Hi
I am having difficulty in understanding a convergence question. I was wondering if someone could please explain how I get started?
Thanks
two random objects converge in probability if in the limit they have the same density or distribution function.
In this case it should be pretty clear that
$\begin{align*}
&\lim \limits_{n \to \infty}~P_{X_n} = \\ \\
&\lim \limits_{n \to \infty}~\left \{ \dfrac{n-2}{n},~\dfrac 1 n,~\dfrac 1 n \right \} = \\ \\
&\{1,~0,~0\}
\end{align*}$
The distribution of $\theta$ is just $P[X=\theta]=1$
which is seen to be identical to the limiting distribution of $X_n$
and thus $X_n$ does converge to $\theta$ in probability
$X_n$ will converge in mean square to $\theta$ if
$\lim \limits_{n\to\infty}~E[(X_n - \theta)^2] = 0$
$\begin{align*}
&\lim \limits_{n\to\infty}~E[(X_n - \theta)^2] = \\ \\
&\lim \limits_{n\to \infty}~\left(\dfrac{n-2}{n}(\theta - \theta)^2 + \dfrac 1 n (\theta - (\theta+1))^2 + \dfrac 1 n (\theta - (\theta+n))^2\right) = \\ \\
&\lim \limits_{n\to \infty}~ \left(\dfrac 1 n (1^2 + n^2)\right) = \\ \\
&\lim \limits_{n\to \infty}~ n + \dfrac 1 n = \infty
\end{align*}$
$X_n$ does not converge to $\theta$ in mean square