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Suppose I have random variables $X,Y,Z$ and I would like to compute the probability that random variable $X$ is smaller than $Y$ and $Z$:

$$ \pi_X \overset{def}{=} Pr(X < Y, X < Z) = \int Pr(x < Y,x < Z)dPr(X\le x) $$ Define $\pi_Y$ and $\pi_Z$ similarly and without loss of generality, let $\pi_X \le \pi_Y \le \pi_Z$. I know that $X,Y$ and $Z$ are positively associated. Suppose that I compute $\pi_X$ "falsely" assuming that $Z \perp Y$ using the formula below:

$$ \pi^{ind}_X = \int Pr(x < Y)Pr(x < Z)dPr(X\le x). $$

I would like to know what condition I need for the the dependence structure among $X,Y$ and $Z$ (e.g., positive correlation?) to claim $\pi_X \ge \pi_X^{ind}$? I.e., the $\pi_X^{ind}$ under the (false) independence assumption always underestimates the chance that $X$ is smaller than both $Y$ and $Z$.

Based on the following small simulation study in R, I am guessing that $\pi_X \ge \pi_X^{ind}$ is true when $X$, $Y$ and $Z$ are "positively" associated in some way. The simulation model assume that $X$, $Y$ and $Z$ are from the multivariate normal as:

$$ \begin{bmatrix} X\\ Y\\ Z \end{bmatrix} \sim N_3\left( \begin{bmatrix} -2\\ 0\\ 2 \end{bmatrix} \sim \begin{bmatrix} \sigma^2&\rho\sigma &\rho\sigma\\ \rho\sigma& 1 &\rho\\ \rho\sigma & \rho & 1\\ \end{bmatrix} \right) $$ where I consider the values $\sigma=0.5,1,1.5,2$. Then I generate large enough samples from this multivariate distribution and computed $\pi_X$ based on the Monte Carlo samples. I also computed $\pi_X^{ind}$ assuming the independence across $X,Y$ and $Z$. I see that regardless of the values of $\sigma$, the $\pi_X$ is always greater than $\pi_X^{ind}$ when $rho > 0$.

 library(mvtnorm)
 get.pi <- function(xyz)
   {
     ## obtain the MC estimate of pi^ind and pi
     xs <- xyz[,1]
     ys <- xyz[,2]
     zs <- xyz[,3]
     num.indp <- num.dp <- 0
     M <- length(xs)
     for (ix in 1 : M)
      {
       x.temp <- xs[ix]
       num.dp <- num.dp + mean(x.temp < ys & x.temp < zs)
       num.indp <- num.indp + mean(x.temp < ys)*mean(x.temp < zs)
      }
    return(list(indp=num.indp/M,dp=num.dp/M)) 
   }


 N <- 10000
 sds1 <- seq(0.5,2,0.5)
 sd2 <- 1
 sd3 <- 1
 rhos <- seq(0,1,0.05)
 mus <- c(-2,0,2)  
 cols <- c("red","green")
par(mfrow=c(2,length(sds1)/2),mar=c(2,2,2,2))
for (isd in 1 : length(sds1))
  {
   sd1 <-sds1[isd]
   res.indP <- res.noIndP <- rep(NA, length(rhos))
   for (irho in 1 : length(rhos))
    {
      rho <- rhos[irho]
      rho12 <- rho
      rho13 <- rho
      rho23 <- rho

      S <- matrix(c(sd1^2,sd1*sd2*rho12,sd1*sd3*rho13,
                   sd1*sd2*rho12,sd2^2,sd2*sd3*rho13,
                   sd1*sd3*rho13,sd2*sd3*rho13,sd3^2),3,3)

      xyz <- rmvnorm(N,mean=mus,sigma=S)
      re <- get.pi(xyz)
      res.indP[irho] <- re$indp
          res.noIndP[irho] <- re$dp
     }
    rang <- c(res.indP,res.noIndP)
    ylim <- c(min(rang),max(rang))
    plot(rhos,res.indP,type="l",col=cols[1],ylim=ylim,
         main=paste("sd1=",sd1,sep=""))
    points(rhos,res.noIndP,type="l",col=cols[2],ylim=ylim)
   }

legend("topright",c("Independent","No assumption"),col=cols,lwd=1)
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  • 3
    $\begingroup$ I believe the answer will vary depending on the precise meaning of "positively associated." What exactly do you mean by this? $\endgroup$ – whuber Nov 14 '14 at 3:18
  • $\begingroup$ @whuber I am actually looking which condition of "positive association" is required to satisfy $\pi_X \ge \pi_X^{ind}$... $\endgroup$ – FairyOnIce Nov 14 '14 at 3:20
  • $\begingroup$ You might want to emphasize that by editing your question. $\endgroup$ – whuber Nov 14 '14 at 3:22
  • $\begingroup$ @whuber Thank you for the advice. I just edited my question. $\endgroup$ – FairyOnIce Nov 14 '14 at 3:25
  • 1
    $\begingroup$ This post in math.SE, math.stackexchange.com/questions/452484/…, may be interesting as a more general treatment of such issues. $\endgroup$ – Alecos Papadopoulos Nov 14 '14 at 10:23

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