# Convergence in $L^1$ of ratio of random variables

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.

• 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? – Michael Apr 9 '18 at 21:41
• 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. – Michael Apr 10 '18 at 0:04
• 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 $b$]. – user2258552 Apr 10 '18 at 2:30
• [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. – user2258552 Apr 10 '18 at 2:35
• 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. – Michael Apr 10 '18 at 2:42

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. $\Box$

• I think the assumption $Y_n \geq 1$ is not reasonable. – Kavi Rama Murthy Apr 10 '18 at 6:48
• The assumption was added in a comment before this answer. I will update the post to make it clear. – user2258552 Apr 10 '18 at 6:57
• Not clear why there is a downvote. – Michael Apr 10 '18 at 13:45
• 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 $n$. – Michael Apr 11 '18 at 1:00

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 $L^{1}$.

• (+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? – Michael Apr 10 '18 at 15:08