let $Y_i$ be a sequence of identically distributed independent random variables with mean 0 and variance 1. If $X$ is independent and gaussian with the same mean and variance then does $\frac{1}{\sqrt{n}}(\sum_{i=1}^n Y_i)$ converge in $L^p$ to $X$? The central limit theorem usually holds in distribution but I would like to know if there is a stronger convergence where the sum converges to a gaussian random variable.
Is there a central limit theorem for $L^p$
probabilityprobability distributionsprobability theorystatistics
Best Answer
Here is a counterexample: assume each $Y_i$ has distribution $\mathcal N(0,1)$, so that $\frac 1{\sqrt n}\sum_{i=1}^nY_i \sim \mathcal N(0,1)$.
Suppose for the sake of contradiction that there is some $X\sim \mathcal N(0,1)$ independent of $(Y_1,\ldots)$ with $\frac 1{\sqrt n}\sum_{i=1}^nY_i \xrightarrow{L^p}X$.
Since $X$ is independent of $(Y_1,Y_2,\ldots)$ the random variable $\frac 1{\sqrt n}\sum_{i=1}^nY_i - X$ has distribution $\mathcal N(0,2)$ thus
$$E\left[\left|\frac 1{\sqrt n}\sum_{i=1}^nY_i - X\right|^p\right]$$
is a non-zero constant. It does not converge to $0$ as $n\to \infty$, a contradiction.