= Solution
Add the following concrete method inside `FiniteDist[T]`. It forms a <pushforward of a finite probability distribution>, adding the masses of outcomes that map to the same result:
``
def map[U](f: T => U): FiniteDist[U] = {
val source = getDom
val weighted = source.map(x => (f(x), getProb(x)))
val target = weighted.map(_._1).distinct
new FiniteDist[U] {
def getDom: List[U] = target
def getProb(y: U): Double =
weighted.filter(pair => pair._1 == y).map(_._2).sum
}
}
``
The input <finite probability distribution> is assumed to satisfy its interface, with stable masses during construction, and `f` must be a total deterministic <pure function> on its finite support. Output values need stable equality consistent with the mathematical equality being modeled; mutating equality-relevant fields afterwards would undermine any finite-support implementation. Computing `weighted` once also prevents later queries from re-evaluating `f` inconsistently.
Every source support element appears once in `weighted`. The `distinct` operation lists each image value once; every such image has a positive mathematical mass because it has a positive-mass preimage. For each $y$, filtering selects exactly those $x$ with $f(x)=y$, so `getProb(y)` is the required sum, with an empty sum of zero off the image. The fibers form a <set partition> of the original support, giving
$$
\sum_y p_2(y)=\sum_y\sum_{x:f(x)=y}p_1(x)=\sum_xp_1(x)=1.
$$
Thus \b[colliding images have their probabilities added, not overwritten]. For a concrete check, mapping a uniform distribution on $\{0,1,2,3\}$ by parity gives probability $1/2$ to each of $0,1$; mapping every source value to one constant gives a singleton domain with total mass one. <Floating-point rounding> can make a computed total differ slightly from one; the proof is for exact mathematical masses, and the representation uses `Double` approximations.
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