Convergence of sequence of random variables to infinity

Question

Suppose $X=X(\alpha)$ is a sequence of non-negative continuous random variables indexed by a continuous parameter $\alpha$ such that $\alpha$ is in $[0,\alpha_0)$ for some finite positive $\alpha_0$.

Throughout, let $\alpha\uparrow \alpha_0$ denote convergence to $\alpha_0$ from the left. Let $Y(\alpha)$ denote $(\alpha_0-\alpha)X(\alpha)$.

The following facts are known about $X(\cdot)$:

1. $X(\alpha_0)=+\infty$ a.s.
2. $Y(\alpha) \rightarrow 1$ in probability as $\alpha \uparrow \alpha_0$.
3. $E(X(\alpha))<+\infty$, $Var(X(\alpha))<+\infty$ for all $\alpha$ in $[0,\alpha_0)$.
4. $E(Y(\alpha))\rightarrow 1$ as $\alpha \uparrow \alpha_0$.
5. $Var(Y(\alpha))\rightarrow 0$ as $\alpha \uparrow \alpha_0$.

Note that 4.-5. imply 2.

Question: Is evidence 1.-5. sufficient to claim that $X(\alpha)\rightarrow+\infty$ in probability as $\alpha \uparrow \alpha_0$?

If yes, will the stronger fact be also valid that $X(\alpha)\rightarrow+\infty$ a.s. as $\alpha \uparrow \alpha_0$?

Answer 1

The answer to the first part of the question is that statement 2 on its own suffices to show the convergence in probability of $X(\alpha)$ to $\infty$ as $\alpha \to \alpha_0$ :

$Y(\alpha) \to 1$ in probability as $\alpha\to\alpha_0$ says that $(\alpha_0 - \alpha)X(\alpha) \to 1$ in probability, so we have

$$\mathbb{P}\left(\frac{1}{2} < (\alpha_0 - \alpha).X(\alpha) < \frac{3}{2}\right) \to 1, \;\text{as}\; \alpha\to\alpha_0.$$

Then for any $M>0$ we can find $\alpha_1 \in [0, \alpha_0)$ so that for all $\alpha \geq \alpha_1$: $M\leq \frac{1/2}{\alpha_0 - \alpha}$.

Putting this together we have

$$ \forall M>0: \mathbb{P}\left( M < X(\alpha) \right) \geq \mathbb{P}\left( \frac{1/2}{\alpha_0 - \alpha_1} < X(\alpha) \right) \geq \mathbb{P}\left( \frac{1/2}{\alpha_0 - \alpha} < X(\alpha) \right) \to 1$$