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Consider$$A_{i}=\int_0^1\int_0^1 \left(p-(1-p)\sqrt{1-w}\exp\left(\frac{wx_{i}^2}{2}\right)\left[1-\frac{1}{\sqrt{w+1}}\exp\left(\frac{-w^2x_{i}^2}{2(w+1)}\right)\right]\right) \cdot \prod_{j\neq i}^m\left[p+(1-p)\sqrt{1-w}\exp\left(\frac{wx_{j}^2}{2}\right)\right]dwdp,$$ where $0 \le w \le 1$, $0 \le p \le 1$ and $x_i \in (0,\infty)\ \forall i = 1, \dots, m$. Note that, $A_{i}$ is expected value and can get values between $-\infty$ and $\infty$.

And, suppose $x_{i}$'s are ordered such that $x_{1}<x_{2}<...<x_{m}$.

I have been trying to show that $A_{m}<A_{m-1}<...<A_{1}$,but I could not show it analytically. Numerically, I have shown that $A_{m}<A_{m-1}<...<A_{1}$ by generating some $x_{i}$'s. Could anyone help me to show it? Thanks in advance.

I have tried to show it for $m=2$, and its clear that $A_{1}>A_{2}$ when $$p>(1-p)\sqrt{1-w}\exp\left(\frac{wx_{i}^2}{2}\right)\left[1-\frac{1}{\sqrt{w+1}}\exp\left(\frac{-w^2x_{i}^2}{2(w+1)}\right]\right)$$ since $$p-(1-p)\sqrt{1-w}\exp\left(\frac{wx_{i}^2}{2}\right)\left[1-\frac{1}{\sqrt{w+1}}\exp\left(\frac{-w^2x_{i}^2}{2(w+1)}\right]\right)\tag{*}$$ is positive. But I couldn't show it when $$p<(1-p)\sqrt{1-w}\exp\left(\frac{wx_{i}^2}{2}\right)\left[1-\frac{1}{\sqrt{w+1}}\exp\left(\frac{-w^2x_{i}^2}{2(w+1)}\right]\right)$$ since $(*)$ is negative, and in this case showing that $A_{1}>A_{2}$ is being more complicated for me.

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    $\begingroup$ Would help to know where this came from. Presumably it's the probability of something. $\endgroup$ – djechlin Nov 14 '15 at 5:10
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    $\begingroup$ Okay.... what event? $\endgroup$ – djechlin Nov 14 '15 at 5:15
  • $\begingroup$ Also is it the case that the parens are a bit misleading? the term (1-p)(sqrt(1-w)exp(wxi/)) is being multiplied against the second expression, not the full grouping? $\endgroup$ – djechlin Nov 14 '15 at 5:17
  • $\begingroup$ yes this term is being multiplied against the second expression not the full grouping.I think now it looks better.the question is too long, I have computed this probailities and now just wondering if I can order them as I stated above.thanks @djechlin $\endgroup$ – Jimmy Dur Nov 14 '15 at 5:29
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    $\begingroup$ Okay, how I'm actually working on this is by trying to figure out what probability event this is modeling. If you're not going to post information on that after I keep asking, I'm not going to work on this. $\endgroup$ – djechlin Nov 14 '15 at 5:30
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Let $f_j(w)= \sqrt{1-w}\exp\left(w\dfrac{x_j^2}2\right),$ then $f'_j(w)=\left(-\dfrac1{2\sqrt{1-w}}+\sqrt{1-w}\dfrac{x_j^2}2\right)\exp\left(w\dfrac{x_j^2}2\right).$
It is maximal when $f'_j(w)=0$ ie at $x_j^2 = \dfrac1{1-w},$ $$f_j(w) \leq f(w),\quad \text{ where}\quad f(w) = \sqrt{1-w}\exp\dfrac w{2(1-w)}.$$ Derivation of $f(w)$ is:$\quad f'(w)=\left(-\dfrac1{2\sqrt{1-w}} +\dfrac{\sqrt{1-w}}{2(1-w)^2}\right)\exp\dfrac w{2(1-w)},$
$$f'(w)=\dfrac w{2(1-w)^2}f(w).$$
$f(w)$ has minimum when $$w = 0,\quad f_{min}(w) = f(0) = 1,$$ and singularity when $$w\to1,\quad \lim\limits_{w\to 1}f(w) = \infty.$$
On the other hand, function $$g_i(w)=\frac{1}{w+1}\exp\left(\frac{-w^2x_{i}^2}{2(w+1)}\right)$$ has maximum at $w\in[0,1]$ when $w=0$, ie $$\max\limits_{w\in[0,1]}\frac{1}{w+1}\exp\left(\frac{-w^2x_{i}^2}{2(w+1)}\right) = 1.$$ So $$A_{i} \leq \int_0^1 \int_0^1 \left(p-(1-p)\min_wf_i(w)\left[1-\frac{1}{w+1}\exp\left(\frac{-w^2x_{i}^2}{2(w+1)}\right)\right]\right) \cdot \prod_{j\neq i}^m (p+(1-p)f(w))dwdp =$$$$\int_0^1 \int_0^1 p(p+(1-p)f(w))^{m-1}dwdp \leq \int_0^1 \int_0^1 pf^{m-1}(w))dwdp = \dfrac12\int_0^1 f^{m-1}(w)dw.$$ $$\boxed{A_{i} \leq \dfrac12\int_0^1 f^{m-1}(w)dw}$$

On the main issue.

$$A_i-A_j = \int_0^1\int_0^1(((p-(1-p)f_i(w)(1-g_i(w))((p+(1-p)f_j(w))- $$$$((p-(1-p)f_j(w)(1-g_j(w))((p+(1-p)f_i(w)))\cdot\prod_{k\neq i,\,k\neq j}^m (p+(1-p)f_k(w))dwdp = $$$$ \int_0^1\int_0^1(1-p)(p(f_i(w)g_i(w))-f_j(w)g_j(w)) + (1-p)f_i(w)f_j(w)(g_i(w)-g_j(w)))\cdot \prod_{k\neq i,\,k\neq j}^m (p+(1-p)f_k(w))dwdp = $$$$ \int_0^1\int_0^1(1-p)(p(h_i(w)-h_j(w)) + (1-p)f_i(w)f_j(w)(g_i(w)-g_j(w)))\cdot \prod_{k\neq i,\,k\neq j}^m (p+(1-p)f_k(w))dwdp, $$ where $h_i(w) = f_i(w)\cdot g_i(w).$

Considering this formula with the functions $$f(w,t) = \sqrt{1-w}\exp wt,\quad g(w,t) = \dfrac1{\sqrt{1+w}}\exp\left(-\dfrac{w^2t}{w+1}\right),\quad h(w,t) = f(w,t)g(w,t) = \sqrt{\dfrac{1-w}{1+w}}\exp\dfrac{wt}{w+1}$$ when $\dfrac{x_j^2}2\to \dfrac{x_i^2}2 = t,$ we can change differences under the integral to according derivations: $$\dfrac{h_i(w)-h_j(w)}{t_i-t_j}\to\dfrac{dh}{dt} = \dfrac w{w+1}h(w,t),$$ $$f_i(w)f_j(w)\dfrac{g_i(w)-g_j(w)}{t_i-t_j}\to f^2(w,t)\dfrac{dg}{dt} = -\dfrac{w^2}{w+1}f(w,t)^2g(w,t) = -\dfrac{w^2}{w+1}f(w,t)h(w,t) = -wf(w,t)\dfrac{dh}{dt}.$$ So $$A_i-A_j \to \dfrac{x_i^2-x_j^2}2\int\limits_0^1\int\limits_0^1(1-p)(p-(1-p)wf(w,t))\dfrac{dh}{dt}\,\prod_{k\neq i,\,k\neq j}^m (p+(1-p)f_k(w))\,dwdp.$$ When $m=2$: $$A_i-A_j \to \dfrac{x_i^2-x_j^2}2\int\limits_0^1\int\limits_0^1(1-p)(p-(1-p)wf(w,t))\dfrac{dh}{dt}\,dwdp = {\dfrac{x_i^2-x_j^2}2\int\limits_0^1\int\limits_0^1((p-p^2)-(1-p)^2w \sqrt{1-w}\exp(wt))\dfrac{w}{1+w}\sqrt{\dfrac{1-w}{1+w}}\exp\left(\dfrac{w}{w+1}t\right)\,dwdp}= {\dfrac{x_i^2-x_j^2}{12}\int\limits_0^1(1-2w \sqrt{1-w}\exp(wt))\dfrac{w}{1+w}\sqrt{\dfrac{1-w}{1+w}}\exp\left(\dfrac{w}{w+1}t\right)\,dw}.$$

Using Wolfram Alpha program, we can see that this expression is greater than $0$ when $t\leq 0.83$ and less than $0$ when $t\geq 0.84$.

When $m=3$: $$A_i-A_j \to \dfrac{x_i^2-x_j^2}2\int\limits_0^1\int\limits_0^1(1-p)(p-(1-p)wf(w,t))(p+(1-p)f(w,t_1))\dfrac{dh}{dt}\,dwdp = \dfrac{x_i^2-x_j^2}2\int\limits_0^1\int\limits_0^ 1(p^2(1-p) - p(1-p)^2wf(w,t) + p(1-p)^2f(w,t))+(1-p)^3wf(w,t)f(w,t_1))\dfrac{dh}{dt}\,dwdp = $$$$\dfrac{x_i^2-x_j^2}2\int\limits_0^ 1\left(\dfrac1{12} - \dfrac7{12}wf(w,t) + \dfrac7{12}f(w,t_1)-\dfrac14wf(w,t)f(w,t_1)\right)\dfrac{dh}{dt}\,dw = $$$$\dfrac{x_i^2-x_j^2}{72}\int\limits_0^ 1(6 - 21wf(w,t) + 21f(w,t_1)-9wf(w,t)f(w,t_1))\dfrac{dh}{dt}\,dw = $$$$\dfrac{x_i^2-x_j^2}{72}\int\limits_0^1((7 - 3wf(w,t))(7 + 3f(w,t_1))-43)\dfrac{dh}{dt}\,dw = $$$$\dfrac{x_i^2-x_j^2}{72}\int\limits_0^1((7 - 3w\sqrt{1-w}\exp(wt))(7 + 3\sqrt{1-w}\exp(wt_1))-43)\dfrac{w}{1+w}\sqrt{\dfrac{1-w}{1+w}}\exp\left(\dfrac{w}{w+1}t\right)\,dw.$$

Using Wolfram Alpha program at $t=4$, we can see that expression is greater than $0$ when $t_1\geq 2.8$ and less than $0$ when $t_1\leq 2.7$.

So the hypothesis about $A_i<A_j$ is wrong.

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    $\begingroup$ Could you explain it by giving more details if its possible?Since I am not very familiar with "notation O",I could not understand what you have done. $\endgroup$ – Jimmy Dur Dec 26 '15 at 9:59
  • $\begingroup$ @Jimmy Dur Now everything is clear and simple) $\endgroup$ – Yuri Negometyanov Dec 26 '15 at 19:25
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    $\begingroup$ Thanks for your time but what you have tried to show is not the answer of my question. And, also you concluded A_{i} <=1/2 which is not true because you made a mistake in the last second line.I need to show that A_{m}<A_{m-1}<...<A_{1} as I indicated above.thanks though. $\endgroup$ – Jimmy Dur Dec 27 '15 at 0:35
  • $\begingroup$ @Jimmy Dur About $A_m \leq A_{m-1} \leq \dots \leq A_1:$ there is a consequence of $f_j(w) \leq f(w) \leq 1$. $\endgroup$ – Yuri Negometyanov Dec 27 '15 at 1:18
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    $\begingroup$ @Jimmy Dur My present is $m=3$ case. Happy New Year! $\endgroup$ – Yuri Negometyanov Dec 31 '15 at 8:22

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