I am trying to formalize the following problem mathematically, and I am failing to do so.
I am in a an island, and there is a (small) unknown number of people whose objective is to find me. They find me when any of them spot me. Given a detailed map of the island, including buildings, I want to pick a position $\vec{x}$ that maximizes the chance that I am not spotted. Let us assume that this is a 2D problem, i.e. no heights and that there is no time here: people do not move.
Assume that each position $\vec{y}$ has or not has direct line of sight to another location $\vec{x}$ and let $r(\vec{x},\vec{y})$ be this indicator function.
My approach so far:
- define the event "I am currently being seen by anyone" to be $s$
- define the event "I am currently being seen by someone at $\vec{y}$" to be $s(\vec{y})$.
- assume that I know $P(s(\vec{y})|\vec{x})$
- assume that if a seeker has a line of sight to me, it has a probability $p$ of spotting me.
- Try to write $P(s|\vec{x})$ as a function of $P(s(\vec{y})|\vec{x})$.
My problem is in step 5. Let me take you through some examples to explain how I am stuck. In the examples below, assume that $r(\vec{x},\vec{y}) = 1$ for all points, but focus on a solution that allows to include more complicated r functions.
Example 1: I know that there will be exactly one seeker on the island exactly at position $\vec{y}_1$.
In this case, $P(s(\vec{y}_1)|\vec{x}) = p$ and all other events have probability zero. Thus, $P(s|\vec{x}) = p$.
Example 2: I know that there will be one seeker on the island but the probability of it being at some place is a normal distribution (I do not know it exactly)
Specifically, say that his position is a normal distribution centered at $\vec{y}_1$ and width $\sigma$, $N(\vec{y}_1,\sigma)$.
In this case, I cannot assign a single position to the event. However, because I know that there is only 1 seeker in the island, I know that the events $s(\vec{y}_i)$ and $s(\vec{y}_j)$ are mutually exclusive for all pairs $i,j$ whose $i\neq j$: if I am found from $\vec{y}_i$ I cannot be found from $\vec{y}_h$.
$P(s(\vec{y})|\vec{x}) = p N(\vec{y}_1,\sigma)$
The event $s$ is just the union of $s(\vec{y})$, $P(s|\vec{x}) = P(\bigcup s(\vec{y})|\vec{x})$. Because each event is mutually exclusive, I write that $$P(\bigcup s(\vec{y})|\vec{x}) = \int P(s(\vec{y})|\vec{x}) dy$$
And I am done.
Example 3: I know that there are 2 seekers on the island and the probability of each of them being at some place is a normal distribution
In this case, $s(\vec{y}_i)$ and $s(\vec{y}_j)$ are not mutually exclusive because 2 people can see me.
In this case, $P(s|\vec{x}) = P(\bigcup s(\vec{y})|\vec{x})$, but I need to consider the different cases ("seeker A sees me and seeker B does not", "seeker B sees me and seeker A does not", and "both seekers see me". I.e. $s = \left(s_1(\vec{y}_1) \land \neg s_2(\vec{y}_2)\right) \lor \left(\neg s_1(\vec{y}_1) \land s_2(\vec{y}_2)\right) \lor \left(s_1(\vec{y}_1) \land s_2(\vec{y}_2) \right)$. The probability would be something like this "integrated over $\vec{y}_1$ and $\vec{y}_2$". I do not really know how to formalize this.
Example 4: I do not know how many seekers there are on the island but the probability of each of them being at some place is a normal distribution
In this case, $s(\vec{y}_i)$ and $s(\vec{y}_j)$ are not mutually exclusive because 2 or more people see me. The event $s$ is now a combination of all possible permutations over all possible events $s(\vec{y})$. And I am lost again in formalizing this.
Is there a procedure or framework to approach this type of problems?
You've made some serious mistakes in the work you've done so far that need to be addressed before looking at the more difficult problems.
First of all, for any continuous probability distribution on where the seekers are, the probability $P(s(y))$ is $0$ for all $y$. For example, if the position of a seeker is normally distributed, the seeker is at position $y$ with probability $0$. You say further down that the events $s(y)$ are pairwise disjoint, which is true, but in fact you can't have an uncountable family of pairwise disjoint events in a probability space unless the probability of all but countably many of them is zero (proving this is a worthwhile exercise). Thus, your statement that you "know $P(s(y))$" is a bit vacuous.
On the basis of mutual disjointness, you then claim
$$P(s)=\int P(s(y))dy$$
Based on what I said in the last paragraph, that integral is the integral of an identically zero function, and so zero. This equality can't be true. I think I probably don't need to tell you this, but you can't just go around replacing finite sums with integral signs and expect to get correct answers! And not just because it's not formally mathematically justified. Even intuitively, the equation above is not a generalization of the formula $P(\bigcup A_i)=\sum P(A_i)$ for disjoint $A_i$.
An integral is not a sum. The integral of $f$ is not the sum of lots of values of $f$ - not even in an intuitive sense. You're forgetting something important. An integral is a sum of values of $f$ multiplied by tiny little areas. Thus your integral is not so much $\sum P(A_i)$ as it is $\sum a_i P(A_i)$, or some such. Whenever we have $F=\int fdx$, $f$ is not the same type of "thing" as $F$. What an equation like that means is that $f$ is the density of $F$ with respect to area, or volume or whatever. It means that $f$ is a ratio of a thing like $F$ with a thing like $x$. If for example the quantities involved have units, then $f$ has units of $F/x$. This is exactly why $P(A)=\int_A p(x)dy$ does work when $p$ is a probability density function. As its name implies, a probability density function is a density: it measures the probability per unit area at a point.
Now let's look at how to actually solve that second example problem. The concept you need here is the notion of conditional probability with respect to a random variable - not to be confused with the significantly simpler notion of conditional probability with respect to an event. This concept is usually built up using the conditional expectation, but here I'll treat it more intuitively.
Now that your intuitions are hopefully straightened out after the preamble about integration, let's allow ourselves to forget that probabilities like $P(s(y))$ are zero and work intuitively. You're right that we can in principle know $P(s(y))$, but what's more important is that we know how to break it down into simpler probabilities.
$$P(\text{the seeker sees me from $y$}) = P(\text{the seeker is at $y$})P(\text{the seeker sees me}\mid\text{the seeker is at $y$})$$
It seems like we can calculate both of the probabilities on the right. $P(\text{the seeker is at $y$})$ is given by the normal distribution, and the conditional probability is some known function of $y$ determined by the geography of the region, like you said. In fact, I think what you might have meant in your point (3) in the question is that you know that conditional probability, not that you know $P(s(y))$ which, although true, is really skipping a step.
Let $A_y = \{\text{the seeker is at $y$}\}$ and let $S$ be the event that the seeker sees us. If we attempt to generalize the law of total probability to the continuous case, we might expect to get:
$$P(S)=\int P(S|A_y) \frac{P(A_y)}{dy}dy$$
That fraction is meant to indicate some sort of density of $P(A_y)$ with respect to $y$, which of course is just the probability density function of the position of the seeker (which is Gaussian). The term $P(S|A_y)$ is just our known function telling us how easy it is to see you from the point $y$, so this integrand is something we can actually write down a formula for. If, say, $P(S|A_y)=\frac1{1+\lVert y-x\rVert}$, where $x$ is our position, then
$$P(S)=\int_{\mathbb R^2} \frac{f(y)}{1+\lVert y-x\rVert}dy$$
where $f$ is the PDF of Gaussian, and we have a perfectly concrete formula.
To prove that this sort of thing is justified, we need to ask what the hell this $P(S|A_y)$ thing is. Note that $P(A_y)=0$, so we can't apply the usual formula for conditional probability. But explaining the formal basis for conditional probability would make this answer three times as long as it already is.
The other problems you describe are more complicated, but the tools used are the same.