Random variables $x,y,z$ have a multivariate normal distribution with mean $\mu_x$ $\mu_y$ and $\mu_z$; and variance and covariance: $\sigma_x^2$,$\sigma_y^2$,$\sigma_z^2$,$\sigma_{xy}$,$\sigma_{xz}$,$\sigma_{yz}$. Please find the conditional variance $\operatorname{Var}[y\mid x,z]$. Also, it is known that $\operatorname{E}[y\mid x]=\beta x+\alpha$ and $\operatorname{E}[y\mid x,z]=\beta_x x+\beta_z z+\alpha_2$

So far, I have used the variance decomposition rule and have gotten: $\operatorname{Var}[y\mid x]=\operatorname{Var}_z[\operatorname{E}[y\mid x,z]] + \operatorname{E}_z[\operatorname{Var}[y\mid x,z]]$, and $\operatorname{Var}[x\mid y] = \operatorname{Var}[y]-\beta^2\operatorname{Var}[x]$, but I am confused with the conditional mean and variance, and don't know how to proceed. Any hint or help will be greatly appreciated, thanks!!!


Though I haven't gone through the whole calculation a good way to start would be by calculating the conditional density of $X|Y,Z$, which is given by: $$f_{X|Y,Z}(x)=\frac{f_{X,Y,Z}(x,Y,Z)}{f_{Y,Z}(Y,Z)}$$

The top density is the joint density of the variables (the basic multinormal density). The bottom one is the multinormal density ignoring $X$, which you can build since you know the variances and covariance of $Y$ and $Z$.

I would expect $f_{X|Y,Z}$ to be the density function of some known distribution (for example $f_{X|Y}$ turns out to be a normal density). This would allow you to quickly infer the variance that you are looking for.

In case of no "easy" distribution coming out at that point you could always take the long road of calculating $V[X|Y,Z]$ by integration:

$$E[X|Y,Z]=\int_{-\infty}^\infty xf_{X|Y,Z}(x)dx$$ $$V[X|Y,Z]=\int_{-\infty}^\infty (x-E[X|Y,Z])^2 f_{X|Y,Z}(x)dx$$


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