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I am working on a problem involving almost sure convergence and Rayleigh distributed random variables, and I need some help to confirm my understanding and approach. The problem is as follows:

Let $X_1, X_2, X_3, \ldots$ be a sequence of random variables such that $X_n \sim \operatorname{Rayleigh}\left(\frac{1}{n}\right)$, i.e., $$ f_{X_n}(x)= \begin{cases}n^2 x \exp \left\{-\frac{n^2 x^2}{2}\right\} & \text { if } x>0 \\ 0 & \text { otherwise. }\end{cases} $$

I need to show that $X_n \xrightarrow{\text { a.s. }} 0$. Here is my approach:

  1. I understand that to prove $X_n \xrightarrow{\text { a.s. }} 0$, I need to show that for any $\epsilon>0$, $$ \sum_{n=1}^{\infty} P\left(\left|X_n\right|>\epsilon\right)<\infty $$

  2. Using the given PDF, I computed the probability: $$ P\left(X_n>\epsilon\right)=\int_\epsilon^{\infty} n^2 x \exp \left\{-\frac{n^2 x^2}{2}\right\} d x \text {. } $$

  3. I made a change of variable $u=n x$, so $d u=n d x$, leading to: $$ P\left(X_n>\epsilon\right)=\int_{n \epsilon}^{\infty} u \exp \left\{-\frac{u^2}{2}\right\} \frac{d u}{n} $$

  4. Recognizing the integral as the tail of a Rayleigh distribution, which can be expressed in terms of the complementary error function erfc, I approximate: $$ P\left(X_n>\epsilon\right) \approx \exp \left\{-\frac{n^2 \epsilon^2}{2}\right\} $$

  5. I then check the series: $$ \sum_{n=1}^{\infty} \exp \left\{-\frac{n^2 \epsilon^2}{2}\right\} $$

Given the rapid decay of the exponential term, this series converges. Therefore, by the Borel-Cantelli Lemma, $X_n \xrightarrow{\text { a.s. }} 0$.

Could someone please verify if this approach and conclusion are correct? Is there any step I should reconsider or any additional insight needed to solidify the proof?

Thank you (Also, I am very new in probability theory)!

bruno
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  • Why approximate after step 2? Just compute the integral directly $$n^{2} x e^{- n^{2} x^{2} / 2} = - (e^{- n^{2} x^{2} / 2})' \implies P(X_{n} > \epsilon) = - e^{- n^{2} x^{2} / 2} \big\lvert_{\epsilon}^{\infty} = e^{- n^{2} \epsilon^{2} / 2}$$ – Matthew Cassell May 24 '24 at 10:59
  • @MatthewCassell can the expression be simplified directly using the fundamental theorem of calculus? – bruno May 24 '24 at 11:03

1 Answers1

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A simple calculation tells us that $$\mathbb{P}( X_1 > x ) = e^{ - x^2/2}.$$ It follows that $$\mathbb{P}( X_n > 1/\sqrt{n} ) = \mathbb{P}( X_1 > \sqrt{n} ) = e^{ - n/2}.$$ In particular, $$\sum_{n \geq 1} \mathbb{P}(X_n \geq 1/\sqrt{n}) < \infty.$$ The Borel Cantelli lemma states that if $(A_n)_{n \geq 1}$ is a sequence of events with $\sum_{n \geq 1} \mathbb{P}(A_n) < \infty$, finitely many of the $A_n$ occur almost-surely.

It follows in our setting that finitely many of the events $\{X_n > 1/\sqrt{n} \}$ occur, almost-surely. In other words, almost-surely there exists a random $n_0$ such that $\{ X_n \leq 1/\sqrt{n} ~ \forall n \geq n_0\}$, which indeed implies that $X_n \to 0$ almost-surely.

(The choice of $X_n \geq 1/\sqrt{n}$ was of no particular significance, we could have also chosen $X_n \geq \log n/n$ or $X_n \geq 1/n^{1/3}$.)

Samuel Johnston
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