1
$\begingroup$

If the sequences ${X_n}$ and $Y_n$ converge both almost surely and in $L^1$ to the constants $a$ and $b \ne 0$, respectively, and $Y_n$ is deterministically bounded below by some constant $0 < c \le b$, does this imply that $\mathbb{E}\left[\frac{X_n}{Y_n}\right]$ converges to $\frac{a}{b}$, i.e. does the ratio of $X_n$ and $Y_n$ converge in $L^1$ to $\frac{a}{b}$? [Follow-up: does $\frac{X_n}{Y_n}$ converge almost surely to $\frac{a}{b}$? Edit: Yes, the latter is a trivial consequence of the continuous mapping theorem.]

Of course, $\frac{X_n}{Y_n}$ converges in probability to $\frac{a}{b}$, but this does not imply convergence in $L^1$.

Intuitively, I think that it should indeed converge in $L^1$, but I am having a tough time taking a crack at this. I've tried using indicator variables, etc., but that does not seem to be the right approach.

I would love to know whether this is true, at least for "nice" sequences of $X_n$'s and $Y_n$'s [in particular, products and sums of independent Gaussians and finite integer higher powers of independent Gaussians]. I don't need the proof as long as someone can point me to a relevant resource, but if a proof is available I would of course love to see that as well.

$\endgroup$
  • $\begingroup$ You will need $b\neq 0$ for any kind of convergence. If $b\neq 0$, then $X_n/Y_n\rightarrow a/b$ almost surely since, with prob 1, we know the numerator converges to $a$ and the denominator to $b$. For the expectation, this ratio question can be simplified when $Y_n=1$ for all $n$. It reduces to the question "If $X_n\rightarrow 0$ with prob 1, does $E[X_n]\rightarrow 0$?" Can you answer this simplified question? $\endgroup$ – Michael Apr 9 '18 at 21:41
  • 1
    $\begingroup$ I didn't notice you had a constraint $X_n$ and $Y_n$ also converge in L1. Well what about the example $X_n=1$ and $Y_n=1+G/n$ for all $n \in \{1, 2, 3, ...\}$, where $G$ is Gaussian zero mean unit variance? Then $E[X_n/Y_n] = E[n/(n+G)]$ is not even defined. In general, the expectation $E[1/Y]$ is either infinity, -infinity, or undefined, if there is an $\epsilon>0$ such that the PDF of $Y$ satisfies $f_Y(y)\geq \epsilon$ over an interval that contains 0. $\endgroup$ – Michael Apr 10 '18 at 0:04
  • 1
    $\begingroup$ You're absolutely right. I am sorry, I should have been more precise. In my case, $Y_n$ is indeed lower bounded by 1, i.e. $f_y(y) = 0$ for all $y < 1$. (I realize it seems that I am just adding arbitrary constraints at this point, but in fact this is the case for the situation I am working on right now.) After thinking about it some more, it would be sufficient to show that $X_n$ converges in $L^2$ to $a$ and $1/Y_n$ converges in $L^2$ to $1/b$ [which are stronger assumptions than in my original post, but the former is true in my case, as is $Y_n$ converging in $L^2$ to $b$]. $\endgroup$ – user2258552 Apr 10 '18 at 2:30
  • 1
    $\begingroup$ [Second comment because I did not have space to write both.] So, the question becomes: if $Y_n$ converges a.s. and in $L^2$ to $b$, and is lower bounded by $k > 0$, does $1/Y_n$ converge in $L^2$ to $b$? I just tried proving a similar statement for convergence in $L^1$ by using the Law of Total Expectation to split into the cases $Y < b-\epsilon$, $|Y-b| < \epsilon$, and $Y>b+\epsilon$, for any sufficiently small fixed $\epsilon$, but did not even end up using the fact of almost sure convergence in my argument, which makes me somewhat skeptical. $\endgroup$ – user2258552 Apr 10 '18 at 2:35
  • 1
    $\begingroup$ No, you are right, total expectation works in that case. If $Z_n$ converges in probability to $z$, and if $Z_n$ is deterministically bounded (as are your $1/Y_n$ values) then $E[Z_n]\rightarrow z$. Almost sure convergence is not needed. $\endgroup$ – Michael Apr 10 '18 at 2:42
1
$\begingroup$

Recall that convergence in L1 implies convergence in probability. The bounded convergence theorem says that if $Z_n$ converges to a constant $z$ in probability, and if there is a finite constant $M$ such that $|Z_n|\leq M$ for all $n$, then $E[|Z_n-z|]\rightarrow 0$ and $E[Z_n]\rightarrow z$. (You indeed prove this via the law of total expectation.)

Claim:

Let $\{X_n\}_{n=1}^{\infty}, \{Y_n\}_{n=1}^{\infty}$ be such that $X_n$ converges to a constant $a$ in L1 (and hence in probability); $Y_n\geq 1$ for all $n$, and $Y_n$ converges in probability to a constant $b\geq 1$. Then $E[\frac{X_n}{Y_n}]\rightarrow a/b$.

Proof:

For all $n \in \{1, 2, 3, ...\}$ we have \begin{align} \frac{X_n}{Y_n} = \frac{a}{Y_n} + \frac{X_n-a}{Y_n} \implies E\left[\frac{X_n}{Y_n}\right] = E\left[\frac{a}{Y_n}\right] + E\left[\frac{X_n-a}{Y_n}\right] \end{align} Since a $|a/Y_n| \leq |a|$ for all $n$, and $a/Y_n$ converges to $a/b$ in probability, we get by the bounded convergence theorem that $E[a/Y_n]\rightarrow a/b$. Hence, it remains only to prove $E[(X_n-a)/Y_n]\rightarrow 0$. But this is true since:

$$ |E[(X_n-a)/Y_n]|\leq E[|X_n-a|/Y_n] \leq E[|X_n-a|] \rightarrow 0 $$ where the final limit holds because $X_n$ converges to $a$ in L1. $\Box$

$\endgroup$
  • $\begingroup$ I think the assumption $Y_n \geq 1$ is not reasonable. $\endgroup$ – Kavi Rama Murthy Apr 10 '18 at 6:48
  • $\begingroup$ The assumption was added in a comment before this answer. I will update the post to make it clear. $\endgroup$ – user2258552 Apr 10 '18 at 6:57
  • $\begingroup$ Not clear why there is a downvote. $\endgroup$ – Michael Apr 10 '18 at 13:45
  • $\begingroup$ The case $Y_n \geq c > 0$ can be treated the same by defining $\tilde{Y}_n = Y_n/c$ and noting that $\tilde{Y}_n\geq 1$ for all $n$. $\endgroup$ – Michael Apr 11 '18 at 1:00
1
$\begingroup$

This is false. Consider $(0,1)$ with Lebesgue measure and take $X_n \equiv 1$, $Y_n=\frac 1 {n^{2}}$ on $(0,\frac 1 n)$, $Y_n =1$ on $[\frac 1 n, 1)$. Then $X_n \to 1$ and $Y_n \to 1$ almost surely. But $E\frac {X_n} {Y_n} \geq \int_0^{1/n} n^{2}dx=n \to \infty$ Note that $E|Y_n -1| \leq \int _0 ^{1/n} |\frac 1 {n^{2}} -1|dx \leq \int _0 ^{1/n} 2dx \to 0$ so $Y_n \to 1$ in $L^{1}$.

$\endgroup$
  • $\begingroup$ (+1). However, counter-examples in this direction were already established in the comments. The question was gradually refined based on those comments. I wonder if you will upvote those comments as well as my answer? $\endgroup$ – Michael Apr 10 '18 at 15:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.