3
$\begingroup$

I'm trying to prove the following proposition, which is another version of Dominated Convergence Theorem in Probability Theory:

Suppose $X_{n}\overset{p}\rightarrow X,$ i.e., $X_{n}\rightarrow X$ in probability, and there is a continous function $g$ with $g(x)>0$ for large $x$ with $\frac{|x|}{g(x)}\rightarrow 0$ as $|x|\rightarrow\infty$ so that $E(g(X_{n}))\leq C<\infty$ for all $n.$ Then $E(X_{n})\rightarrow E(X).$

My attempt is based in the convergence of $\frac{|x|}{g(x)}$ to $0.$ So, for an $\epsilon=1$ we have $|x|< g(x)$ for $x$ large. Then $E(|X_{n}|)<E(g(X_{n}))\leq C.$ Then $X_{n}$ are integrable and because of continuity of $g$ $g(X_{n})\rightarrow g(X)$ in probability. So $E(g(X_{n}))\rightarrow E(g(X)),$ that is because of Dominated Convergence Theorem with almost sure convergence.

But I don't know how to use this to prove the convergence of $E(X_{n})$ to $E(X).$

Any kind of help is thanked in advance.

$\endgroup$
2
$\begingroup$

For any $\epsilon$ and $R$, \begin{align} \left|X_n-X\right|&=\left|X_n-X\right|\mathbf 1\{\left|X_n-X\right|\leqslant \epsilon\}+ \left|X_n-X\right|\mathbf 1\{\left|X_n-X\right|\gt \epsilon\}\\ &\leqslant \epsilon+\left|X_n\right|\mathbf 1\{\left|X_n-X\right|\gt \epsilon\}+\left|X\right|\mathbf 1\{\left|X_n-X\right|\gt \epsilon\}\\ &\leqslant \epsilon+\left|X_n\right|\mathbf 1\{\left|X_n\right|\gt R\}+ R\mathbf 1\{\left|X_n-X\right|\gt \epsilon\}+\left|X\right|\mathbf 1\{\left|X\right|\gt R\} +R\mathbf 1\{\left|X_n-X\right|\gt \epsilon\}\\ \tag{*}\left|X_n-X\right|&\leqslant \epsilon+\left|X_n\right|\mathbf 1\{\left|X_n\right|\gt R\}+ 2R\mathbf 1\{\left|X_n-X\right|\gt \epsilon\}+\left|X\right|\mathbf 1\{\left|X\right|\gt R\}. \end{align} Note that $$\mathbb E\left[\left|X_n\right|\mathbf 1\{\left|X_n\right|\gt R\}\right] \leqslant \sup_{t\geqslant R}\frac t{g(t)} \mathbb E\left[g\left(X_n\right)\right]\leqslant C\sup_{t\geqslant R}\frac t{g(t)},$$ hence, integrating on both sides of (*), we get $$\mathbb E\left[\left|X_n-X\right|\right] \leqslant \epsilon+2R\cdot \mathbb P\left(\left\{\left|X_n-X\right|\gt \epsilon\right\}\right)+\mathbb E\left[\left|X\right|\mathbf 1\{\left|X\right|\gt R\}\right].$$ Using the convergence in probability of $\left(X_n\right)_{n\geqslant 1}$ to $X$,we derive that for any $R$ and $\epsilon$, $$\mathbb E\left[\left|X_n-X\right|\right] \leqslant \epsilon+\mathbb E\left[\left|X\right|\mathbf 1\{\left|X\right|\gt R\}\right].$$ To conclude, note that $\lim_{R\to +\infty}\mathbb E\left[\left|X\right|\mathbf 1\{\left|X\right|\gt R\}\right]=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.