Continuous mapping theorem: Difference between revisions

Revision as of 13:49, 10 February 2016

Revision as of 01:35, 17 September 2016

Line 72:

\operatorname{Pr}\Big(\lim_{n\to\infty}g(X_n) = g(X)\Big)

&\geq \operatorname{Pr}\Big(\lim_{n\to\infty}g(X_n) = g(X),\ X\notin D_g\Big) \\

&\geq \operatorname{Pr}\Big(\lim_{n\to\infty}X_n = X,\ X\notin D_g\Big) \\ = 1.

\end{align}</math>,

&\geq \operatorname{Pr}\Big(\lim_{n\to\infty}X_n = X\Big) - \operatorname{Pr}(X\in D_g) = 1-0 = 1.

because the intersection of two almost sure events is almost sure.

\end{align}</math>

By definition, we conclude that ''g''(''X_n'') converges to ''g''(''X'') almost surely.