# Verify that Convergence in Probability Proof is Correct

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 $$Y = \sqrt{X}$$.

$$\textbf{Proof}$$.

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.

• 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. – Michael Jul 12 at 16:13
• 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. – Michael Jul 12 at 16:26