Suppose I have the following Bayesian Network:

Bayesian Network

It's given by the following probability distributions:

$$\begin{aligned}X_1&\sim \mathcal N(\mu, \delta^2)\\ \forall i, 2\leq i\leq n: X_i|X_{i-1}&\sim \mathcal N(x_{i-1}, \lambda^2)\\ \forall i, 1\leq i\leq n: Z_i|X_i&\sim\mathcal N(x_{i}, \sigma_i^2) \end{aligned}$$

In the above, $Z_i$ is actually the mean of various observations drawn from $X_i$, such that the distribution of each draw is also normal and has mean $x_i$ and variance $\sigma^2$ (and therefore $\sigma_i^2$ is actually just $\sigma^2 / I_i$, where $I_i$ is the number of observations/draws made).

If I define:

$$\begin{aligned} \lambda_0^2 &= 0 \\ \delta_0^2 &= \delta^2 \\ \mu_0 &= \mu \\ \forall i, 1 \leq i \leq n : \lambda_i^2 &= \lambda^2 \\ \forall i, 1 \leq i \leq n : \delta_i^2 &= \left(\frac{I_i}{\sigma^2} + \frac {1}{\lambda_{i-1}^2 + \delta_{i-1}^2}\right)^{-1} \\ \forall i, 1 \leq i \leq n : \alpha_i &= \frac {I_i}{I_i + \sigma^2 / \left(\lambda_{i-1}^2 + \delta_{i-1}^2\right)} \\ \forall i, 1 \leq i \leq n : \mu_i &= \alpha_i * z_i + (1 - \alpha_i) * \mu_{i-1} \end{aligned}$$

Then it follows that:

$$\forall i, 1 \leq i \leq n : X_i | \pmb Z_{j \leq i} \sim \mathcal N (\mu_i, \delta_i^2)$$

Where $\pmb Z_{j \leq i}$ is the vector containing all the observations of the $Z_j$ up to and including $Z_i$. Therefore, if I want to reason about the hidden variable $X_n$ based on a vector of observations $\pmb Z_{j \leq n}$, my best point estimate for its value is the $\mu_n$ given by the above recursive relations.

What I want to estimate, however, is not $X_n$ but rather $X_{n + 1}$. The definition of the problem, in which the $X_i$ are performing a random walk, means that:

$$\mathbb E[X_{n+1}|\pmb Z_{j \leq n}] = \mathbb E[X_{n}|\pmb Z_{j \leq n}] = \mu_n$$

So this is sufficient for my purposes.

Now suppose that, for $\Delta^2 = \delta^2$, I have a similar Bayesian network for $A_i$ and $B_i$ with:

$$\begin{aligned}A_1&\sim \mathcal N(\nu, \Delta^2)\\ \forall i, 2\leq i\leq n: A_i|A_{i-1}&\sim \mathcal N(a_{i-1}, \lambda^2)\\ \forall i, 1\leq i\leq n: B_i|A_i&\sim\mathcal N(a_{i}, \sigma^2 / J_i) \end{aligned}$$

Then I can similarly define $\nu_i$ and $\Delta_i$ such that:

$$\forall i, 1 \leq i \leq n : A_i | \pmb B_{j \leq i} \sim \mathcal N (\nu_i, \Delta_i^2)$$

And once again, it is the case that:

$$\mathbb E[A_{n+1}|\pmb B_{j \leq n}] = \mathbb E[A_{n}|\pmb B_{j \leq n}] = \nu_n$$

If I were to combine the two populations $Z_n$ and $B_n$ to find their aggregate estimated mean, it would be given by:

$$\frac {I_n * \mu_n + J_n * \nu_n}{I_n + J_n}$$

What if, however, I want to estimate the mean of the next step of this chain, i.e. of the population of observations given by $Z_{n+1}$ and $B_{n+1}$? Is it the same? Do I have to also have a model for the evolution of the $I_i$ and $J_i$?


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