilled for ${\varvec{m}}_{k,n}$, i.e., provided that . Because $\nabla g_{k,k}(\pmb {\mu }_{k})=(-\,k\mu _{k-1},0,\ldots ,0,1)'$, we get $[\nabla g_{k,k}(\pmb {\mu }_{k})]'({\varvec{m}}_{k,n}-\pmb {\mu }_{k})=-\,k\mu _{k-1}m_{1,n}+(m_{k,n}-\mu _k)$. Since ${\mathbb {E}}(m_{j,n})=\mu _j$ for all n and j, we get ${\mathbb {E}}[-\,k\mu _{k-1}m_{1,n}+(m_{k,n}-\mu _k)]=0$. Also,

$$\begin{aligned}&{\textsf {Var}}[-\,k\mu _{k-1}m_{1,n}+(m_{k,n}-\mu _k)]\\&\quad = { k^2 \mu _{k-1}^2 {\textsf {Var}}(m_{1,n}) + {\textsf {Var}}(m_{k,n})-2k\mu _{k-1}\mathsf {Cov}(m_{1,n},m_{k,n}) } \\&\quad = k^2 \mu _{k-1}^2 \frac{\sigma ^2}{n} + \frac{\mu _{2k}-\mu _k^2}{n}-2k\mu _{k-1}\frac{\mu _{k+1}}{n} \\&\quad = \frac{1}{n}\left( k^2 \mu _{k-1}^2 \sigma ^2 + \mu _{2k}-\mu _k^2-2k\mu _{k-1}\mu _{k+1}\right) \\&\quad = \frac{1}{n}\left[ \mu _{2k}-\mu _k^2+k\mu _{k-1} \left( k\sigma ^2\mu _{k-1}-2\mu _{k+1}\right) \right] = \frac{v_k^2}{n} =0, \end{aligned}$$

because $v_k^2=0$ by the assumed singularness. Therefore, $[\nabla g_{k,k}(\pmb {\mu }_{k})]'({\varvec{m}}_{k,n}-\pmb {\mu }_{k})=0$ with probability one and, thus, in a trivial sense. Now, a simple calculation, since $\nabla g_{k,k}(\pmb {\mu }_{k})=(-\,k\mu _{k-1},0,\ldots ,0,1)'$, shows that

$$\begin{aligned} \mathbf {H}_k(\pmb {\mu }_{k})= \left( \begin{array}{cccccc} k(k-1)\mu _{k-2} &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -\,k &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots \\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ -k &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 \end{array} \right) , \end{aligned}$$

i.e.,

$$\begin{aligned} \mathbf {H}_2(\pmb {\mu }_2)= \left( \begin{array}{cc} 2 &{}\quad 0 \\ 0 &{}\quad 0 \end{array} \right) , \quad \mathbf {H}_3(\pmb {\mu }_3)= \left( \begin{array}{ccc} 0 &{}\quad -\,3 &{}\quad 0\\ -\,3 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 \end{array} \right) , \quad \mathbf {H}_4(\pmb {\mu }_4)\!=\! \left( \begin{array}{cccc} 12\sigma ^2 &{}\quad 0 &{}\quad -\,4 &{}\quad 0\\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0\\ -\,4 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \end{array} \right) , \end{aligned}$$

etc. Applying (22), we see that $n(M_{k,n}-\mu _k)$ converges weakly to the distribution of $\frac{1}{2}{\varvec{W}}'_k\mathbf {H}_k(\pmb {\mu }_k){\varvec{W}}_k =\frac{1}{2}k(k-1)\mu _{k-2} W_1^2 - k W_1 W_{k-1}$, while, by (1), the distribution of $(W_1,W_{k-1})'$ is given by (24). $\square$