Convergence in Probability of Empirical Median I'm stuck with this one.
Let $X_1,...X_n$ be an i.i.d. sequence of random variables with CDF F. The empirical CDF of $X_i$ is defined
$$
\hat F_n(x) = \frac{1}{n} \sum_{1 \leq i \leq n} I\{X_I \leq x \}.
$$
Note that for $x \in \Re$, $\hat F_n(x) \xrightarrow{p} F(x)$. Also, define the smallest median of P as
$$
\theta_0 = \inf \{ x \in \Re : F(x) \geq .5 \}.
$$
Suppose that the median is unique, i.e., for any $\epsilon > 0$, $P(X_i \leq \theta_0 + \epsilon) > .5$. Define an estimator $\hat \theta_n$ of $\theta_0$ by
$$
\hat \theta_0 = \inf \{x \in \Re : \hat F_n(x) \geq .5 \}.
$$
Show that $\hat \theta \xrightarrow{p} \theta_0$.
 A: We wish to show that $\hat \theta_n \xrightarrow{P} \theta_0$. By definition,
this states that given $\epsilon > 0$
$$
P(|\hat \theta_n - \theta_0| > \epsilon) \rightarrow 0 \text{ as } n \rightarrow \infty.
$$
Given $X_1,..., X_n$ are iid with finite $n$, we see that if we were to
reorder $\{X_i\}$ in ascending order, the definition of
$\hat \theta_n$ gives
$\hat \theta_n = X_{\lceil n/2 \rceil}$.
From this we have that
$$
\hat F_n(\hat \theta_n) = \left \{
\begin{array}{ll}
1/2 & \text{ if n is even} \\
\frac{1}{2} + \frac{1}{2n} & \text{ if n is odd.}
\end{array}
\right.
$$
Using this, we can calculate
\begin{align}
P(|\hat \theta_n - \theta_0| > \epsilon) 
    &\leq P(|\hat \theta_n - \theta_0| \geq \epsilon) \\
    &= P(\hat \theta_n \leq \theta_0 - \epsilon) + P(\theta_n + 
    \epsilon \leq \hat \theta_n) \\
    &= P(\hat F_n(\hat \theta_n) \leq \hat F_n(\theta_0 - \epsilon)) + 
    P(\hat F_n(\theta_n + \epsilon) \leq \hat F_n(\hat \theta_n)) 
\end{align}
where the last equality is due to the fact that $\hat F_n(x)$ is non-decreasing.
Notice that because $\hat F_n(x)$ is non-decreasing and not strictly
increasing, it can only preserve weak inequalities.
We continue by analyzing the terms individually. First, consider the term
$$
P(\hat F_n(\hat \theta_n) \leq \hat F_n(\theta_0 - \epsilon)).
$$
When $n$ is even,
\begin{align*}
P(1/2 \leq \hat F_n(\theta_0 - \epsilon))
    &= P(1/2 - F(\theta_0 - \epsilon)) \leq \hat F_n(\theta_0 - \epsilon) - 
    F(\theta_0 - \epsilon))\\
    &\rightarrow 0 \text{ as } n \rightarrow \infty
\end{align*}
where this last fact follows from the fact that 
$1/2 - F(\theta_0 - \epsilon) > 0 $ (we can see this from the defintion of the median) and that 
$\hat F_n(x) \xrightarrow{p} F(x)$, as we demonstrated previously.
Similarly, when $n$ is odd,
\begin{align*}
P(1/2 + \frac{1}{2n} \leq \hat F_n(\theta_0 - \epsilon))
    &= P(1/2 + \frac{1}{2n} - F(\theta_0 - \epsilon)) \leq \hat F_n(\theta_0 - \epsilon) - 
    F(\theta_0 - \epsilon))\\
    &\rightarrow 0 \text{ as } n \rightarrow \infty.
\end{align*}
Now consider the term
$$
P(\hat F_n(\theta_n + \epsilon) \leq \hat F_n(\hat \theta_n)).
$$
Similar to before, we have
\begin{align*}
P(\hat F_n(\theta_n + \epsilon) \leq \hat F_n(\hat \theta_n)) 
    &= P(-\hat F_n(\theta_n + \epsilon) \geq -\hat F_n(\hat \theta_n)) \\
    &= P(F(\theta_n + \epsilon) -\hat F_n(\theta_n + \epsilon) \geq 
    F(\theta_n + \epsilon) -\hat F_n(\hat \theta_n)) \\
    &\rightarrow 0 \text{ as } n \rightarrow \infty.
\end{align*}
This last fact comes again from the fact that that $\hat F_n(x) \xrightarrow{p} F(x)$ and that $F(\theta_n + \epsilon) -\hat F_n(\hat \theta_n) > 0$, which we have because the uniqueness of the median gives that for $\epsilon > 0$ we have $F(\theta_0 + \epsilon) > .5$.
These two results, applied back give
\begin{align*}
P(|\hat \theta_n - \theta_0| > \epsilon) 
    &\leq P(|\hat \theta_n - \theta_0| \geq \epsilon) \nonumber \\
    &\rightarrow 0 \text{ as } n \rightarrow \infty
\end{align*}
and thus
$$
\hat \theta_n \xrightarrow{p} \theta_0.
$$
