Convergence in L2 (up to a constant) implies convergence in probability?

Suppose we have a sequence of random variables ${X_n}$ such that $mathbb{E} [|X_n|^2] = a_n + c$ where $a_n to 0$ is a decreasing sequence of positive real numbers and $c > 0$ is a constant. Then can we say anything about the convergence in probability of $X_n$?

Mathematics Asked by user3294195 on November 18, 2021

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No. Consider the sequence of random variables $X_n = (1 + frac{1}{n})$ with the uniform measure on $[0,1]$. Well $mathbb{E}(|X_n|^2) = frac{1}{n^2} + frac{2}{n} + 1$ where $c = 1$ and $a_n = frac{1}{n^2} + frac{2}{n}$ fulfill your properties. Now consider the sequence of random variables $Z_n = (-1)^n X_n$. You get that $Z_n$ have the same properties, but $Z_n$ does not converge in measure/probability to any random variable; the two candidates being $1$ and $-1$.

Answered by Andrew Shedlock on November 18, 2021

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