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Suppose that the sequence of r.v $\{X_{n}\}_{n\geq 1}$ has all the moments, and $X_{n}\stackrel{D}{\longrightarrow}X\sim N(0,\sigma)$. Assume that $E\left\{(X_{n})^{K}\right\} \stackrel{n}{\longrightarrow} E(X^{K})$, where $K\geq 1$ is an integer number. Can we say that $E\left\{(X_{n})^{K+1}\right\} \stackrel{n}{\longrightarrow} E(X^{K+1})?$

Clarifications: The simbol $\stackrel{D}{\longrightarrow}$ represents Convergence in Distribution.

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No, this is not true. Let $X \sim N(0,1)$ and define $Y_n$ to be independent of $X$, such that $Y_n = \sqrt{n}$ with probability $1/n$ and 0 otherwise. Set $X_n = X+ Y_n$. Since $Y_n \to 0$ in $L^1$, we have $X_n \to X$ in $L^1$ and hence also in distribution; in particular, $E X_n^1 \to E X^1 = 0$. But by independence $E[X_n^2] = E[X^2] + E[Y_n^2] = 1+1 = 2$ for all $n$, whereas $EX^2 = 1$.

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  • $\begingroup$ This also works, mutatis mutandis, for any K and any distribution. $\endgroup$
    – Ori Gurel-Gurevich
    Commented Jan 23, 2015 at 9:09
  • $\begingroup$ I understand your argument. In general, we can't affirm the convergence in $(K + 1)$-th order moment from the convergence of $K$-th order moment. I've been wondering about the Theorem 25.12's Corollary in Billingsley's book (Probability and Measure, English version, third edition, 1995, page 338), which requires $\sup_n[|X_n|^{k+\varepsilon}]<\infty$ plus convergence in distribution to get the $K$-th order moment convergence. In my case, the $X_n$'s Moment generating function exists in $|t|<t_0$ for each $n$. $\endgroup$
    – S. W. M
    Commented Jan 23, 2015 at 14:28
  • $\begingroup$ In addition, I've already checked the convergence of three first moments. The combination of these facts made me hopeful about the convergence of all moments, at least in this situation. Thanks for the reply. $\endgroup$
    – S. W. M
    Commented Jan 23, 2015 at 14:51

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