TITLE: Question on limit in probability of the ratio of max to min of 2 sequences of non-ive, continuous iid random variables with support $[0, \infty).$
QUESTION [0 upvotes]: For each $ m \ge 1$, let $X_m$ and $Y_m$ be two non-negative iid random variables with the same distribution. (The distributions of $X_m$ may change with different $m$.)
**Assume that their support of $X_m$ is $[0, \infty),$ so we're not considering point mass situation.
Denote by $\to_{p}$ the convergence in probability.
Suppose $$\frac{\max(X_m,Y_m)}{\min(X_m,Y_m)} \to _{p}1\ \text{ as }\ m \to \infty.$$ Does this imply $$\frac{X_m}{\mathbb{E}X_m}\to _{p}1\ \text{ as }\ m \to \infty?$$
The reason I think it's true is that I feel:
Assumption I: $$\frac{\max(X_m,Y_m)}{\min(X_m,Y_m)} \to _{p}1\ \text{ as }\ m \to \infty \iff \frac{X_m}{Y_m} \to_{p} 1 \text{ as }\ m \to \infty  $$ (I need to check this!).
If the above Assumption I is correct, then we can look at Equation (18)-(20) of this link, where they calculated the approximation for $Var[\frac{X_m}{Y_m}], $ which clearly becomes almost equal to $\frac{Var[X_m]}{\mathbb{E}X_m^2}.$ Now since by Assumption I, $\frac{X_m}{Y_m} \to _{p}1\ \text{ as }\ m \to \infty,$ this means $Var[\frac{X_m}{Y_m}] \to 0,$ (EDIT: this is clearly wrong) but by the above link, this also means: $\frac{Var[X_m]}{\mathbb{E}X_m^2} \to 0, m \to \infty,$ partially establishing my claim, partially because the Assumption I still remains to be proved (but I think it's true!).

REPLY [3 votes]: The answer is no, and the additional support/non-discreteness condition is of no help.
Consider e.g. the following modification of the counterexample in Matt F.'s comment. Suppose that $P(U_m=0)=1/m=P(U_m=m^2)$ and $P(U_m=1)=1-2/m$ for $m\ge2$. Let $V_m$ be independent of $U_m$ and have the exponential distribution with mean equal $1/m$. Let $X_m:=U_m+V_m$. Then the distribution of $X_m$ is absolutely continuous and its support is $[0,\infty)$.
Moreover, $U_m\to1$ and $V_m\to0$ (in probability, as $m\to\infty$), and hence $X_m\to1$ and $$\frac{\max(X_m,Y_m)}{\min(X_m,Y_m)}\to1.$$
On the other hand, $EX_m>EU_m>m\to\infty$, so that $X_m/EX_m\to0$.

$\newcommand\de\delta\newcommand\ep\varepsilon$
Here is the good news: you desired result will hold under the additional assumption that
$$EU_m^p\le C \tag{1}$$
for some real $p>1$, some real $C>0$, and all $m$, where
$$U_m:=X_m/EX_m.$$
Introduce also
$$V_m:=Y_m/EY_m=Y_m/EX_m,$$
so that $U_m$ and $V_m$ are iid (for each $m$) and
$$R_m:=\frac{\max(X_m,Y_m)}{\min(X_m,Y_m)}=\frac{\max(U_m,V_m)}{\min(U_m,V_m)}\to1. \tag{2}$$
We need to show that $U_m\to1$. Suppose the contrary. Then, passing to a subsequence, we see that without loss of generality at least one of the following two cases takes place:
Case 1: $P(U_m>c)>\de$ for some real $c>1$, some real $\de>0$, and all $m$
Case 2: $P(U_m<b)>\de$ for some $b\in(0,1)$, some real $\de>0$, and all $m$
Consider first Case 1. Then $0=E(U_m-1)=E(U_m-1)1(U_m>1)-E(1-U_m)1(U_m<1)$, whence
$$(c-1)\de\le(c-1)P(U_m>c)=E(c-1)I(U_m>c)\le E(U_m-1)1(U_m>c)\le E(U_m-1)1(U_m>1)=E(1-U_m)1(U_m<1)\le E1(U_m<1)=P(U_m<1),$$
which implies
$$P(R_m>c)\ge P(U_m>c,V_m<1)=P(U_m>c)P(V_m<1)=P(U_m>c)P(U_m<1)\ge\de(c-1)\de>0$$
for all $m$, which contradicts (2).
Consider now Case 2 (condition (1) is only needed for this case). Then again $0=E(U_m-1)=E(U_m-1)1(U_m>1)-E(1-U_m)1(U_m<1)$, whence
$$(1-b)\de\le(1-b)P(U_m<b)=E(1-b)1(U_m<b)\le E(1-U_m)1(U_m<b)
\le E(1-U_m)1(U_m<1)=E(U_m-1)1(U_m>1)\le EU_m\,1(U_m>1)
\le(EU_m^p)^{1/p}P(U_m>1)^{1/q}\le C^{1/p}P(U_m>1)^{1/q},$$
by Hölder's inequality (with $q:=1/(1-1/p)$) and (1).  So,
$P(U_m>1)\ge\ep:=((1-b)\de/C^{1/p})^q<\infty$ and hence
$$P(R_m<b)\ge P(U_m<b,V_m>1)=P(U_m<b)P(V_m>1)=P(U_m<b)P(U_m>1)\ge\de\ep>0$$
for all $m$, which again contradicts (2).
Thus, the assumption $U_m\not\to1$ leads to a contradiction in either case. This means that $U_m\to1$.
{"set_name": "stack_exchange", "score": 0, "question_id": 375577}