Suppose $\{a_n\}$ is a sequence of real numbers which converges to $a$, and $\{X_n\}$ is a sequence of random variables which converges to $X$ in probability. Prove that $a_nX_n \to aX$ in probability.

Through some work I deduced that $$P(|a_nX_n-aX| > \epsilon) \leq P(|a_n||X_n-X|>\epsilon/2)+P(|X||a_n-a|>\epsilon/2).$$

We know $|a_n|$ is bounded and so $P(|a_n||X_n-X|>\epsilon/2) \to 0$. However, it doesn't seem like we know anything about $|X|$ so I'm not sure what to do about the second term above. Any thoughts?

  • $\begingroup$ Note that $P(|X|> \epsilon/(2|a_n-a|))= P(|X|> k_n)$ where $k_n \to \infty$ and $X$ is a single random variable. Hence, this probability goes to $0$. [$P(|X|=\infty) = 0$] $\endgroup$ – Atbey Mar 8 '18 at 6:59

We can safely assume that $P(|X| = \infty) = 0$. Then $P\left(|X| > \frac{\epsilon}{2|a_n - a|}\right)$. If we ever have $|a_n - a| = 0$ then we'd have $P(0 > \epsilon/2) = 0$. Aside from those cases, though, $\frac{\epsilon}{2|a_n - a|} \to \infty$, and since the cdf $F(x)$ of any finite random variable converges to 1 as $x \to \infty$, we can say $P\left(|X| > \frac{\epsilon}{2|a_n - a|}\right) = 1 - P\left(|X| \leq \frac{\epsilon}{2|a_n - a|}\right) = 1 - F_{|X|}\left(\frac{\epsilon}{2|a_n - a|}\right) \to 0$ as $n \to \infty$.

This is what you needed. The rest of your argument checks out.

  • 1
    $\begingroup$ Another way is to note $\lim_n P(X > k_n) = P(\cap_n\{|X| > k_n\}) = P(|X| = \infty) = 0$. $\endgroup$ – Atbey Mar 8 '18 at 7:11

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