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Let $(X_i)_{i\geq 0}$ be i.i.d. nonnegative random variables with continuous density function $f$. Let \begin{align} \mu_n = \mathbb{E}[X_{(n)}-X_{(n-1)}] \end{align} be the expected difference between the largest and second largest of the first $n$ random variables.

Question: Can we show that $\mu_n$ is decreasing in $n$?

We can assume the density $f$ to be decreasing for $x>0$ and the mean of the order statistics to be well-defined; in particular, assume $\mathbb{E}[X_i]<\infty$.

Edit: My intuition tells me that $\mu_n$ is generally decreasing in $n$, but I have been unable to prove it. I am more interested in simple conditions under which the result is true, rather than a specific counterexample where it is not.

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    $\begingroup$ You haven't even assumed $\mathbb E[X_i]$ exists, so I see no reason to think even that $\mu_n$ exists. $\endgroup$ Dec 11, 2023 at 6:14
  • $\begingroup$ @RobertIsrael Thank you for pointing that out. I have updated the question. $\endgroup$
    – svonimir
    Dec 11, 2023 at 6:22
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    $\begingroup$ Do you want $\mu_{n+1} < \mu_n$ for all $n \geq 2$, or do you just want some $m$ such that it holds for all $n \geq m$? $\endgroup$
    – Brian Tung
    Dec 11, 2023 at 7:28
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    $\begingroup$ Looks like this question had been posed already and answered in the negative by the existence of non-"ES short" distributions? $\endgroup$ Dec 11, 2023 at 20:33

2 Answers 2

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It turns out that the claim is false. For example, assume the common CDF $F(\cdot)$ of $X_i$'s takes the form

$$ F(t) = (1-p) (1 - (1-t)^m) + p t^m, \qquad 0 \leq t \leq 1$$

where $p \in (0, 1)$ and $m \geq 1$ are prescribed parameters. It is designed so that $X_i$'s approximately have Bernoulli distribution with parameter $p$ for $m$ large enough. So, if $\tilde{X}_i$'s are i.i.d. random variables having $\operatorname{Ber}(p)$ distribution, then

\begin{align*} \mu_n := \mathbb{E}[X_{(n)} - X_{(n-1)}] &\approx \mathbb{E}[\tilde{X}_{(n)} - \tilde{X}_{(n-1)}] \\ &= \mathbb{P}(\tilde{X}_{(n)} = 1 \text{ and } \tilde{X}_{(n-1)} = 0) \\ &= n p(1-p)^{n-1} =: \tilde{\mu}_n. \end{align*}

Since $\tilde{\mu}_n$ is not monotone in $n$, we can expect that the same is true for $\mu_n$ if $m$ is large. Indeed, below is the graph of $\mu_n$ (blue dots) and $\tilde{\mu}_n$ (orange dashed curve) corresponding to $m = 10$ and $p = \frac{1}{10}$:

enter image description here

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Edit: As kindly pointed out by Sangchul Lee, the computations do not suffice as a proof and only the correct parts shall be kept:

The maximum $X_{(n)}$ (and second order $X_{(n-1)}$ too) still has finite expectation: $$ E \left[X_{(n)} \right] = E \left[\max_{0 \leq i \leq n} X_i \right] \leq E \left[\sum_{i=0}^n X_i \right] = \sum_{i=0}^n E[X_i] < \infty.$$ So we can actually write $\mu_n = E[X_{(n)}] - E[X_{(n-1)}]$ and use the general relation $E[Y] = \int_0^{\infty} 1-F_Y(x) dx$. Here, by independence $$F_{X_{(n)}}(x) = P[ \forall \, 0 \leq i \leq n: X_i \leq x ] = F(x)^{n+1},$$ for $F:=F_{X_0}$. Similarly (since the called events are disjoint) \begin{align} F_{X_{(n-1)}}(x) &= P[\max_{0 \leq i \leq n}Xi \leq x] + P[ \exists 0 \leq i_0 \leq n: \, \forall i_0 \neq i \leq n: X_i < x, X_{i_0} > x] \\ &= F(x)^{n+1} + (n+1)F(x)^n(1-F(x)).\end{align}

More generally, the distribution function of $X_{(k)}$ is $\sum_{j=k+1}^{n+1} \binom{n+1}{j} F(x)^j [1-F(x)]^{n+1-j}$, as also pointed out by Sangchul Lee to be found on Wikipedia.

So, in total \begin{align*}\mu_n &= E[X_{(n)}] - E[X_{(n-1)}] \\ &= \int_0^{\infty} 1-F(x)^{n+1} dx - \int_0^{\infty} 1-[F(x)^{n+1} + (n+1)F(x)^n(1-F(x))] dx \\ &= \int_0^{\infty} (n+1)F(x)^n(1-F(x)) \, dx \end{align*}

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    $\begingroup$ (+1) Your approach seems promising, but I am not sure if your formula for the CDF of $X_{(n)}$ and $X_{(n-1)}$ are correct. In OP's setting, $X_{(n)}$ and $X_{(n-1)}$ are the largest and second largest elements in the first $n$ variables, respectively. Then according to Wikipedia article, we have $$F_{X_{(n)}}(x)=F(x)^n\qquad\text{and}\qquad F_{X_{(n-1)}}(x)=F(x)^n+n F(x)^{n-1}(1-F(x)).$$ $\endgroup$ Dec 11, 2023 at 8:11
  • $\begingroup$ @SangchulLee That is correct, thank you. I edited the post and do not arrive at the conclusion anymore. $\endgroup$
    – Alex
    Dec 11, 2023 at 9:42
  • $\begingroup$ @Alex Nice work. I guess the question becomes, under what simple conditions is your expression for $\mu_n$ decreasing in $n$? $\endgroup$
    – svonimir
    Dec 12, 2023 at 3:24
  • $\begingroup$ @svonimir I think you should ask the follow-up question in a separate post $\endgroup$ Dec 12, 2023 at 4:41

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