Covariance Matrix of a Vector Random Variable w/ Components Related

[CAVision]

Homework Statement

Given X=ZU+Y
where
(i) U,X,Y, and Z are random variables
(ii) U~N(0,1)
(iii) U is independent of Z and Y
(iv) f(z) = $\frac{3}{4}$ z² if 1 $\leq z$ $\leq 2$, f(z)=0 otherwise
(v) f_Y|Z=z(y) = ze^-zy (i.e. Y depends conditionally on Z)
(vi) define vector W=[X Y Z]^T

Q: What is the covariance matrix K_WW?

Homework Equations

K_WW = E[(W-E[W])(W-E[W])^T]
K_WW = $\begin{pmatrix}K_{XX} & K_{XY} & K_{XZ} \\ K_{YX} & K_{YY} & K_{YZ} \\ K_{ZX} & K_{ZY} & K_{ZZ} \end{pmatrix}$

$V[X] = E[(X-E[X])^2] = E[X^2] - E^2[X]$
$V[X] = E_Z[V[X|Z=z]] + V_Z[E[X|Z=z]]$ (Total Variance Theorem)

The Attempt at a Solution

I can calculate using $K_{ZZ} = E[Z^2] - E^2[Z]$, and the definition of expectation.
However, I'm having difficulty calculating the other terms.

For example, if I use the Total Variance Theorem for K_XX I get the following:
(using Total Variance Theorem)
$V[X] = E_Z[V[X|Z=z]] + V_Z[E[X|Z=z]]$
(using X=ZU+Y)
$V[X] = E_Z[V[ZU+Y|Z=z]] + V_Z[E[ZU+Y|Z=z]]$
(using Z=z)
$V[X] = E_Z[V[zU+Y|Z=z]] + V_Z[E[zU+Y|Z=z]]$
(using V[aX+bY|Z=z]=a²V[X|Z=z]+b²V[Y|Z=z]+2abCov[X,Y|Z=z])
$V[X] = E_Z[z^2V[U|Z=z] + V[Y|Z=z] + 2zCov[U,Y|Z=z]] +V_Z[E[zU+Y|Z=z]]$
(using E[aX+bY|Z=z]=aE[X|Z=z] +bE[Y|Z=z])
$V[X] = E_Z[z^2V[U|Z=z] + V[Y|Z=z] + 2zCov[U,Y|Z=z]] +V_Z[zE[U|Z=z] + zE[Y|Z=z]]$
(using U,Z are independent, for U|Z=z terms drop the Z=z)
[itex]V[X] = E_Z[z^2V + V[Y|Z=z] + 2zCov[U,Y|Z=z]] +V_Z[zE + zE[Y|Z=z]] [/itex]
(using E= 0 and V=1 which is given by U~N(0,1) )
$V[X] = E_Z[z^2 + V[Y|Z=z] + 2zCov[U,Y|Z=z]] +V_Z[0 + zE[Y|Z=z]]$

Now, I can find $V[Y|Z=z]$ and $E[Y|Z=z]]$, but I'm not sure how to find Cov[U,Y|Z=z]].

From my understanding U independent of Y means Cov[U,Y]=0, but not necessarily that Cov[U,Y|Z=z] = 0. Let me know if this last statement is incorrect.

I'm I approaching this correctly?
Is there an easier way?

 

[mighty2000]

Your approach is correct so far. To find Cov[U,Y|Z=z], you can use the definition of conditional covariance:

Cov[X,Y|Z=z] = E[(X-E[X|Z=z])(Y-E[Y|Z=z])|Z=z]

Since U and Y are independent, E[U|Z=z] = 0 and E[Y|Z=z] = E[Y], so the above equation becomes:

Cov[U,Y|Z=z] = E[(U-0)(Y-E[Y])|Z=z]

Now, using the definition of expectation, we can rewrite this as:

Cov[U,Y|Z=z] = E[U(Y-E[Y])|Z=z]

And since U is independent of Y, we can pull U out of the expectation:

Cov[U,Y|Z=z] = E[U|Z=z]E[(Y-E[Y])|Z=z]

Again, since U is independent of Y, E[U|Z=z] = E = 0, so we are left with:

Cov[U,Y|Z=z] = 0

Therefore, using the Total Variance Theorem, we can simplify our equation for V[X] to:

V[X] = E_Z[z^2 + V[Y|Z=z]] + V_Z[zE[Y]]

Now, to find E[Y] and V[Y|Z=z], we can use the definition of conditional expectation and variance:

E[Y] = E[E[Y|Z=z]] = E[Z] = 3/4

V[Y|Z=z] = E[(Y-E[Y|Z=z])^2|Z=z] = E[(Y-3/4)^2|Z=z] = E[Y^2|Z=z] - (3/4)^2

Using the definition of expectation, we can find E[Y^2|Z=z]:

E[Y^2|Z=z] = ∫y^2ze-zy dy = z∫y^2e-zy dy = z(2/z^3) = 2/z^2

Therefore, V[Y|Z=z] = 2/z^2 - (3/4)^2

Now, we can plug these values into our equation for V[X] and simplify to get:

V[X] = 3/4 + z^2(2/z^2 - (3/4)^2) + z