How to find to which random variable the sequence converges in probability. Is there almost everywhere convergence?

Question

Given a sequence of random variables $\{\xi_n\}$ \begin{equation*} \xi_n(\omega) = \begin{cases} -1 &\text{with probability $\frac{1}{2n}$}\\ 3 &\text{with probability $\frac{1}{3n}$}\\ 10 &\text{otherwise} \end{cases} \end{equation*}

1. I need to find to which random variable the sequence converges in probability.

$\frac{1}{2n} \rightarrow 1$, as $n \rightarrow \infty$

Similarly for $\frac{1}{3n}$:

$\frac{1}{3n} \rightarrow 1$, as $n \rightarrow \infty$

$\Rightarrow P(|\xi_n - \xi| \geq \epsilon) \rightarrow 0$

2. And define is there almost everywhere convergence or not.

Sequence $\{\xi_n\}$ converges almost everywhere to a random variable $\xi$ as $n \rightarrow \infty$ if $P(\{\omega|\xi_n(\omega) \rightarrow \xi(\omega) \}) \rightarrow 1$.

The answer is yes, because:

$\frac{1}{2n} \rightarrow 1$, as $n \rightarrow \infty$

and

$\frac{1}{3n} \rightarrow 1$, as $n \rightarrow \infty$

$\Rightarrow P(\{\omega|\xi_n(\omega) \rightarrow \xi(\omega) \}) \rightarrow 1$

Are my assumptions correct?

Answer 1

Note $$E[\xi_n]=-\frac{1}{2n}+\frac{1}{n}+10\bigg(1-\frac{1}{2n}-\frac{1}{3n}\bigg)=10-11\frac{1}{2n}-7\frac{1}{3n}\to 10$$ $$E[|\xi_n-E[\xi_n]|^2]=(-1-E[\xi_n])^2\frac{1}{2n}+(3-E[\xi_n])^2\frac{1}{3n}+(10-E[\xi_n])^2\bigg(1-\frac{1}{2n}-\frac{1}{3n}\bigg)\to 0$$ We then employ Chebyscev's inequality and Minkowski's inequality: $$\begin{aligned}P(|\xi_n-10|>\varepsilon)^{1/2}&\leq \varepsilon^{-1}E[|\xi_n-10|^2]^{1/2}\leq\\ &\leq \varepsilon^{-1}(E[|\xi_n-E[\xi_n]|^2]^{1/2}+|E[\xi_n]-10|)\to 0\end{aligned}$$