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](http://www.stat.cmu.edu/%7Ehseltman/files/ratio.pdf), 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!).



Question on limit in probability of the ratio of max to min of 2 sequences of non-ive, continuous iid random variables with support $[0, \infty).$
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 $X\_m/EX\_m\to0$.




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$\newcommand\de\delta\newcommand\ep\varepsilon$


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 $U\_m\to1$.


