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

MathOverflow Asked by Learning math on January 16, 2021

1 AnswersThe 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 $mge2$. 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, $U_mto1$ and $V_mto0$ (in probability, as $mtoinfty$), and hence $X_mto1$ and $$frac{max(X_m,Y_m)}{min(X_m,Y_m)}to1.$$

On the other hand, $EX_m>EU_m>mtoinfty$, so that $X_m/EX_mto0$.

$newcommanddedeltanewcommandepvarepsilon$

Here is the good news: you desired result will hold under the additional assumption that $$EU_m^ple C tag{1}$$ 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 $U_mto1$. 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 $bin(0,1)$, 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 $$(c-1)dele(c-1)P(U_m>c)=E(c-1)I(U_m>c)le E(U_m-1)1(U_m>c)le E(U_m-1)1(U_m>1)=E(1-U_m)1(U_m<1)le E1(U_m<1)=P(U_m<1),$$ which implies $$P(R_m>c)ge P(U_m>c,V_m<1)=P(U_m>c)P(V_m<1)=P(U_m>c)P(U_m<1)gede(c-1)de>0$$ 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 $$(1-b)dele(1-b)P(U_m<b)=E(1-b)1(U_m<b)le E(1-U_m)1(U_m<b) le E(1-U_m)1(U_m<1)=E(U_m-1)1(U_m>1)le EU_m,1(U_m>1) le(EU_m^p)^{1/p}P(U_m>1)^{1/q}le C^{1/p}P(U_m>1)^{1/q},$$ by Hölder's inequality (with $q:=1/(1-1/p)$) and (1). So, $P(U_m>1)geep:=((1-b)de/C^{1/p})^q<infty$ and hence $$P(R_m<b)ge P(U_m<b,V_m>1)=P(U_m<b)P(V_m>1)=P(U_m<b)P(U_m>1)gedeep>0$$ for all $m$, which again contradicts (2).

Thus, the assumption $U_mnotto1$ leads to a contradiction in either case. This means that $U_mto1$.

Correct answer by Iosif Pinelis on January 16, 2021

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