Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $X_1, X_2, \ldots$ converge in probability to a constant $c$, then does $1-X_1, 1-X_2, \ldots$ converge in probability to $1-c$? Is there a way to show this is true / is there an already existent theorem for this?

share|improve this question
2  
This should be obvious from the definition. If $X_i$ is within $\epsilon$ of $c$, then $1-X_i$ will be within $\epsilon$ of $1-c$. –  ShawnD Oct 16 '11 at 7:24
add comment

2 Answers 2

up vote 6 down vote accepted

If $X_n$ converges in probability to a constant $c$, then this means that

$$\forall \epsilon \in \mathbb{R}^+_0 : \lim_{n \to \infty} \mathbb{P}(|X_n-c|>\epsilon) = 0 \; .$$

If you rewrite:

$$|X_n - c| = |c-X_n| = |c-1+1-X_n| = |(1-X_n)-(1-c)| \; ,$$

then you see that $1-X_n$ also converges in probability to $1-c$.

share|improve this answer
add comment

If $g$ is any continuous function, and $X_n \to X$ in probability, then $g(X_n) \to g(X)$ in probability. This is sometimes called the continuous mapping theorem.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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