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How to Spot Dominated Convergence in a Probability Solution

Original question

Why does the solution use the dominated convergence theorem in this probability exercise, and is it necessary?
Dominated convergence is useful when a solution moves a limit through an expectation. Check that the random variables converge almost surely and that a single integrable random variable bounds their absolute values. Another theorem may justify the same step if its own hypotheses hold, such as monotone convergence for nonnegative increasing approximations.

Find the line that moves a limit through expectation

Suppose Xn→XX_n\to X almost surely and ∣Xn∣≤Y|X_n|\le Y for all nn almost surely, with E[Y]<∞E[Y]<\infty. Then ∣X∣≤Y|X|\le Y almost surely and ∣Xn−X∣≤2Y|X_n-X|\le2Y. The dominated convergence theorem gives E∣Xn−X∣→0E|X_n-X|\to0. In particular,

∣E[Xn]−E[X]∣≤E∣Xn−X∣→0.|E[X_n]-E[X]|\le E|X_n-X|\to 0.
The natural proof move is to search for the envelope before exchanging the limit and expectation. A fixed deterministic bound works on a probability space. For truncated integrable variables, the absolute value of the original variable often works. If the approximations are nonnegative and increase to the limit, monotone convergence may be simpler because it also covers an infinite limiting expectation. Uniform integrability can supply a different route when a simple pointwise envelope is unavailable.
The dominating variable can depend on the outcome, but it must work for every nn and have finite expectation. A bound that grows with nn is not enough. In the clipped-payoff problem, ∣X∣|X| works because clipping moves the realised value towards zero. Establish convergence at each realised finite XX first. Then use the same integrable bound to control the contribution from rare, very large values. This is exactly the step that pointwise convergence on its own does not justify.
Let ZZ be standard normal and X=3Z−1X=3Z-1. Define the capped and floored payoff Xn=max⁡(−n,min⁡(X,n))X_n=\max(-n,\min(X,n)). Prove that E[Xn]→−1E[X_n]\to-1. For any realised finite XX, the clipping eventually has no effect. Also ∣Xn∣≤∣X∣≤3∣Z∣+1|X_n|\le |X|\le3|Z|+1. The envelope is integrable, since E∣Z∣≤E[Z2]=1E|Z|\le\sqrt{E[Z^2]}=1. Dominated convergence gives E[Xn]→E[X]=3E[Z]−1=−1E[X_n]\to E[X]=3E[Z]-1=-1. The signed truncations need not be increasing along paths, so the most immediate monotone-convergence argument would require separately handling the positive and negative parts.

For a contrasting sequence, take one uniform UU on (0,1)(0,1) and Yn=n1{U<1/n}Y_n=n\mathbf1_{\{U<1/n\}}. For every U>0U>0 it is eventually zero. Its expectation is nP(U<1/n)=1nP(U<1/n)=1 at every nn. There is no integrable common envelope. Indeed, if one existed, dominated convergence would give zero as the limit of these expectations. The spike example pinpoints the exact line of a solution that needs justification.

A similar step appears in option pricing when you remove a numerical payoff cap. If the uncapped payoff is integrable, it can dominate the capped approximations. The payoff distribution and the measure under which the expectation is taken both matter.

References and next steps