Convergence in probability: Does squeeze theorem apply? Does squeeze theorem apply in convergence in probability?
My statistics reference (where it talks about convergence in probability and its condition) does not cite it (but does seem to apply it), but it seems like it could apply.
What does the formulation look like for probabilities?
 A: Another version, just because I like the approach
$f_n\stackrel{p}{\to} L$ if and only if every subsequence $f_{n_k}$ has a subsubsequence $f_{n_{k_i}}$ that converges almost surely to $L$
Almost-sure convergence clearly satisfies the squeeze theorem.
If $X_n\leq Y_n\leq Z_n$ and $X_n, Z_n\stackrel{p}{\to} L$, then every subsequence has a subsubsequence where the convergence is almost-sure and thus along that subsubsequence $Y_n\stackrel{as}{\to} L$. But that is just to say $Y_n\stackrel{p}{\to}L$.
As an additional note: both these proofs actually only need $P(X_n\leq Y_n\leq Z_n)\to 1$.  You might argue that's the 'real' squeeze theorem in probability.
A: $f_n \xrightarrow{p} L$ means, for any $\epsilon > 0$, that
\begin{equation*}
\lim_{n \to \infty} \mathbb{P}\left( \left|f_n - L \right| < \epsilon \right) = 1.
\end{equation*}
If you have $X_n \leq Y_n \leq Z_n$ for every appropriate $n$ then $X_n - L \leq Y_n - L\leq Z_n - L$ for any $L$.  Take each of those differences to be nonnegative for simplicity; then for any $\epsilon > 0$,
\begin{equation*}
\left\{\left|Z_n - L\right| < \epsilon\right\} \subseteq \left\{\left|Y_n - L\right| < \epsilon\right\} \subseteq \left\{\left|X_n - L\right| < \epsilon\right\}.
\end{equation*}
You thus have that $\mathbb{P}(|Z_{n} - L| < \epsilon) \leq \mathbb{P}(|Y_n - L| < \epsilon) \leq \mathbb{P}(|X_n - L| < \epsilon)$, and the standard squeeze theorem gives you the result you're looking for.
