# Understanding the Opening Notation from *Statistical Decision Theory & Bayesian Analysis*

I am having significant trouble understanding the notation used in the opening pages of Statistical Decision Theory & Bayesian Analysis. The book assumes that the reader has some basic familiarity with probability theory (which I do). Section 1 below outlines the book notation. Section 2 outlines my actual questions.

I'm sorry this post is so long, but I suspect that all of my questions are related to some core misunderstanding that can (hopefully) be explained/pointed out rather simply.

## 2 Questions

Let us assume, using the notation above, that $X : \mathscr{X} \rightarrow \mathbb{R}$ is a continuous random variable.

1. Does the phrase from the book "$A ( A \subset \mathscr{X})$" mean anything more than the simple claim $A \subset \mathscr{X}$?

2. Do we have that

$$P_\theta (A) = P_\theta (X \in A) = P_\theta (X^{-1}(\mathbb{R}) \cap A)?$$

3. Expanding on (2), does the notation

$$P_\theta(A)$$

imply that

$$\mathscr{X} = (\ldots \times \Theta \times \ldots)$$

so that with the notation

$$\mathscr{X}_\theta = \{ (\ldots, \theta, \ldots) \mid \theta \in \Theta \} \subset \mathscr{X}$$

we have that $P_\theta(A)$ can be further expanded as

$$P_\theta(A) = P(\mathscr{X}_\theta \cap A)?$$

4. The subscript on the $\sum$ in the snippet shows that $x \in A$, and we know that $A \subset \mathscr{X}$ (where $\mathscr{X}$ is the sample space). But isn't this definition meaningless unless $x \in X$? That is, don't we have that $f : \text{Codomain}(X) \rightarrow [0,1]$? Clearly, $A$ needn't be a subset of $\text{Codomain}(X) = \mathbb{R}$.

EDIT: I have added the questions below, since they use the same notation as above:

1. What on earth does

$$P_\theta(\delta_1(X) = \delta_2(X) )$$

mean? Is it short-hand for

$$P_\theta(\delta_1(X)) = P_\theta(\delta_2(X) ) = 1?$$

If so, would that mean we have a paramaterized probability distribution defined on the sample space $\mathscr{A}$?

2. Wrapping up, it seems we so far have at least two sample spaces: the first being $\mathscr{X} = X = \{x\}$, and the second being $\Theta$ (and, per the question above, perhaps even a third on $\mathscr{A})$. Is this the case?

Let us assume, using the notation above, that $X : \mathscr{X} \rightarrow \mathbb{R}$ is a continuous random variable.

That is not how they are using $X$ here.   Their notation is not fully compatible with the usual measure theoretical definitions.

They are using the measurable space as the sample space; that is using the values of the random variable as the outcomes.

Thus their definition of "random variable" is simply to use $X$ to denote the realised outcome.   So $X\in\mathscr X$ and $\mathscr X\subseteq \Bbb R^n$

1. Does the phrase from the book "$A ( A \subset \mathscr{X})$" mean anything more than the simple claim $A \subset \mathscr{X}$?

They do mean "... the event $A$, where $A\subset\mathscr X$, ...".

1. Do we have that $$P_\theta (A) = P_\theta (X \in A) = P_\theta (X^{-1}(\mathbb{R}) \cap A)?$$

No. Usually when we write $P(X\in B)$ where $B$ was a subset of the measurable space we'd mean $P(X^{-1}(B))$, and $X^{-1}(B)$ would be the preimage of $B$, a subset of the sample space. They're not doing it that way.

They just mean $P_\theta(X\in A)$, the probability that the realised outcome is in $A$, can be written as simple $P_\theta(A)$, the probability measure for event $A$.

1. Expanding on (2), does the notation $P_\theta(A)$ ....

No.   It is just the probability measure for the set of outcomes $A$ evaluated with parameter $\Theta$ equal to $\theta$.

1. The subscript on the $\sum$ in the snippet shows that $x \in A$, and we know that $A \subset \mathscr{X}$ (where $\mathscr{X}$ is the sample space). But isn't this definition meaningless unless $x \in X$? That is, don't we have that $f : \text{Codomain}(X) \rightarrow [0,1]$? Clearly, $A$ needn't be a subset of $\text{Codomain}(X) = \mathbb{R}$.

Like I said, they are not distinguishing between the sample space and the measureable space. An outcome is the measure, and event $A$ is a set of values for the random variable.

What they are doing is basically:

Let $\mathscr X=\{1,2,3,4,5,6\}$ be the sample space for the roll of a biased die, and "random variable" $X$ be the result of a roll of that die. The distribution of $X$ follows a parametised probability mass function, $f_{X\mid\Theta}(x\mid\theta)$, for some parameter.   Let $A$ be the event for rolling an even result, so $A=\{2,4,6\}$. $$P_\theta(X\in A) {= P_\theta(\{2,4,6\}) \\= \sum_{x\in\{2,4,6\}} f_{X\mid\Theta}(x\mid\theta)}$$

1. What on earth does $P_\theta(\delta_1(X) = \delta_2(X) )$ mean?

It is the probability that the two rules will produce the same decision based on the same random outcome.   Basically, the probability that two random variables realise the same value.

Notice that $\delta_1, \delta_2$ map the sample space, $\mathscr X$, to a measurable space $\mathscr A$, as is the more usual definition of a random variable.

1. Wrapping up, it seems we so far have at least two sample spaces: the first being $\mathscr{X} = X = \{x\}$, and the second being $\Theta$ (and, per the question above, perhaps even a third on $\mathscr{A})$. Is this the case?

No, you have the one sample space, though of course this may be mapped to various measurable spaces.

The sample space is associated with different possible probability measures, a familiy of parametised functions.   The parameter, $\Theta$, is not itself a random variable, but an unknown value.   However a measure for the likelihood that the paramter has some value when given particular evidence can be evaluated, as if this were a probability function, by Bayesian rules.

• Just to clarify though: the notation used in this book is entirely different than the notation used in modern probability theory (as shown throughout, i.e., wikipedia), right? – user1770201 Apr 2 '18 at 17:56
• In particular, the statements $P_\theta$ and $f(x|\theta)$ are being used completely different than they are here, right? That is: I was interpreting $f(x \mid \theta)$ as a conditional probability, but it's NOT a conditional probability, right? – user1770201 Apr 2 '18 at 17:59
• I edited my question post to include some more content. I'm going to give you bounty afterwards for answering everything. Thanks for all your help on this long post. – user1770201 Apr 2 '18 at 19:32