Is the following conditional expectation property true for four random variables?

Question

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.

Where is this exercise/question from? – Zhanxiong Jul 29 '22 at 20:17 — Zhanxiong, Jul 29 '22 at 20:17

score 0 · Answer 1 · answered Aug 09 '22 at 02:08

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(pA$X)
  Ey <- sum(pA$Y)
  Ez <- sum(pA$Z)
  Ew <- sum(pA$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$Z+1])
  stopifnot(Ex == EExgz)
E(X)E(Y)
ExEy <- Ex * Ey
E(XY)
Exy <- sum(pA$XA$Y)
E(E(X|Z)Y)
EExgz_y <- sum(pExgz[A$Z+1]A$Y)
E(X)E(W)
ExEw <- Ex * Ew
E(XW)
Exw <- sum(pA$XA$W)
E(E(X|Z)W)
EExgz_w <- sum(pExgz[A$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$W))
  stopifnot(EExgzw == Ex)
E(E(X|Z,W) Y)
EExgzw_y <- sum(pmapply(function(z,w) Exgzw[z+1,w+1], z=A$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)}

How should I use your constructions to help solve the original problem? I am still at lost. — Zhanxiong, Aug 09 '22 at 03:43
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. — R Carnell, Aug 09 '22 at 12:15
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". — Zhanxiong, Aug 09 '22 at 15:04

Is the following conditional expectation property true for four random variables?

1 Answers1

E(X|Z)

E(E(X|Z))

E(X)E(Y)

E(XY)

E(E(X|Z)Y)

E(X)E(W)

E(XW)

E(E(X|Z)W)

E(X|Z,W)

E(E(X|Z,W))

E(E(X|Z,W) Y)

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(X,Y,Z,W) = P(Z)P(X|Z)P(Y|Z,X)*P(W|X,Y,Z)