This is a slightly modified version of the problem found in Berger & Casella "Statistical Inference" (5.32). I think my proof is correct, but I'd like to be sure I did not mess up anything.

Let $X_i$ be a sequence of random variables such that $\mathbb{P}[X_i > 0] = 1 \ \forall i$, and the sequence converges in probability to a random variable $X$ which is also positive almost surely. Prove that $Y_i = \sqrt{X_i}$ converges in probability to $Y = \sqrt{X}$.


First, we have $|\sqrt{X_n} - \sqrt{X}| \leq \sqrt{|X_n - X|}$.

Fix $\epsilon > 0$.

From that, we can write: $$\mathbb{P}[|\sqrt{X_n} - \sqrt{X}| < \epsilon] \geq \mathbb{P}[|\sqrt{X_n} - \sqrt{X}| < \sqrt{\epsilon}] \geq \mathbb{P}[|\sqrt{|X_n - X|} < \sqrt{\epsilon}] = \mathbb{P}[|X_n - X| < \epsilon]$$

Since $\mathbb{P}[|X_n - X| < \epsilon] = 1$ eventually $\mathbb{P}[|\sqrt{X_n} - \sqrt{X}| < \epsilon] = 1$ eventually, completing the proof.

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    $\begingroup$ Note that if $0<\epsilon<1$ then $\sqrt{\epsilon}>\epsilon$, which means the inequality after “from that we can write” is flipped. To fix that, don’t bother to convert to the square root of epsilon. $\endgroup$ – Michael Jul 12 at 16:13
  • $\begingroup$ Also, your use of the word “eventually” at the end is not correct. You need to use a limit. You may also want to justify your very first inequality of the proof, though it is indeed correct. $\endgroup$ – Michael Jul 12 at 16:26

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