Question

The Conditional Covariance Formula. The conditional covariance of $X$and $Y$, given $Z$is defined by$Cov(X,Y\mid Z)\equiv E\left[\right(X-E[X\mid Z]\left)\right(Y-E[Y\mid Z])\mid Z]$

a) Show that$Cov(X,Y\mid Z)=E[XY\mid Z]-E[X\mid Z]E[Y\mid Z]$

b) Prove the conditional covariance formula $\mathrm{Cov}(X,Y)=E\left[\mathrm{Cov}\right(X,Y\mid Z\left)\right]+\mathrm{Cov}\left(E\right[X\mid Z],E[Y\mid Z\left]\right)$

c) Set $X=Y$in part (b) and obtain the conditional variance formula.

Short answer

a) It has been shown that$\mathrm{Cov}(X,Y\mid Z)=E[XY\mid Z]-E[X\mid Z]E[Y\mid Z]$

b) The conditional covariance formula has been proved

$\mathrm{Cov}(X,Y\mid Z)=E[XY\mid Z]-E[X\mid Z]E[Y\mid Z]$

c) The conditional variance formula is$\mathrm{Var}\left(\mathrm{Y}\right)=\mathrm{E}\left[\mathrm{Var}\right(\mathrm{Y})\mid \mathrm{Z}]+\mathrm{Var}\left[\mathrm{E}\right(\mathrm{Y}\mid \mathrm{Z}\left)\right]$

Step-by-step solution

Given Information (Part a) 

Show that$Cov(X,Y\mid Z)\equiv E\left[\right(X-E[X\mid Z]\left)\right(Y-E[Y\mid Z])\mid Z]$

Explanation (Part a) 

We are given that,

$\begin{array}{r}\mathrm{Cov}(X,Y\mid Z)=E[X-E(X\mid Z\left)\right(Y-E(Y\mid Z)\mid Z\left)\right]\end{array}$

$\begin{array}{r}\Rightarrow \mathrm{Cov}(X,Y\mid Z)=E[XY-XE(Y\mid Z)-YE(X\mid Z)+E(X\mid Z)\cdot E(Y\mid Z)\mid Z]\end{array}$

$=E(XY\mid Z)-E\left[XE\right(Y\mid Z)\mid Z]-E\left[YE\right(X\mid Z)\mid Z]+E\left[E\right(X\mid Z\left)E\right(Y\mid Z)\mid Z]$

$=E[XY\mid Z]-E(X\mid Z)·E(Y\mid Z)-E(Y\mid Z)·E(X\mid Z)+E[X\mid Z]·E[Y\mid Z]$

$=E[XY\mid Z]-E[X\mid Z]·E[Y\mid Z]$

Final Answer (Part a)

It has been shown that$\mathrm{Cov}(X,Y\mid Z)=E[XY\mid Z]-E[X\mid Z]E[Y\mid Z].$

Given Information (Part b) 

The conditional covariance formula$=\mathrm{Cov}(X,Y)=E\left[\mathrm{Cov}\right(X,Y\mid Z\left)\right]+\mathrm{Cov}\left(E\right[X\mid Z],E[Y\mid Z\left]\right)$

Explanation (Part b) 

b) $\text{R.H.S.}=E\left[\mathrm{Cov}\right(X,Y\mid Z\left)\right]+\mathrm{Cov}\left[E\right(X\mid Z),E(Y\mid Z\left)\right]$

Using Result of part (a)

$\begin{array}{r}=E\left[E\right(XY\mid Z)-E(X\mid Z\left)E\right(Y\mid Z\left)\right]+E\left[E\right(X\mid Z)\cdot E(Y\mid Z)\mid Z]-E\left[E\right(X\mid Z)\mid Z]\cdot E\left[E\right(Y\mid Z)\mid \end{array}$

$\begin{array}{r}=E\left(XY\right)-E\left(X\right)E\left(Y\right)+E\left(X\right)\cdot E\left(Y\right)-E\left(X\right)\cdot E\left(Y\right)\end{array}$

role="math" localid="1647526707691" $=E\left(XY\right)-E\left(X\right)\cdot E\left(Y\right)$

$=\mathrm{Cov}(X,Y)$

Final Answer (part b) 

Therefore, the conditional covariance formula $\mathrm{Cov}(X,Y)=E\left[\mathrm{Cov}\right(X,Y\mid Z\left)\right]+\mathrm{Cov}\left(E\right[X\mid Z],E[Y\mid Z\left]\right)$has been proved.

Given Information (Part c) 

Set$X=Y$in part (b)

Explanation (Part c) 

c) Putting $X=Y$in result of part (b)

role="math" $\mathrm{Cov}(X,X)=E\left[\mathrm{Cov}\right(X,X)\mid Z]+\mathrm{Cov}\left[E\right(X\mid Z),E(X\mid Z\left)\right]$

$\Rightarrow \mathrm{Var}\left(X\right)=E\left[\mathrm{Var}\right(X)\mid Z]+\mathrm{Var}\left[E\right(X\mid Z\left)\right]$

Similarly$\mathrm{Var}\left(\mathrm{Y}\right)=\mathrm{E}\left[\mathrm{Var}\right(\mathrm{Y})\mid \mathrm{Z}]+\mathrm{Var}\left[\mathrm{E}\right(\mathrm{Y}\mid \mathrm{Z}\left)\right]$

Final Answer 

Therefore, the conditional variance formula is$\mathrm{Var}\left(Y\right)=E\left[\mathrm{Var}\right(Y)\mid Z]+\mathrm{Var}\left[E\right(Y\mid Z\left)\right]$.