Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be a random variable defined over $0-1$ and let $X_1 . . . X_4$ be the order statistics of $4$ draws from $X$ (where $X_1$ is the minimum).

I am looking to see if I can identify any general conditions for $X$ such that $$\mathbb{E(}X_4 - X_3) < 3\mathbb{E}(X_2 -X_1)$$

Intuitively, the validity of the inequality seems like it would be a property of the skewness of $X$. For example, in response to this question (where the inequality in the first part of that question reduces to the one presented here) Sasha showed that the inequality holds for the Beta distribution(a,2) when $a \ > 0.512761$. As $\beta$ increases, I believe the critical value of $a$ increases (for example, when $\beta = 3$, $a > .6017$) but I believe in all cases the distribution has to be extremely positively skewed for it not to hold. However, skewness alone can't be the only factor, as $Skew(Be(.512761,2)) = 1.21989$ and $Skew(Be(.6017,3)) = 1.3617$.

Does anyone have any ideas? I am looking for some way to say "the inequality holds so long as $X$ has ___ properties."

share|improve this question
    
You might be interested in L-moments, which are defined in terms of expectations of order statistics. One can construct measures of skewness, kurtosis and scale out of these. –  Sasha Aug 15 '11 at 18:00
    
Thanks for the tip, I will start reading. –  Jand Aug 15 '11 at 18:10

1 Answer 1

up vote 2 down vote accepted

Let us start with an elementary remark: for every real numbers $u\le v$, $$ \int\limits_{-\infty}^{+\infty} [u\le x\le v]\,\text{d}x=v-u. $$ Thus, for every random variables $U$ and $V$ such that $U\le V$ almost surely, integrating this with respect to the distribution of $(U,V)$ yields $$ E(V-U)=\int\limits_{-\infty}^{+\infty} P(U\le x\le V)\,\text{d}x. $$ Coming to the OP's question, we first note that the event $[X_1\le x\le X_2]$ is realized if and only if exactly one value from the sample is smaller than $x$ (and the other three are greater than $x$). Thus, $$ P(X_1\le x\le X_2)=4F(x)(1-F(x))^3. $$ Likewise, the event $[X_3\le x\le X_4]$ is realized if and only if exactly one value from the sample is greater than $x$ (and the other three are smaller than $x$). Thus, $$ P(X_3\le x\le X_4)=4(1-F(x))F(x)^3. $$ One sees that, for every positive $a$, $E(X_4-X_3)<aE(X_2-X_1)$ if and only if $$ \int\limits_{-\infty}^{+\infty} F(x)(1-F(x))(a(1-F(x))^2-F^2(x))\,\text{d}x>0. $$ For $a=3$, the condition reads $ \displaystyle\int\limits_{-\infty}^{+\infty} F(1-F)(3-6F+2F^2)>0. $

share|improve this answer
    
Thank you, thank you, thank you!! –  Jand Aug 17 '11 at 1:14
    
You are welcome (thrice). –  Did Aug 17 '11 at 11:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.