Question About Bayesian stats( from a DSP estimation theory book)

from "Fundamentals of Statistical Signal Processing, Volume I: Estimation Theory"

It is a fundamental rule of estimation theory that the use of prior knowledge will lead to a more accurate estimator. For example, if a parameter is constrained to lie in a known interval, then any good estimator should produce only estimates within that interval. In Example 3.1 it was shown that the MVU estimator of $$A$$ is the sample mean $$\bar{x}$$. However, this assumed that $$A$$ could take on any value in the interval $$-\infty. Due to physical constraints it may be more reasonable to assume that $$A$$ can take on only values in the finite interval $$-A_0 \leq A \leq A_0$$. To retain $$\hat{A}=\bar{x}$$ as the best estimator would be undesirable since $$\hat{A}$$ may yield values outside the known interval. As shown in Figure 10.1a, this is due to noise effects. Certainly, we would expect to improve our estimation if we used the truncated sample mean estimator $$\check{A}=\left\{\begin{array}{cc} -A_0 & \bar{x}<-A_0 \\ \bar{x} & -A_0 \leq \bar{x} \leq A_0 \\ A_0 & \bar{x}>A_0 \end{array}\right.$$ which would be consistent with the known constraints. Such an estimator would have the PDF \begin{aligned} & p_{\check{A}}(\xi ; A)=\quad \operatorname{Pr}\left\{\bar{x} \leq-A_0\right\} \delta\left(\xi+A_0\right) \\ &+p_{\hat{A}}(\xi ; A)\left[u\left(\xi+A_0\right)-u\left(\xi-A_0\right)\right] \\ &+\operatorname{Pr}\left\{\bar{x} \geq A_0\right\} \delta\left(\xi-A_0\right) \end{aligned}

where u(x) is the unit step function. This is shown in Figure 10.1b.

I have seen a few different definitions of the unit step function:

1. $$u(t)= \begin{cases}0 & t<0 \\ 1 & t>0\end{cases}$$

2. $$u(t)= \begin{cases}0 & t<0 \\ 1 & t \geq 0\end{cases}$$

My confusion is with this equation

\begin{aligned} & \operatorname{Pr}\left\{\bar{x} \leq-A_0\right\} \delta\left(\xi+A_0\right) \\ & \quad+p_{\hat{A}}(\xi ; A)\left[u\left(\xi+A_0\right)-u\left(\xi-A_0\right)\right] \\ & \quad+\operatorname{Pr}\left\{\bar{x} \geq A_0\right\} \delta\left(\xi-A_0\right) \end{aligned}

I understand that $$u\left(\xi+A_0\right)-u\left(\xi-A_0\right)$$ is a rectangular impulse that "windows" the gaussian pdf between $$-A_0$$ and $$A_0$$; it looks like the following

I dont understand this part of the equation \begin{aligned} \operatorname{Pr}\left\{\bar{x} \leq-A_0\right\} \delta\left(\xi+A_0\right) \quad ,\operatorname{Pr}\left\{\bar{x} \geq A_0\right\} \delta\left(\xi-A_0\right) \end{aligned}

Why are they using a Pr{} as opposed to $$p_{\hat{A}}(\xi ; A)$$? why are they multiplying it with shifted dirac deltas?

one thing thats specifically unclear to me is why this truncation even creates those two terms. Is it due to possibly using the 1st definition, where $$u(\xi-A_0)$$ is undefined at $$A_0$$?

• $Pr()$ would refer to the probability of the event in $()$; $p()$ refers to the probability density function. Commented May 28 at 2:47

This truncation approach makes the distribution have different specification in the interior than in the boundaries of the interval $$[-A_0, A_0]$$. The interior can be represented by $$u(\xi+A_0)-u(\xi-A_0)\in(-A_0,A)$$, and here the truncated distribution maintains the same density $$p$$ of the (non-truncated) sample mean. The boundary points are $$\{-A_0,A_0\}$$, and they can be modeled by degenerate distributions $$\delta(\xi+A_0)$$ and $$\delta(\xi+A_0)$$, representing point-events for an absolutely continuous variable. These points have "masses" $$\mathbb{P}(\bar{x}\leq -A_0)$$ and $$\mathbb{P}(\bar{x}\geq A_0)$$, respectively.