Notes:Probability of an RV being less than another

[ilmath]\newcommand{\P}[2][]{\mathbb{P}#1{\left[{#2}\right]} } \newcommand{\Pcond}[3][]{\mathbb{P}#1{\left[{#2}\!\ \middle\vert\!\ {#3}\right]} } \newcommand{\Plcond}[3][]{\Pcond[#1]{#2}{#3} } \newcommand{\Prcond}[3][]{\Pcond[#1]{#2}{#3} }[/ilmath]

[ilmath]\newcommand{\E}[1]{ {\mathbb{E}{\left[{#1}\right]} } } [/ilmath][ilmath]\newcommand{\Mdm}[1]{\text{Mdm}{\left({#1}\right) } } [/ilmath][ilmath]\newcommand{\Var}[1]{\text{Var}{\left({#1}\right) } } [/ilmath][ilmath]\newcommand{\ncr}[2]{ \vphantom{C}^{#1}\!C_{#2} } [/ilmath]

This was solved/worked out, see: Probability of i.i.d random variables being in an order and not greater than something

Solution

• [math]\P{X\le Y\le Z}\eq\sum_z\sum_{y\le z}\sum_{x\le y}\P{X\eq x\cap Y\eq y\cap Z\eq z} [/math] - duh! - silly me!
  • Integral form is obvious.

SILLY STUFF

Notes

Here I investigate:

1. [ilmath]\P{X\le Y} [/ilmath], and
2. [ilmath]\P{X\le Y\le Z} [/ilmath] which of course is short for [ilmath]\P{\big(X\le Y\big)\cap\big(Y\le Z\big)} [/ilmath]

Case 1:

• [math]\P{X\le Y}\eq\sum_{y\in S_Y}\Big( \P{Y\eq y}\times\Pcond{X\le Y}{Y\eq y}\Big) [/math]

  [math]\eq\sum_{y\in S_y}\Big(\P{Y\eq y}\times\P{X\le y} \Big)[/math]

  • So: [math]\P{X\le Y}\eq\sum_{y\in S_Y}\Big( f_Y(y)\times F_X(y)\Big) [/math] where [ilmath]f_Y[/ilmath] is the probability mass function of [ilmath]Y[/ilmath] and [ilmath]F_X[/ilmath] is the cumulative probability function of [ilmath]X[/ilmath]
    • Note that there are bounds on [ilmath]X[/ilmath] hiding in here, as another way to write this is:
      • [math]\P{X\le Y}\eq\sum_{y\in S_y}\left( f_Y(y)\times \left[\sum_{x\in S_X,\ x\le y}\P{X\eq x}\right]\right)[/math]
  • The "integral form" is obviously:
    • [math]\P{X\le Y}\eq\int_{S_y}\Big(f_Y(y)\!\ F_X(y)\Big)\mathrm{d}y[/math] from the infinitesimal-style abuse of notation: [ilmath]\P{Y\eq y}\mathrm{d}y\cdot \P{X\le y} [/ilmath]
      • Which may be written as:
        • [math]\P{X\le Y}\eq\int_{S_y}\left(f_Y(y)\cdot\left[\int_{x\in S_x,\ x\le y}f_X(x)\right]\mathrm{d}x\right)\mathrm{d}y[/math]

Case 2:

• [math]\P{X\le Y\le Z}\eq\sum_{z\in S_Z}\P{Z\eq z}\cdot\Pcond{X\le Y\le Z}{Z\eq z} [/math]

  [math]\eq\sum_{z\in S_Z}\P{Z\eq z}\cdot\Pcond{X\le Y}{Y\le z} [/math]

  [math]\eq\sum_{z\in S_Z}\left(\P{Z\eq z}\cdot\frac{\P{X\le Y\le z} }{\P{Y\le z} }\right)[/math]

  [math]\eq\sum_{z\in S_Z}\left(\frac{\P{Z\eq z} }{\P{Y\le z} }\left[\sum_{y\in S_Y,\ y\le z}\P{Y\eq y}\cdot\Pcond{X\le Y}{Y\eq y}\right]\right)[/math]

  [math]\eq\sum_{z\in S_Z}\left(\frac{\P{Z\eq z} }{\P{Y\le z} }\sum_{y\in S_Y,\ y\le z} \P{Y\eq y}\cdot\P{X\le y} \right) [/math]

Integral form... coming soon!