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Is there a smart way to show that one variable stochastically dominates the other without knowledge of the CDF? Thanks so much!

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    $\begingroup$ What do you do know about these variables? $\endgroup$
    – user603
    Feb 8 '15 at 3:59
  • $\begingroup$ Hey @user603 For one of them I know everything like density cdf and characteristic function for the other only characteristic function and maybe density. $\endgroup$
    – Hirek
    Feb 8 '15 at 10:04
  • $\begingroup$ Can you assume finite first moment for both? $\endgroup$
    – user603
    Feb 8 '15 at 11:18
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    $\begingroup$ Hey @user603 yes, I can. In fact the only reason why I don't know the CDF is because the integral is really hard, i.e. it's /int etr(matrix+Id)det(matrix+Id)0F1(scalar,matrix+Id) which messes up the conventional integrals worked out in Muirhead (1982). The entire problem is otherwise well-behaved, continuous, differentiable etc. all the good stuff just the integral of the density is not tractable but I have the CF so I am trying to get around the integral. $\endgroup$
    – Hirek
    Feb 8 '15 at 13:07
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    $\begingroup$ I have the CF of the intractable part already. The other is just chi-squared against which I want to compare this. Thanks @user603 $\endgroup$
    – Hirek
    Feb 8 '15 at 13:36
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Let $X$ and $Y$ be two real random variables with distributions $F$ and $G$ respectively

Lemma 1 in [0] establishes that if $E(X)$ and $E(Y)$ both exist, then:

$$X \underset{s}{>} Y\implies E(X)\geq E(Y)$$

Furthermore, suppose that $X$ and $Y$ are non negative random variables with $F(0)=G(0)$ and that they have densities $f$ and $g$ on $(0,\infty)$ with $F(0)=G(0)$. Then, in their lemma 2, the authors show that either one of the following two conditions implies $F \underset{s}{>} G$:

  1. The densities $g$ crosses $f$ only once and from above,
  2. $f[F^{-1}(t)]\leq g[G^{-1}(t)],\;\forall t\in (F(0),1)$.

    • [0] S. W. Dharmadhikari and K. Joag-dev (1983). Mean, Median, Mode III. Statistica Neerlandica, Volume 37, Issue 4, pages 165–168.
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    $\begingroup$ Ah thanks so much! So I can use the densities indeed. That is neat. I was also wondering if you can also use the characteristic functions of sorts? There is a related question of mine here stats.stackexchange.com/questions/136683/… @user603 Again, thank you so much! $\endgroup$
    – Hirek
    Feb 8 '15 at 14:56
  • $\begingroup$ But 2 requires one to compute the inverse of the cumulative distributions and also the two conditions are sufficient but not equivalent, is that correct? $\endgroup$ Sep 8 '16 at 1:30

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