Monotone and Dominated Convergence Theorems

Suppose a sequence of random variables $X_n$ converges to some $X$ as $n$ grows. It is tempting to assume $E[X_n]$ converges to $E[X]$ — that you can just swap the limit and the expectation. This is false in general, and the gap between "limit of the averages" and "average of the limit" is not a technicality; it is where entire pricing models go wrong if ignored. Monotone and dominated convergence are the two conditions under which the swap is actually legal.

The analogy: a shrinking spike

Picture a sequence of narrow spikes on a graph, each one taller and thinner than the last, always covering exactly one unit of area, marching off toward a single point. At every fixed location the spike's height eventually goes to zero — the sequence of heights converges to the flat zero function. But the area under each spike stays at 1 forever; it never goes to zero. So the limit of the areas is 1, while the area of the limit (the flat zero function) is 0. Swapping limit and integral silently lost all the area. Monotone and dominated convergence are exactly the two situations where this kind of vanishing-area trick cannot happen, so the swap is guaranteed safe.

Writing it down

Monotone convergence theorem (MCT). If $0 \leq X_1 \leq X_2 \leq \cdots$ is a non-negative, non-decreasing sequence of random variables converging pointwise to $X$, then

$$\lim_{n\to\infty} E[X_n] = E\left[\lim_{n\to\infty} X_n\right] = E[X] .$$

In words: if the sequence only ever climbs (never dips) and stays non-negative, the expectation of where it ends up equals the limit of the expectations along the way — no area can leak out because nothing is allowed to shrink.

Dominated convergence theorem (DCT). If $X_n \to X$ pointwise and there exists a single fixed random variable $Y$ with $E[Y] < \infty$ and $|X_n| \leq Y$ for every $n$ (every term in the sequence, forever, stays under one shared "umbrella"), then again

$$\lim_{n\to\infty} E[X_n] = E[X] .$$

In words: as long as an integrable umbrella $Y$ bounds every term in the sequence, no term can sneak off to build up unbounded area the way the spike example did — the umbrella caps how much area could possibly hide.

Worked example 1: MCT on a simple increasing sequence

Let $X_n = \min(X, n)$ for a non-negative random variable $X$ with $E[X] = 5$ (a "capped" version of $X$, capped at level $n$). As $n$ grows, $X_n$ only ever increases (raising the cap can only let more value through, never less) and $X_n \to X$ pointwise. Since $0 \leq X_1 \leq X_2 \leq \cdots$, MCT applies directly: $\lim_n E[X_n] = E[X] = 5$. Concretely, with $n = 1$ the cap throws away most of the upside so $E[X_1] < 5$; with $n = 100$ almost nothing is capped so $E[X_{100}] \approx 5$; the sequence of expectations climbs monotonically toward 5, exactly tracking the theorem.

Worked example 2: DCT rescuing a shrinking-tail calculation

Let $X_n = X \cdot \mathbb{1}(X \leq n)$ for $X$ exponential with mean 2, so $E[X] = 2$. As $n \to \infty$, $X_n \to X$ pointwise (eventually $n$ exceeds any fixed outcome, so the indicator turns on and stays on). Does $E[X_n] \to E[X]$? Check the DCT condition: $|X_n| \leq X$ for every $n$ (capping the indicator only ever removes mass, never adds it), and $E[X] = 2 < \infty$, so $X$ itself is a valid umbrella. DCT applies: $\lim_n E[X_n] = E[X] = 2$. Contrast this with the spike example above, where no finite umbrella exists (the spikes get arbitrarily tall) — that is precisely why the swap failed there and succeeds here.

What this means in practice

Every time a pricing model computes an expected payoff as $n \to \infty$ time steps, or a risk model takes a limit of a discretised sum of losses, it is implicitly relying on MCT or DCT to justify swapping a limit with an expectation. Monte Carlo pricing, martingale convergence arguments, and the justification for Fubini-style reordering all lean on these theorems. Without them, "more simulation paths converges to the true price" would be an assumption, not a proven fact.