Can the ratio of the two smallest element of an iid sample converge to 1 if the support of $X$ is positive?

We have: $$\mathbb P(X \leq 0)=0$$ and $$\mathbb P (X \leq a)>0$$ for any $$a>0$$.

• Doesn't something like the exponential of a standard Gaussian already provide a counter example? Then getting closer to zero gets exponentially less likely, so $X^{(2)}-X^{(1)}$ wil converge to zero but I don't think $X^{(2)}/X^{(1)}$ converges to $1$. – quarague Apr 11 at 12:24
• Sorry, I meant if there exists a random variable with the given support such that the ratio still converges to 1 - I did not properly write that down; 1000x sorry – user657766 Apr 11 at 13:20

The answer is no. Here is a counterexample. Lets $$X_1, X_2, \dots, X_n$$ random samples from an exponential distribution with parameter $$\lambda$$.the order statistics $$X_{(1)}, X_{(2)}$$ each have distribution \begin{align*} X_{(1)} &= \frac{Z_1}{\lambda n} \\ X_{(2)} &= \frac{1}{\lambda}\left(\frac{Z_1}{n}+\frac{Z_2}{n-1}\right) \end{align*} where the $$Z_1, Z_2$$ are iid standard exponential random variables (see Wikipedia). Therefore $$\frac{X_{(2)}}{X_{(1)}} = \frac{Z_1}{Z_1+Z_2n/(n-1)} \rightarrow \frac{Z_1}{Z_1+Z_2} \neq0$$

• I actually meant if it is possible to construct a random variable $X$ with the given support as indicated, such that the ratio will converge to 1; Sorry, I corrected it – user657766 Apr 11 at 13:17

You're right for most cases, however this depends on the distribution as well.

If $$X_1, \dots, X_n \sim \Bbb{P}_\theta$$ where $$Var_\theta(X_i) = 0$$, this doesn't hold true, because $$\forall i,j \in \{1,\dots,n\}\quad\Bbb{P}(X_i = X_j) = 1 \rightarrow \Bbb{P}\left(\lim_{n \to \infty}\frac{X_i}{X_j}=1\right)=1$$

But for most cases, where $$Var_\theta(X_i) \ne 0$$, you can prove this indirectly:

Let's assume that $$\exists \lim_{n \to \infty}\frac{X^{(2)}}{X^{(1)}} := A \in \Bbb{R}^+_0 \cup \{+\infty\}$$ We can assume that $$A \ge 0$$, because $$X^{(2)} \ge X^{(1)}$$.

If

$$\lim_{n \to \infty}\frac{X^{(2)}}{X^{(1)}} = A \stackrel{\text{then }}{\longrightarrow}\lim_{n \to \infty}\Bbb{E}\left(\frac{X^{(3)}}{X^{(2)}}\right) = A$$

Since $$\forall \varepsilon, N_1 \in \Bbb{R}^+ \exists, N_2 \in \Bbb{N}^+$$ such that from a sample of $$X_1,\dots,X_{N_1}$$ $$A - \varepsilon < \frac{X^{(2)}}{X^{(1)}} < A + \varepsilon,$$ We can assure that from a sample of $$X_1,\dots,X_{N_2}$$, where $$N_2$$ is sufficiently large: $$A - \varepsilon < \frac{X^{(3)}}{X^{(2)}} < A + \varepsilon$$ Since as $$n \to \infty$$, $$\lim_{n \to \infty}\left(\frac{X^{(3)}}{X^{(2)}} - \frac{X^{(2)}}{X^{(1)}}\right) = 0$$

Now we'll have to prove $$2$$ cases separately:

Case $$1$$: $$A = 1$$

From our previous argument it follows that

$$\lim_{n \to \infty}\frac{X^{(2)}}{X^{(1)}} = \lim_{n \to \infty}\frac{X^{(3)}}{X^{(2)}} = \dots = \lim_{n \to \infty}\frac{X^{(n)}}{X^{(n-1)}} = 1$$

$$\Rightarrow \lim_{n \to \infty}\frac{X^{(n)}}{X^{(1)}} = 1$$

$$\Rightarrow \forall i \in \{1,\dots, n\} \quad \sup {Ran}_{\theta} X_i = \inf {Ran}_{\theta} X_i$$

$$\Rightarrow \forall i \in \{1,\dots, n\}\quad Var(X_i) = 0$$

Which is a contradiction, becuase in this case we assumed that $$Var(X_i) \ne 0$$.

Case $$2$$: $$A \in \Bbb{R}^+_0 \backslash \{1\}$$

Again, we get

$$\lim_{n \to \infty}\frac{X^{(2)}}{X^{(1)}} = \lim_{n \to \infty}\frac{X^{(3)}}{X^{(2)}} = \dots = \lim_{n \to \infty}\frac{X^{(n)}}{X^{(n-1)}} = A$$

Which is this case means:

$$\lim_{n \to \infty}\frac{X^{(2)}}{X^{(1)}} = \lim_{n \to \infty}\frac{1}{A}\frac{X^{(3)}}{X^{(1)}} = \dots =\lim_{n \to \infty}\frac{1}{A^n}\frac{X^{(n)}}{X^{(1)}}$$

$$\Rightarrow \lim_{n \to \infty}\frac{X^{(n)}}{X^{(1)}} = \lim_{n \to \infty}A^n\frac{X^{(2)}}{X^{(1)}} = \\ = 0, +\infty \text{ or does not exist (depending on the value of } A)$$

It can't be $$0$$, since then $$\Bbb{P}(X_n = 0) = 1 \rightarrow \forall i \in \{1\dots n\}\quad \Bbb{P}(X_i = 0) = 1 \rightarrow Var(X_i) = 0$$ Which is a contradiction again.

It can' be $$+ \infty$$ either, because that can only occur if $$X_n \ne 0$$ and $$X_{n-1} = 0$$, but as we've already discussed,

$$\lim_{n \to \infty}\frac{X^{(n)}}{X^{(n-1)}} = \lim_{n \to \infty}\frac{X^{(n-1)}}{X^{(n-2)}} (= + \infty)$$

Which also means that $$X_{n-1} \ne 0$$ and $$X_{n-2} = 0$$, and we get a contradiction from $$X_{n-1} = 0$$ and $$X_{n-1} \ne 0$$.

The only option in this case is for the limit not to exist, which isn't technically a contradiction, but it further proves our point.

Result:

Both cases lead to contradiction, which means that the given limit does not exist.

• I still need a couple of minutes to work through your proof; Thank you so much for your answer already - I highly appreciate it – user657766 Apr 11 at 13:23
• Just a little remark; If $\mathbb P (X \in (0,a])>0$ for any fixed $a$, then e.g. variance 0 is simply not possible and also there is no "smallest" value that $X$ can take but the minimum will just converge to zero, getting smaller and smaller – user657766 Apr 11 at 13:24
• @Johannes That's true, I glossed over the fact that $\mathbb P (X \in (0,a])>0$, but the proof is still valid, since we're trying to contradict the fact that the variance is $0$. – Daniel P Apr 11 at 13:26
• I either did not properly understand your proof or it is wrong; Imagine I have a random variable $X$ with supp$(X)=(1,\infty)$ and $\mathbb P(X \in (1,a])>0$ for any fixed $a>1$ (so everything shifted up by one) Then obviously both, the smallest and the second smallest realization will converge to one and hence the ratio will also converge to 1; However, if I understood your proof correctly, I could also show with the same arguments that this does not hold; – user657766 Apr 11 at 13:51
• The basic idea for it to work must be that the minimum converges to zero; If it converges to some fixed number $a$ with $a \neq 0$, then the ratio will indeed converge to 1... – user657766 Apr 11 at 13:53