Convergence in Probability with a case environment

Question

Suppose $\left(X_{j}\right)_{j\ \geq\ 1}$ is a sequence of independent random variables such that $$X_{j} = \begin{cases} j^{3}, &\text{with probability}\ \frac{1}{j}; \\[1mm] 1, &\text{with probability}\ 1 - \frac{1}{j}. \end{cases} $$

Show that $\lim\limits_{j \to \infty}X_{j} = 1$ in probability, and that $E\left\{X_{j}^{2}\right\} \geq j$.

I suppose that what is throwing me of is the case environment. Please help !.

Answer 1

$X_j$ converging to $1$ in probability means that for any $\epsilon > 0$ we have

$$P( \lvert X_j - 1 \rvert > \epsilon ) \rightarrow 0$$

But $\lvert X_j - 1 \rvert$ is only possibly non-zero with probability $\frac{1}{j}$, so what does this tell us?

For the other part, $X_j^2 \ge 0$ and hence we have

$$E(X_j^2) \ge E(j^6 1_{X_j = j^3}) = j^6 \frac{1}{j} = j^5$$

which is bigger than $j$.

Answer 2

For the first part you can just use the definition of limit "in probability" http://en.wikipedia.org/wiki/Convergence_of_random_variables and for the second part you can see that E(X_j^2) = (1^2)*(1-1/j) + ((j^3)^2)*(1/j) = 1 - 1/j + j^5 >= j^5 >= j when j >= 1.