Here is the calculation to complete the proof of the arcsin law from class last Thursday. Recall
that we were trying to prove the following theorem.
Theorem 0.1. Let {B(t)}t≥0 be a standard Brownian motion. Let x ∈ (0, 1) be fixed. Then we
have for all t > 0,
√
2
P(no zeroes of B ∈ (xt, t)) = arcsin x.
π
Proof. Recall that in class we conditioned on B(xt) = y and observed that
ˆ |y|
2
2
P(no zeroes of B ∈ (xt, t) | B(xt) = y) = p
e−z /2t(1−x) dz.
2πt(1 − x) 0
Multiplying by the density of B(xt) and integrating we got
ˆ ∞ ˆ y
2
2
2
2
√
P(no zeroes of B ∈ (xt, t)) = p
e−z /2t(1−x) dz e−y /2xt dy
2πt(1 − x) 2πxt 0
0
where
´ ∞ we also used
´ ∞the fact that the density of B(xt) is symmetric and hence replaced the integral
over
y
to
2
−∞
0 .
First let us try to get rid of the variablep
t as we know that the√answer will not depend on t. To
this end we do a change of variable u = z/ t(1 − x) and v = y/ xt. This reduces the integral on
the RHS above to
ˆ ˆ √
2 ∞ v x/(1−x) −(u2 +v2 )/2
e
dudv.
π 0
0
At this point we use another trick that is very useful in manipulating with bivariate normal distribution: we move to polar coordinates.
Set u = r cos p
θ and v = r sin θ. We calculate that the range of the integral now is r ∈ (0, ∞) and
√
0 ≤ θ ≤ arctan x/(1 − x) = arcsin x. Also the Jacobian for this change of variable is given by
dudv = rdrdθ. By doing the change of variables we see that the two variables are separated out
and we end up with
ˆ
ˆ arcsin √x
ˆ ∞
√
2
2 ∞ −r2 /2
2
re
dr
dθ = arcsin x
re−r /2 dr.
π 0
π
0
0
By doing another change of variable s = r2 /2, it is very easy to see that the final integral is equal
to 1, which gives the required result.
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