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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?

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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
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 $1-c$.

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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.

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