It is known from the continuous-mapping theorem that if $X_n \to X$ in probability then $g(X_n) \to g(X)$.
I wonder if there is an elementary proof for the case that $g(x) = x^2$, i.e. showing that $X^2_n \to X^2$.
One approach is to use the fact that $X_n \to X$ in probability if and only for even subsequence $\{n_k\}$ there is a subsubequence $\{n_{k_j}\}$ such that $X_{n_{k_j}} \to X$ almost surely, so we can conclude that for this subsubsequence we have $X^2_{n_{k_j}} \to X^2$ almost surely and so $X^2_n \to X^2$ in probability.
Is there an even more elementary proof (using e.g. Markov, Cauchy-Schwartz , etc.) that directly shows that $\mathbb{P}[|X_n^2 – X^2| > \epsilon] \to 0$?
Best Answer
Let $\epsilon >0, \eta >0$ ($\epsilon <1$) and choose $M>1$ such that $P\{|X| >M\} <\eta$. Then $|X_n-X| \leq \frac {\epsilon} M$ and $|X| \leq M$ together imply that $|X_n^{2}-X^{2}|=|X_n-X|(|X_n-X|+2|X|) \leq \frac {\epsilon} M (1+2M)<3\epsilon$. Hence $P\{|X_n^{2}-X^{2}| >3\epsilon\} \leq P\{|X_n-X| > \frac {\epsilon} M\}+P\{|X| >M\}$. For $n$ sufficiently large we get $P\{|X_n^{2}-X^{2}| >3\epsilon\} <2\eta$. This proves that $X_n^{2} \to X^{2}$ in probability.