Reflection principle

3.2

Why is the chance that a path ever reaches level a exactly twice the chance that it ends above a?

Goal: reflect a path after its first crossing and confirm by Monte Carlo that P(M(T) ≥ a) = 2·P(W(T) ≥ a).

Simulation controls

Level (barrier) aThe principle holds for every level a > 0

a1.00

0.22.5

Play the path

t1.00

0.001.00

Simulated paths2,000

1003000

Random seed7

19999

Empirical vs theory

P(M(1) ≥ a) empirical

0.297

2·P(W(1) ≥ a) empirical

0.279

2(1 − Φ(a)) theory

0.317

P(W(1) ≥ a) empirical

0.140

About half of the crossing paths end below the level — the terminal probability misses them, but reflection pairs each with a path ending above. The grid maximum slightly underestimates.

Try this

Predict first: once a path reaches the level, how likely is it to end above it? Count over the next few crossing paths.
Set a close to 0: why does P(M(1) ≥ a) approach 1 while P(W(1) ≥ a) only approaches ½?
Drop the paths to 100: which pulls the estimate further from theory — sampling noise or the grid?

A path and its reflection after the first crossing

τₐ = 0.372

Up to τₐ the two paths coincide; afterwards the reflected path is 2a − W(t). By symmetry both are equally likely, so every crossing path that ends below a is paired with one that ends above it.

P(M(1) ≥ a) as a function of the level

P(M(1) ≥ a) empirical2·P(W(1) ≥ a) empirical2(1 − Φ(a))

The empirical tail of the running maximum matches twice the terminal probability and the theoretical 2(1 − Φ(a)).