# Convergence in distribution versus convergence of moments

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 ?

• To help your intuition, let $X$ have the limiting distribution and let $Y$ have infinite expectation. Consider the sequence of variables $(1-1/n)X+(1/n)Y.$ – whuber Mar 4 at 18:22
• Assume the mean and variance are bounded, I'll update my question. – ArnoV Mar 5 at 8:36
• 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. – whuber Mar 5 at 14:37