Does $\frac{X_n}{Y_n}\xrightarrow[]{p}0$ if $Y_n$ converges in probability to infinity?

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Let $X_n$ and $Y_n$ be random variables such that $X_n\xrightarrow[]{p} \alpha$, where $\alpha$ is a constant, and $Y_n$ diverges in probability to infinity.

When can we say that:

$$\frac{X_n}{Y_n}\xrightarrow[]{p}0$$

I know that is $b_n$ is some nonrandom sequence that diverges then the ratio $\frac{X_n}{b_n}$ converges in probability to zero, but I'm not sure if this is enough if the sequence in the denominator is random.

Here's my attempt:

We know that $Y_n$ diverges to infinity in probability which implies for every $\delta>0$, $\mathbb{P}(\vert Y_n\vert\leq \delta)\xrightarrow[]{}0$.

Since we have:

$\mathbb{P}(\frac{1}{\vert Y_n\vert}> \epsilon) = \mathbb{P}(\frac{1}{\epsilon}> \vert Y_n\vert)\rightarrow 0$ as $n\rightarrow \infty$, assuming $\vert Y_n\vert$ almost surely not equal to zero.

This shows $Z_n = \frac{1}{Y_n}\xrightarrow[]{p} 0$ and $X_n\xrightarrow[]{p} \alpha$, so by Slutsky's theorem we have:

$$Z_nX_n = \frac{1}{Y_n}X_n \xrightarrow[]{p} 0$$