Let $\{X_n\}_{n\geq 1}$ be a sequence of random variables s.t $X_n \to a$ in probability, where $a>0$ is a fixed constant. I'm trying to show the following: $$\sqrt{X_n} \to \sqrt{a}$$ and $$\frac{a}{X_n}\to 1$$ both in probability. I'm here to see if my logic was sound. Here's my work


For the first part, we have $$|\sqrt{X_n}-\sqrt{a}|<\epsilon \impliedby |X_n-a|<\epsilon|\sqrt{X_n}+\sqrt{a}|=\epsilon|(\sqrt{X_n}-sqrt{a})+2\sqrt{a}|$$ $$\leq \epsilon|\sqrt{X_n}-\sqrt{a}|+2\epsilon\sqrt{a}<\epsilon^2+2\epsilon\sqrt{a}$$ Notice that $$\epsilon^2+2\epsilon\sqrt{a}>\epsilon\sqrt{a}$$ It follows then that $$P(|\sqrt{X_n}-\sqrt{a}|\leq \epsilon)\geq P(|X_n-a|\leq \epsilon\sqrt{a})\to 1 \;\;as\;n\to\infty$$ $$\implies \sqrt{X_n}\to\sqrt{a} \;\;in\;probability$$

For the second part, we have $$|\frac{a}{X_n}-1|=|\frac{X_n-a}{X_n}|<\epsilon \impliedby |X_n-a|<\epsilon|X_n|$$ Now, since $X_n \to a$ as $n \to \infty$, we have that $X_n$ is a bounded sequence. In other words, there exists a real number $M<\infty$ s.t $|X_n|\leq M$. Thus, $$|X_n-a|<\epsilon|X_n|\impliedby |X_n-a|<\epsilon M$$ Looking at it in probability, we have $$P(|\frac{a}{X_n}-1|>\epsilon)=P(|X_n-a|>\epsilon|X_n|)\leq P(|X_n-a|>\epsilon M)\to 0 \;\;as\;n\to\infty$$

I'm pretty confident in the first one, but am pretty iffy on the second. Was my logic sound?

  • 6
    $\begingroup$ Consider the sequence $X_n$ where $\Pr(X_n=a)=1-1/n$ and $\Pr(X_n=n)=1/n$. It seems to me that since $1-1/n\to 1$ this sequence converges to $a$ in probability, but clearly it is unbounded since $\sup(X_n)=\max(a,n)\to\infty$. $\endgroup$
    – whuber
    Commented Apr 7, 2015 at 17:41
  • 2
    $\begingroup$ Continuous mapping theorem? $\endgroup$ Commented Apr 9, 2015 at 19:55

3 Answers 3


The details of the proofs matter less than developing appropriate intuition and techniques. This answer focuses on an approach designed to help do that. It consists of three steps: a "setup" in which the assumption and definitions are introduced; the "body" (or a "crucial step") in which the assumptions are somehow related to what is to be proven, and the "denouement" in which the proof is completed. As in many cases with probability proofs, the crucial step here is a matter of working with numbers (the possible values of random variables) rather than dealing with the much more complicated random variables themselves.

Convergence in probability of a sequence of random variables $Y_n$ to a constant $a$ means that no matter what neighborhood of $0$ you pick, eventually each $Y_n-a$ lies in this neighborhood with a probability that is arbitrarily close to $1$. (I won't spell out how to translate "eventually" and "arbitrarily close" into formal mathematics--anybody interested in this post already knows that.)

Recall that a neighborhood of $0$ is any set of real numbers containing an open set of which $0$ is a member.

The setup is routine. Consider the sequence $Y_n = a/X_n$ and let $\mathcal{O}$ be any neighborhood of $0$. The objective is to show that eventually $Y_n-1$ will have an arbitrarily high chance of lying in $\mathcal{O}$. Since $\mathcal{O}$ is a neighborhood, there must be an $\epsilon \gt 0$ for which the open interval $(-\epsilon, \epsilon)\subset \mathcal{O}$. We may shrink $\epsilon$ if necessary to ensure $\epsilon \lt 1$, too. This will assure that subsequent manipulations are legitimate and useful.

The crucial step will be to connect $Y_n$ with $X_n$. That requires no knowledge of random variables at all. The algebra of numeric inequalities (exploiting the assumption $a\gt 0$) tells us that the set of numbers $\{Y_n(\omega)\,|\, Y_n(\omega) - 1 \in (-\epsilon, \epsilon)\}$, for any $\epsilon \gt 0$, is in one-to-one correspondence with the set of all $X_n(\omega)$ for which

$$\frac{a}{1+\epsilon} \lt X_n(\omega) \lt \frac{a}{1-\epsilon}.$$


$$X_n(\omega)-a \in \left(-\frac{a\epsilon}{1+\epsilon}, \frac{a\epsilon}{1-\epsilon}\right) = \mathcal{U}.$$

Since $a\ne 0$, the right hand side $\mathcal{U}$ indeed is a neighborhood of $0$. (This clearly shows what breaks down when $a=0$.)

We are ready for the denouement.

Because $X_n \to a$ in probability, we know that eventually each $X_n-a$ will lie within $\mathcal{U}$ with arbitrarily high probability. Equivalently, $Y_n-1$ will eventually lie within $(-\epsilon,\epsilon) \subset \mathcal{O}$ with arbitrarily high probability, QED.

  • $\begingroup$ I apologize for such a late best answer. It's been a busy week. Thank you so for much this!!! $\endgroup$ Commented Apr 19, 2015 at 4:46

For the first part, take $x,a,\epsilon>0$, and note that $$ \begin{align} |\sqrt{x}-\sqrt{a}|\geq\epsilon &\Rightarrow |\sqrt{x}-\sqrt{a}|\geq\epsilon \frac{\sqrt{a}}{\sqrt{a}} \Rightarrow |\sqrt{x}-\sqrt{a}|\geq\frac{\epsilon\sqrt{a}}{\sqrt{x}+\sqrt{a}} \\ &\Rightarrow |(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})|\geq\epsilon\sqrt{a}\Rightarrow |x-a|\geq\epsilon\sqrt{a}. \end{align} $$ Hence, for any $\epsilon>0$, defining $\delta=\epsilon\sqrt{a}$, we have $$ \Pr(|\sqrt{X_n}-\sqrt{a}|\geq\epsilon)\leq\Pr(|X_n-a|\geq\delta)\to 0, $$ when $n\to\infty$, implying that $\sqrt{X_n}\stackrel{\Pr}\to\sqrt{a}$.

For the second part, take again $x,a,\epsilon>0$, and cheat from Hubber's answer (this is the key step ;-) to define $$ \delta = \min\left\{ \frac{a\epsilon}{1+\epsilon},\frac{a\epsilon}{1-\epsilon}\right\}. $$ Now, $$ \begin{align} |x-a|<\delta &\Rightarrow a-\delta<x<a+\delta \Rightarrow a - \frac{a\epsilon}{1+\epsilon} < x < a + \frac{a\epsilon}{1-\epsilon} \\ &\Rightarrow \frac{a}{1+\epsilon} < x < \frac{a}{1-\epsilon} \Rightarrow 1-\epsilon<\frac{a}{x}<1+\epsilon \Rightarrow \left| \frac{a}{x}-1\right|<\epsilon. \end{align} $$ The contrapositive of this statement is $$ \left| \frac{a}{x}-1\right|\geq\epsilon \Rightarrow |x-a|\geq\delta. $$

Therefore, $$ \Pr\!\left(\left|\frac{a}{X_n}-1\right|\geq\epsilon\right)\leq\Pr(|X_n-a|\geq\delta)\to 0, $$ when $n\to\infty$, implying that $\frac{a}{X_n}\stackrel{\Pr}\to 1$.

Note: both items are consequences of a more general result. First of all remember this Lemma: $X_n\stackrel{\Pr}{\to}X$ if and only if for any subsequence $\{n_i\}\subset\mathbb{N}$ there is a subsequence $\{n_{i_j}\}\subset\{n_i\}$ such that $X_{n_{i_j}}\to X$ almost surely when $j\to\infty$. Also, remember from Real Analysis that $g:A\to\mathbb{R}$ is continuous at a limit point $x$ of $A$ if and only if for every sequence $\{x_n\}$ in $A$ it holds that $x_n\to x$ implies $g(x_n)\to g(x)$. Hence, if $g$ is continuous and $X_n\to X$ almost surely, then $$ \Pr\!\left(\lim_{n\to\infty}g(X_n)=g(X)\right)\geq\Pr\!\left(\lim_{x\to\infty}X_n=X\right)=1, $$ and it follows that $g(X_n)\to g(X)$ almost surely. Moreover, $g$ being continuous and $X_n\stackrel{\Pr}{\to}X$, if we pick any subsequence $\{n_i\}\subset\mathbb{N}$, then, using the Lemma, there is a subsequence $\{n_{i_j}\}\subset\{n_i\}$ such that $X_{n_{i_j}}\to X$ almost surely when $j\to\infty$. But then, as we have seen, it follows that $g(X_{n_{i_j}})\to g(X)$ almost surely when $j\to\infty$. Since this argument holds for every subsequence $\{n_i\}\subset\mathbb{N}$, using the Lemma in the other direction, we conclude that $g(X_n)\stackrel{\Pr}{\to}g(X)$. Hence, to answer your question you can just define continuous functions $g(x)=\sqrt{x}$ and $h(x)=a/x$, for $x>0$, and apply this result.

  • $\begingroup$ Zen thank you for you answer. This was very clear! $\endgroup$ Commented Apr 19, 2015 at 4:53

We are given that

$$\lim_{n \rightarrow \infty}P(|X_n-\alpha| > \epsilon) = 0$$

and we want to show that

$$\lim_{n \rightarrow \infty}P\left(\left|\frac {\alpha}{X_n}-1\right| > \epsilon\right) = 0$$

We have that

$$\left|\frac {\alpha}{X_n}-1\right| = \left|\frac {1}{X_n}(\alpha-X_n)\right| = \left|\frac {1}{X_n}\right|\left|X_n-\alpha\right|$$

So equivalently, we are examining the probability limit

$$\lim_{n \rightarrow \infty}P\left(\left|\frac {1}{X_n}\right|\left|X_n-\alpha\right| > \epsilon\right) = \;?\;0$$

We can break the probability into two mutually exclusive joint probabilities

$$P\left(\left|\frac {1}{X_n}\right|\left|X_n-\alpha\right| > \epsilon\right) =\\ P\left(\left|\frac {1}{X_n}\right|\left|X_n-\alpha\right| > \epsilon,|X_n| \geq 1 \right) + P\left(\left|\frac {1}{X_n}\right|\left|X_n-\alpha\right| > \epsilon,|X_n| < 1 \right)$$

For the first element we have the series of inequalities

$$P\left(\left|\frac {1}{X_n}\right|\left|X_n-\alpha\right| > \epsilon,|X_n| \geq 1 \right) \leq P\big[\left|X_n-\alpha\right| > \epsilon,|X_n| \geq 1 \big] \leq P\big[\left|X_n-\alpha\right| > \epsilon \big]$$

The first inequality comes from the fact that we are considering the region where $|X_n|$ is higher than unity and so its reciprocal is smaller than unity. The second inequality because a joint probability of a set of events cannot be greater than the probability of a subset of these events.
The limit of the rightmost term is zero (this is the premise), so the limit of the leftmost term is also zero. So the first element of the probability that interests us is zero.

For the second element we have

$$P\left(\left|\frac {1}{X_n}\right|\left|X_n-\alpha\right| > \epsilon,|X_n| < 1 \right) = P\left(\left|X_n-\alpha\right| > \epsilon|X_n|,|X_n| < 1 \right)$$

Define $\delta \equiv \epsilon\cdot \max |X_n|$. Since here $|X_n|$ is bounded, it follows that $\delta$ can be made arnitrarily small or large, and so it is equivalent to $\epsilon$. So we have the inequality

$$P\big[\left|X_n-\alpha\right| > \delta,|X_n| < 1 \big] \leq P\big[\left|X_n-\alpha\right| > \delta \big]$$

Again, the limit on the right side is zero by our premise, so the limit on the left side is also zero. Therefore the second element of the probability that interests us is also zero. QED.


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