The setup for computing the conditional (co)variance

Question

Please tell me if my procedure is outlined correctly for the following:

Given that we have

Show that

$$Var(\beta_{igj}|\mu_{gj})=\gamma_{gg}^j+\sigma^2_{gj}$$ where $\gamma_{gg}^j$ i the $g$th diagonal element of $\Gamma_j$.

Question

My question is only if I am setting up my calculation correctly, not to perform the calculation (but if one knows one a derivation, please provide a reference). Am I correct to assume, that I should

1. Calculate the density of $$\beta_{igj},\alpha_{igj},\sigma^2_{gj}| \mu_{gj}\sim p_1$$ which is done by multiplying the densities from lines 1,3 and 4
2. Integrate $p_1$ with respect to $\alpha_{igj},\sigma^2_{gj}$, i.e. integrate these parameters out, ataining the result $p$
3. Compute the variance of $\beta_{igj}$ with respect to this distribution ($p$)

Edit in light of the discussion

Using the tower property, we have (we suppress the dependence of $\mu$)

$$E_{\beta_{igj}}\left[(\beta_{igj}-E(\beta_{igj}))^2\right]=E_{\alpha_{gj},\sigma^2_{gj}}\left[E_{\beta_{igj}}\left[(\beta_{igj}-E(\beta_{igj}))^2| \alpha_{gj},\sigma^2_{gj}\right]\right]$$

but we also have

$$E_{\beta_{igj}}\left[(\beta_{igj}-E(\beta_{igj}))^2| \alpha_{gj},\sigma^2_{gj}\right]=\sigma^2_{gj}$$

This seems like now I must compute

$$E_{\alpha_{gj},\sigma^2_{gj}}\left[\sigma^2_{gj}\right]$$

This is the point where I am stuck now. How can I make the dependence on $\alpha$ and explicit? How do I proceed?Perhaps (wlog assume $\mu=0$)

$$E_{\alpha_{gj},\sigma^2_{gj}}\left[E_{\beta_{igj}}\left[(\beta_{igj}-\alpha_{gj})^2\right]\right]=E_{\alpha_{gj},\sigma^2_{gj}}\left[E_{\beta_{igj}}\left[\beta_{igj}^2-2\alpha_{gj}\beta_{igj}+\alpha_{gj}^2\right]\right]$$

Looking at only the inner expectation for now

$$E_{\beta_{igj}}\left[\beta_{igj}^2-2\alpha_{gj}\beta_{igj}+\alpha_{gj}^2\right]=E_{\beta_{igj}}\left[\beta_{igj}^2\right]-\alpha_{gj}^2$$

Inserting this into the outer expectation, we have

$$E_{\alpha_{gj},\sigma^2_{gj}}\left[E_{\beta_{igj}}\left[\beta_{igj}^2\right]-\alpha_{gj}^2\right]=E_{\alpha_{gj},\sigma^2_{gj}}\left[E_{\beta_{igj}}\left[\beta_{igj}^2\right]\right]-E_{\alpha_{gj},\sigma^2_{gj}}\left[\alpha_{gj}^2\right]$$

But it doesn't seem I have accomplished much.

Answer 1

For a function, $f$, of $\beta_{igj}$ you can use the Tower property of conditional expectations to calculate $$ \mathbb{E}\left[f(\beta_{igj})\right] = \mathbb{E}\left[ \mathbb{E}\left[ f(\beta_{igj})|\alpha_{igj}, \sigma^2_{gj}\right] \right], $$ where the dependence on $\mu_{gj}$ is left unstated. Now you just need to marginalise out $\Gamma_j$ to get the distribution $p(\alpha_{ij} ) = \int p(\alpha_{ij}, \Gamma_j) \operatorname{d}\Gamma_j$ and the result should follow on shortly after.

---

Looking at your question you are asking to show that $\operatorname{Var}(\beta_{igj}|\mu_{gj}) = \gamma_{gg}^j + \sigma_{gj}^2$, the fact that $\sigma_{gj}$ appears on the right-hand side suggests it hasn't been marginalised out, i.e. you are looking at $\operatorname{Var}(\beta_{igj}|\mu_{gj}, \sigma_{gj}^2)$ where only $\alpha$ and the those variables below it in the model hierarchy have been marginalized out.

If we condition on both $\mu_{gj}$ and $\sigma_{gj}^2$ so that $\mathbb{E}\left[\cdot\right] = \mathbb{E}\left[\cdot|\mu_{gj},\sigma_{gj}^2\right]$ then you get (I have dropped the subscripts for now but it is straight forward to put that back in) \begin{align} \mathbb{E}\left[\beta^2\right] &= \mathbb{E}\left[\mathbb{E}\left[ \beta^2|\alpha\right]\right] = \mathbb{E}\left[(\mu+\alpha)^2\right] + \sigma^2 = \mu^2 + \mathbb{E}\left[\alpha^2\right]+\sigma^2. \end{align} You should check the steps involved there and then repeat for $\mathbb{E}\left[\beta\right]$ and I think you will be able to show the result you want.