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Suppose we have that a random variable sequence $(X_n)_n$ converges in distribution to a law with mean $\bar{\mu}$ and variance $\bar{\sigma}^2$, or formally $X_n \stackrel{d}{\to} \mathcal{L}(\bar{\mu}, \bar{\sigma})$. Further assume the moments of the sequence and the limiting distribution are bounded: $\bar{\mu}, \bar{\sigma} < \infty$ and $\mu_n, \sigma_n < \infty$.

Do we have that $\mathrm{E}[X_n] = \mu_n \to \bar{\mu}$ and $\mathrm{Var}[X_n] = \sigma_n^2 \to \bar{\sigma}^2$ holds ?

My intuition is as follows: for $n$ large, we have that $X_n$ is approximately distributed as $ \mathcal{L}(\bar{\mu}, \bar{\sigma})$, so when we compute its expectation $\mu_n$ we should find approximately $\bar{\mu}$.

Is there a formal statement which corroborates or invalidates this intution of mine ?

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  • $\begingroup$ To help your intuition, let $X$ have the limiting distribution and let $Y$ have infinite expectation. Consider the sequence of variables $(1-1/n)X+(1/n)Y.$ $\endgroup$ – whuber Mar 4 at 18:22
  • $\begingroup$ Assume the mean and variance are bounded, I'll update my question. $\endgroup$ – ArnoV Mar 5 at 8:36
  • $\begingroup$ Your update doesn't work: all it says is to assume all variances are finite. As a counterexample, let $Y$ have unit variance and consider the sequence of mixture distributions of $X$ and $nY$ with weight $1/n$ on $nY.$ Each of these mixtures has finite variance but their variances diverge. To block that, you need to assume there exist (finite) numbers $N$ and $M$ for which $n\ge N$ implies $\sigma_n\le M.$ By varying this counterexample you can construct sequences $(X_n)$ where none of the moments converges yet $(X_n)$ converges in distribution. $\endgroup$ – whuber Mar 5 at 14:37

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