3
$\begingroup$

Given random variables $W,X,Y,Z$ satisfying $E(XY)=E(E(X|Z)Y)$ and $E(XW)=E(E(X|Z)W)$, must it hold that $E(XY)=E(E(X|Z,W)Y)$?

I tried the case with $X,Z,W$ being jointly gaussian, and the case with $X,Z$ being jointly gaussian with mixing parameter $W$ etc, in both cases $E(XY)=E(E(X|Z,W)Y)$ holds.

However, when I tried showing it generally using orthogonal projection of conditional expectation, the given orthogonality statements doesn't seem to imply anything.

Improve this question
$\endgroup$
1
  • $\begingroup$ Where is this exercise/question from? $\endgroup$
    – Zhanxiong
    Commented Jul 29, 2022 at 20:17

1 Answer 1

Reset to default
0
$\begingroup$

I wasn't able to come up with anything definitive, but I did come up with this toy example that might be helpful to you.

In the first example, all three hold: $E(XY)=E(E(X|Z)Y)$, $E(XW)=E(E(X|Z)W)$, and $E(XY)=E(E(X|Z,W)Y)$

In the second example, none of the three hold.

require(dplyr)

calc <- function(A, p) {
  stopifnot(sum(p) == 1)
  
  Ex <- sum(p*A$X)
  Ey <- sum(p*A$Y)
  Ez <- sum(p*A$Z)
  Ew <- sum(p*A$W)
  # E(X|Z)
  Exgz <- with(A, c(sum(X[Z==0]*p[Z==0])/sum(p[Z==0]), sum(X[Z==1]*p[Z==1])/sum(p[Z==1])))
  # E(E(X|Z))
  EExgz <- sum(p*Exgz[A$Z+1])
  stopifnot(Ex == EExgz)
  
  # E(X)E(Y)
  ExEy <- Ex * Ey
  # E(XY)
  Exy <- sum(p*A$X*A$Y)
  # E(E(X|Z)Y)
  EExgz_y <- sum(p*Exgz[A$Z+1]*A$Y)
  
  # E(X)E(W)
  ExEw <- Ex * Ew
  # E(XW)
  Exw <- sum(p*A$X*A$W)
  # E(E(X|Z)W)
  EExgz_w <- sum(p*Exgz[A$Z+1]*A$W)
  
  # E(X|Z,W)
  Exgzw <- with(A, matrix(c(sum(X[Z==0 & W==0]*p[Z==0 & W==0])/sum(p[Z==0 & W==0]), 
                            sum(X[Z==1 & W==0]*p[Z==1 & W==0])/sum(p[Z==1 & W==0]),
                            sum(X[Z==0 & W==1]*p[Z==0 & W==1])/sum(p[Z==0 & W==1]),
                            sum(X[Z==1 & W==1]*p[Z==1 & W==1])/sum(p[Z==1 & W==1])
  ), nrow = 2, ncol = 2))
  # E(E(X|Z,W))
  EExgzw <- sum(p*mapply(function(z,w) Exgzw[z+1,w+1], z=A$Z, w=A$W))
  stopifnot(EExgzw == Ex)
  # E(E(X|Z,W) Y)
  EExgzw_y <- sum(p*mapply(function(z,w) Exgzw[z+1,w+1], z=A$Z, w=A$W)*A$Y)
  
  cov_xy <- with(A, sum(p*(X - mean(Ex))*(Y - mean(Ey))))
  cov_xz <- with(A, sum(p*(X - mean(Ex))*(Z - mean(Ez))))
  cov_xw <- with(A, sum(p*(X - mean(Ex))*(W - mean(Ew))))
  cov_yz <- with(A, sum(p*(Z - mean(Ez))*(Y - mean(Ey))))
  cov_yw <- with(A, sum(p*(W - mean(Ew))*(Y - mean(Ey))))
  cov_zw <- with(A, sum(p*(Z - mean(Ez))*(W - mean(Ew))))
  
  return(list(Exy=Exy, ExEy=ExEy, EExgz_y=EExgz_y,
              Exw=Exw, ExEw=ExEw, EExgz_w=EExgz_w,
              EExgzw_y=EExgzw_y,
              cov=matrix(c(cov_xy,0,0,cov_xz,cov_yz,0,cov_xw,cov_yw,cov_zw), ncol=3, dimnames = list(c("X","Y","Z"), c("Y","Z","W")))))
}

################################################################################

A <- expand.grid(X=c(0,1), Y=c(0,1), Z=c(0,1), W=c(0,1))

p_x <- data.frame(X = c(0, 0, 1, 1),
                  Z = c(0, 1, 0, 1),
                  p_x = c(0.3, 0.4, 0.7, 0.6))

A <- A %>% dplyr::inner_join(p_x, by = c("X", "Z"))

A$p_y <- ifelse(A$Y == 0, 0.15, 0.85)
A$p_z <- ifelse(A$Z == 0, 0.8, 0.2)
A$p_w <- ifelse(A$W == 0, 0.1, 0.9)

# P(X,Y,Z,W) = P(W)*P(Z|W)*P(Y|W,Y)*P(X|Z,W,Y)
#  = P(W)*P(Z)*P(Y)*P(X|Z)
p <- with(A, p_x * p_y * p_z * p_w)

calc(A, p)
#> $Exy
#> [1] 0.578
#> 
#> $ExEy
#> [1] 0.578
#> 
#> $EExgz_y
#> [1] 0.578
#> 
#> $Exw
#> [1] 0.612
#> 
#> $ExEw
#> [1] 0.612
#> 
#> $EExgz_w
#> [1] 0.612
#> 
#> $EExgzw_y
#> [1] 0.578
#> 
#> $cov
#>              Y             Z             W
#> X 6.505213e-19 -1.600000e-02 -1.301043e-18
#> Y 0.000000e+00  1.734723e-18  1.084202e-19
#> Z 0.000000e+00  0.000000e+00 -1.084202e-19



################################################################################

A <- expand.grid(X=c(0,1), Y=c(0,1), Z=c(0,1), W=c(0,1))

A$p_z <- ifelse(A$Z == 0, 0.8, 0.2)

p_x <- data.frame(X = c(0, 0, 1, 1),
                  Z = c(0, 1, 0, 1),
                  p_x = c(0.3, 0.4, 0.7, 0.6))

A <- A %>% dplyr::inner_join(p_x, by = c("X", "Z"))

p_y <- data.frame(Y = c(0, 0, 0, 0, 1, 1, 1, 1),
                  X = c(0, 0, 1, 1, 0, 0, 1, 1),
                  Z = c(0, 1, 0, 1, 0, 1, 0, 1),
                  p_y = c(0.1, 0.2, 0.3, 0.4, 0.9, 0.8, 0.7, 0.6))

A <- A %>% dplyr::inner_join(p_y, by = c("Y", "X", "Z"))

A$p_w <- c(0.01, 0.02, 0.03, 0.04, 0.05, 0.06, 0.07, 0.08, 0.99, 0.98, 0.97, 0.96, 0.95, 0.94, 0.93, 0.92)

# P(X,Y,Z,W) = P(Z)*P(X|Z)*P(Y|Z,X)*P(W|X,Y,Z)
p <- with(A, p_x * p_y * p_z * p_w)

calc(A, p)
#> $Exy
#> [1] 0.464
#> 
#> $ExEy
#> [1] 0.50592
#> 
#> $EExgz_y
#> [1] 0.5072
#> 
#> $Exw
#> [1] 0.65232
#> 
#> $ExEw
#> [1] 0.6530176
#> 
#> $EExgz_w
#> [1] 0.653616
#> 
#> $EExgzw_y
#> [1] 0.5073216
#> 
#> $cov
#>          Y       Z           W
#> X -0.04192 -0.0160 -0.00069760
#> Y  0.00000 -0.0128 -0.00287808
#> Z  0.00000  0.0000 -0.00598400

Created on 2022-08-08 by the reprex package (v2.0.1)

Improve this answer
$\endgroup$
3
  • $\begingroup$ How should I use your constructions to help solve the original problem? I am still at lost. $\endgroup$
    – Zhanxiong
    Commented Aug 9, 2022 at 3:43
  • $\begingroup$ I am also at a loss. I was searching for a situation where the first two conditions were true, but the third is not so that I could understand the underlying reason. I was exploring the covariance between the RVs to see if that could be the cause. I agree this isn't a complete answer, but I shared in the hope it would help. This is an interesting question. $\endgroup$
    – R Carnell
    Commented Aug 9, 2022 at 12:15
  • $\begingroup$ Thanks for your interest and experiment. I tried to solve it in the same way as you. It is indeed a simple question (in appearance) but profound. To my surprise, it didn't draw much attention even after a bounty :). Probably it takes too long time to "search". $\endgroup$
    – Zhanxiong
    Commented Aug 9, 2022 at 15:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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