1. ## convergences in probability

Hi

I am having difficulty in understanding a convergence question. I was wondering if someone could please explain how I get started?

Thanks

2. ## Re: convergences in probability

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

3. ## Re: convergences in probability

The way you've written it makes a lot of sense. I just couldn't make sense of it from the lecture notes.

Thank you