Conditional Variance formula $\mathbb{V}[\mu_t - \mu_{t|t-1} | F_{t-1}]$

Question

I am working on linear state space models of the form $$\mu_{t+1} = \mu_t + e_t, \ \ \ \ e_t \sim N(0, \sigma_e^2)$$ $$y_{t} = \mu_t + \eta_t, \ \ \ \ \eta_t \sim N(0, \sigma_{\eta}^2)$$

where $\mu_t$ is a pure random walk

In the formula in the title, restated below for clarity, the term $\mu_{t|t-1}$ is the conditional mean of some state $\mu$ at time $t$ conditioned on all observations up until time $t-1$. So it can be written $\mathbb{E}[\mu_t | F_{t - 1}]$ where $F_{t-1}$ is the set of observations $\{y_1, .., y_{t-1}\}$

I am trying to work out here, if the conditional mean, $\mu_{t|t-1}$, is in fact a constant?

$$\mathbb{V}[\mu_t - \mu_{t|t-1} | F_{t-1}]$$

The text I am reading, Financial Time Series Analysis by Tsay, states that the above formula for the variance is actually $$\mathbb{V}[\mu_t | F_{t-1}]$$

Could someone confirm for me please?

Btw this crops up on page 561 in Eq 2.11 of Financial Time Series Analysis by Tsay

Answer 1

If you're working with a martingale, then the expectation of the next value in the sequence is equal to the present value. see here

Since the previous value is part of the filtration, if the sequence of RVs is a martingale, then $\mathbb{E} [X_{t+1}|X_t] = X_t$ and $\mathbb{E} [X_{t+1}|X_t,F] = x_t$, where $x_t$ is the value $X_t$ took.

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We use the fact that

1. $y_{t|t-1} = \mathbb{E}[y_t|F] = \mathbb{E}[\mu_t+\eta_t|F) = \mathbb{E}[\mu_t|F] = \mu_{t|t-1}$

2. $\mu_{t|t-1}=y_{t|t-1}=\mathbb{E}[y_t|y_{t-1},F] = y_{t-1}$ which is constant given it's part of the history F.

3. $Var [\mu_t - \mu_{t|t-1} | F_{t-1}]=Var [\mu_t - y_{t-1} | F_{t-1}]=Var [\mu_t | F_{t-1}]$