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

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!).

MathOverflow Asked by Learning math on January 16, 2021

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 $$mge2$$. 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: $$P(U_mde$$ 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 $$(1-b)dele(1-b)P(U_m1)le EU_m,1(U_m>1) le(EU_m^p)^{1/p}P(U_m>1)^{1/q}le C^{1/p}P(U_m>1)^{1/q},$$ by Hölder's inequality (with $$q:=1/(1-1/p)$$) and (1). So, $$P(U_m>1)geep:=((1-b)de/C^{1/p})^q and hence $$P(R_m1)=P(U_m1)=P(U_m1)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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