Expectation of truncated normal X conditional on truncated normal Y

#1
I am trying to derive:

[math]E(X|a \leq Y \leq b)[/math]

where [math]c \leq X \leq d [/math], [math]X[/math] and [math]Y[/math] are (doubly truncated) Gaussians with the same mean and different variance, and [math]a < c < d < b[/math] are the truncation points.

To start off, I wrote:

[math]E(X|a \leq Y \leq b) = \int_c^d x f_{X|c \leq Y \leq d}(x)dx [/math]

where

[math]f_{X|c \leq Y \leq d}(x) = \frac{f_{X}(x)}{Pr(c \leq Y \leq d)} =
\frac{\frac{\frac{1}{\sigma_x}\phi (\frac{x-\mu_x}{\sigma_x})}{\Phi(\frac{d-\mu_x}{\sigma_x}) - \Phi(\frac{c-\mu_x}{\sigma_x})}}{\Phi(\frac{d-\mu_y}{\sigma_y}) - \Phi(\frac{c-\mu_y}{\sigma_y})}, \forall c \leq Y \leq d
[/math]

At this point I'm absolutely stuck. Is what I wrote correct? Is there any other way to derive the result more directly?

I've attached a Pdf as well for you to see the formulas more clearly, if needed.

Any help and advice would be GREATLY appreciated.
 

BGM

TS Contributor
#2
Several questions need to clarify:

1. [math] X [/math] is truncated normal on [math] (c, d) [/math] and [math] Y [/math] is on [math] (a, b) [/math] ?

2. Are they independent? Or the untruncated X and Y are dependent, say bivariate normal? If they are independent, then the conditional expectation will be the same as the ordinary expectation.

3. They have the same mean, or the untruncated version of them having the same mean?
 
#3
Hi BGM,

1. Yes, [math]X[/math] is truncated normal with truncation points [math](c, d)[/math], while [math]Y[/math] is truncated normal with truncation points [math](a,b)[/math], where [math] a<c<d<b[/math].

2. No, [math]X[/math] and [math]Y[/math] are not independent. In fact, [math]Y = X + Z [/math], where [math]Z[/math] is also truncated normal with truncation points [math](c, d)[/math], and [math]X[/math] and [math]Z[/math] are independent.

3. [math]X[/math] and [math]Y[/math] have the same mean. More specifically: the truncated distribution of [math]X[/math] has mean zero. The truncated distribution of [math]Z[/math] also has mean zero. Since [math]X[/math] and [math]Z[/math] are independent, the density of [math]Y[/math] can be reasonably approximated by a truncated Gaussian with mean [math]\mu_Y[/math] equal to [math]\mu_X+\mu_Z=0[/math].