Proving Brownian Bridge Convergence with Time Change

Summary

The document considers a Brownian bridge defined by an SDE with drift pulling the process toward a terminal value. Its solution contains a stochastic integral whose contribution must vanish as time approaches the endpoint for the process to converge to that value. The proof applies the Dambis-Dubins-Schwarz theorem to represent the integral martingale as a Brownian motion evaluated at its quadratic variation.

The quadratic variation is computed explicitly and diverges as the endpoint approaches. A change of variables then reduces the desired limit to showing that Brownian motion divided by a linearly growing time parameter converges almost surely to zero, a consequence of the law of large numbers for Brownian motion. This establishes the terminal convergence in the stated representation. The argument is an almost-sure limit proof; it does not discuss numerical approximation, trading applications, or extensions to other bridge processes.

Key ideas

• The Brownian bridge solution reduces terminal convergence to a stochastic-integral term vanishing at the endpoint.
• Dambis-Dubins-Schwarz represents the integral martingale as Brownian motion under a time change.
• The martingale’s quadratic variation determines the time change and diverges near the endpoint.
• A variable substitution reduces the limit to Brownian motion divided by a growing time parameter.
• The law of large numbers implies the required almost-sure limit.

Tags

# Brownian Bridge general case


# Brownian Bridge general case












The SDE for the Brownian bridge is the following:

$dY_t=\frac{b-Y(t)}{1-t}dt+dW(t)$

with solution:

$Y(t)=Y(0)(1-t)+bt+(1-t)\int_0^t \dfrac{dW(s)}{1-s}$

Can someone help me on proving that $$\lim_{t\rightarrow 1^-} Y(t)=b$$ using the Dambis-Dubins-Schwarz theorem and the law of large numbers?

## Answer by LucaMac (score 3, accepted)

https://quant.stackexchange.com/a/65427

We need to show that $$\lim_{t\to1^-} (1-t)\int_0^t\frac1{1-s}dW_s \stackrel{\text{a.s.}}= 0.$$

$(M_t)_{t<1}=\Big(\int_0^t\frac1{1-s}dW_s\Big)_{t<1}$ is a martingale, so we can use Dambis-Dubins-Schwarz and say that $M_t = B_{\langle M\rangle_t}$ for a Brownian motion $B$ (with a different filtration of course).

However, $\langle M\rangle_t = \int_0^t \frac1{(1-s)^2}ds = \int_{1-t}^1\frac1{s^2}ds = \frac1{1-t}-1 = \frac t{1-t}.$

This means that we are left to show that $$\lim_{t\to1^-}(1-t)B_{\frac t{1-t}} \stackrel{\text{a.s.}}=0.$$ If we denote $u:=\frac t{1-t}$, we obtain $t = \frac u{1+u}$ and $1-t = \frac1{1+u}$, thus we must show that $$\lim_{u\to\infty} \frac{B_u}{u+1} \stackrel{\text{a.s.}}=0,$$ which is well-known.

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