In this note we give a mathematical translation of the monadic treatment to random variables in this amazing stackexchange answer, which is my (figuratively) sixth attempt to understanding monads. We show the meanings of return, bind / (>>=), join, fmap / (<$>) and ap / (<*>) and the monad laws in this concrete example. See Hackage for a clear and concise definition of monads. For notational consistency, we use capital letters for random variables in Haskell code, as is the convention in probability theory. We also use == to stand for derivation in Haskell code. For any state space $$S$$, let $$R(S)$$ be the space of random variables on $$S$$. In the following $$S, T, P$$ are notations of state spaces, and $$X, Y, Z$$ random variables. In the stackexchange answer (we ignore the psuedorandomness and pretend everything is truly random), • return x is defined as the atomic random variable $$\delta_x$$ • For $$f: S \to R(T)$$, X >>= f is the random variable $$\expe_X f(X)$$. It can be thought of in terms of conditional random variables. Say $$X \in R(S)$$ and $$Y \in R(T)$$ are two random variables, and that $$Y$$ given $$X = x$$ is $$f(x)$$. Then X >>= f is the (unconditional) random variable $$Y$$. • join can be derived from (>>=): Let $$X \in R(R(S))$$ be a random random variable, then join X = X >>= id $$= X >>= id_{R(S)} = \expe X$$ is a random variable on $$S$$, obtained by averaging over the random variables $$X$$ can take values in. • X >> Y is simply $$Y$$. Deriving fmap / (<$>):

• And for $$g: S \to T$$, let $$\mu_X$$ be the distribution of $$X$$, then fmap g X = X >>= (return . g) $$= \expe_X (\delta_{g(\cdot)})(X) = \int \delta_{g(x)} \mu_X(d x) = g(X)$$ which agrees with the definition of the randomMap function in that answer.

As for ap / (<*>):

• Let $$F \in R(T^S)$$ and $$X \in R(S)$$, then F <*> X is defined as F <*> X = do {f <- F; x <- X; return (f x)} == do{f <- F; fmap f X} == F >>= (\f -> fmap f X) which is equal to $$\expe_F F(X)$$, a random variable on $$T$$ by averaging over $$F$$.

Now we translate the monad laws:

• For $$x \in S$$ and $$f: S \to R(T)$$, return x >>= f == f x: $$LHS = \delta_x >>= f = \expe_{X \sim \delta_x} f(X) = f(x) = RHS$$.
• For $$X \in R(S)$$, X >>= return == X: $$LHS = \expe_X \delta_X = X = RHS$$.
• For $$X \in R(S), f: S \to R(T), g: T \to R(P)$$, X >>= (\x -> f x >>= g) == (X >>= f) >>= g: Suppose $$Y \in R(T)$$ and $$Z \in R(P)$$ such that $$(Y|X = x) = f(x)$$ and $$(Z|Y = y) = g(y)$$, then \begin{align} LHS &= \expe_X (\expe_{f(\cdot)} g \circ f (\cdot)) (X) = \expe_X \expe_{Y|X = \cdot} g (Y|X = \cdot) = \expe_X (Z|X = \cdot) (X) = Z\\ RHS &= \expe_{\expe_X f(X)} g(\expe_X f(X)) = \expe_Y g(Y) = Z. \end{align} In the LHS we average first over $$Y$$ then over $$X$$ whereas in the RHS we average over $$X$$ then $$Y$$.