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In this link and in other questions here, I found that the SDE for a Brownian Bridge, $X_t$ from $0$ to $1$, is

$$ d X_t = \frac{-X_t}{1-t} d t\ + \ dW_t $$

But what is the SDE for a Brownian Bridge, $Y_t$, from $0$ to $T$. Is it $$ d Y_t = \frac{-Y_t}{T-t} d t\ + \ dW_t \ ? $$

If so, how would I go about deriving such an expression given the SDE for $X_t$?

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  • $\begingroup$ What is your definition for a Brownian bridge from $0$ to $T$? The link given doesn't say. $\endgroup$
    – fourierwho
    Jun 29, 2018 at 18:29
  • $\begingroup$ Gaussian, $Y_0 = Y_T = 0$ with expectation zero and variance $t (T-t)$. See also here. $\endgroup$
    – Tohiko
    Jun 30, 2018 at 7:40

1 Answer 1

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Observe that if $Y$ is a Brownian bridge on $[0,1]$, then the process $W = (W_{t})_{t \in [0,T]}$ given by $W_{t} = \sqrt{T} Y_{\frac{t}{T}}$ is a Brownian bridge on $[0,T]$. Moreover, we compute: \begin{align*} W_{t} &= -\sqrt{T} \int_{0}^{\frac{t}{T}} \frac{Y_{s}}{1 - s} \, ds + \sqrt{T} B_{\frac{t}{T}} \\ &= - \sqrt{T} \int_{0}^{t} \frac{Y_{T^{-1}u}}{1 - T^{-1}u} \, \frac{du}{T} + \sqrt{T} B_{\frac{t}{T}} \\ &= - \int_{0}^{t} \frac{W_{u}}{T - u} \, du + \sqrt{T} B_{\frac{t}{T}} \\ &\overset{d}= -\int_{0}^{t} \frac{W_{u}}{T - u} \, du + B_{t}. \end{align*} Therefore, \begin{equation*} dW_{t} = - \frac{W_{t}}{T - t} \, dt + dB_{t}. \end{equation*}

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  • $\begingroup$ I edited the post. You may want to check the $\sqrt{T}$ factor you're getting in the drift. $\endgroup$
    – fourierwho
    Jul 2, 2018 at 17:34
  • $\begingroup$ Ah yes, you are right. Thanks for pointing that out. $\endgroup$
    – Tohiko
    Jul 5, 2018 at 6:16

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