Show that $dX_t=\frac{X_t}{1-t}dt+dW_t$ can be written as $X_t=(1-t)\int_{0}^{t}\frac{1}{1-s}dW_s$

Question

I have been reading the following link on Brownian Bridge definition, and came across the following statement (bullet point 9 in the link above):

Suppose $W_t$ is a standard Brownian motion, define $X_1=1$, then, for $h \in [0,1]$, the process $X_t$ is a Brownian Bridge:

$$X_t=(1-t)\int_{h=0}^{h=t}\frac{1}{1-h}dW_h \tag{1}$$

I can actually understand the proof of this statement presented in the link above and do not have an issue with the claim that $X_t$ above is a Brownian Bridge. However, the author then goes on to state that:

"In differential form, the above can be written as:"

$$dX_t=\frac{X_t}{1-t}dt+dW_t \tag{2}$$

I actually am unable to connect the differential form with the equation (1) given for $X_t$.

When I rewrite the differential form in the "long-hand" notation, I get ($X_0:=0$):

$$X_t=\int_{h=0}^{h=t}\frac{X_h}{1-h}dh+\int_{h=0}^{h=t}dW_h=\int_{h=0}^{h=t}\frac{X_h}{1-h}dh+W_t$$

The above is clearly not the same as the former definition of $X_t$ given in equation (1). I am thinking there might be some Ito's lemma application to a smartly defined function $F(X_t,t)$, that I haven't been able to figure out (I've tried playing around with variants of $F(X_t,t):=X_te^t$, but to no avail).

Is there a way to "solve" the differential equation (2) into (1), or has the author made a typo?

Edit: having read the answer linked in the comment below, and in the spirit of my own answer on notation to another question here, I have attempted to rewrite the linked answer using the long hand notation (because I struggle with interpreting some of the steps of the short-hand notation answer):

I am still getting a wrong answer. Could you please help me spot where I am going wrong?.

The "trick" in the linked aswer seems to be applying Ito's lemma to a function $F(W_t,t):=\frac{W_t}{1-t}$. The derivatives are:

$$\frac{\partial F}{\partial t}=\frac{-W_t}{(1-t)^2}, \frac{\partial F}{\partial W_t}=\frac{1}{1-t}, \frac{\partial^2 F}{\partial t^2}=0$$

We also have that:

$$W_t=W_0+\int_{h=0}^{h=t}a(W_h,h)_{=0}dh+\int_{h=0}^{h=t}b(W_h,h)_{=1}dW_h$$ so that:

$$\frac{W_t}{1-t}=\int_{h=0}^{h=t}\left(\frac{\partial F}{\partial t}+\frac{\partial F}{\partial W}*a(W_h,h)_{=0}+\frac{\partial^2 F}{\partial W^2}_{=0}*b(W_h,h)\right)dh+\int_{h=0}^{h=t}\frac{\partial F}{\partial W}b(W_h,h)_{=1}dW_h=\\=\int_{h=0}^{h=t}\left(\frac{-W_h}{(1-h)^2}\right)dh+\int_{h=0}^{h=t}\left(\frac{1}{1-h}\right)dW_h$$

Multiplying through by $1-t$ then gives:

$$W_t=(1-t)\int_{h=0}^{h=t}\left(\frac{-W_h}{(1-h)^2}\right)dh+X_t$$

Therefore we have:

$$X_t=(1-t)\int_{h=0}^{h=t}\left(\frac{W_h}{(1-h)^2}\right)dh+W_t$$

Focusing on the term $(1-t)\int_{h=0}^{h=t}\left(\frac{W_h}{(1-h)^2}\right)dh$, we can write:

$$\int_{h=0}^{h=t}\left((1-t)\int_{s=0}^{s=h}\frac{1}{1-h}dW_s\right)\frac{1}{1-h}dh$$

Note that the term in the bracket above, i.e. $\left((1-t)\int_{s=0}^{s=h}\frac{1}{1-h}dW_s\right)$ in fact does not equal to $X_h$ (as defined in equation (1)), so in fact we do not have that:

$$X_t=\int_{h=0}^{h=t}\frac{X_h}{1-h}dh+W_t$$

Answer 1

Let $Y_{t} = \int_{0}^{t} \frac{1}{1-s}\, dW_{s}$. Next take a look at

$$X_{t} = (1-t) \int_{0}^{t} \frac{1}{1-s}\, dW_{s} = (1-t)Y_{t}$$

and differentiate using It^o's lemma

\begin{align*} dX_{t} &= -Y_{t}\, dt + (1-t)\, dY_{t} + d[ 1-t, Y_{t} ] \\ &= - Y_{t}\, dt + (1-t)\cdot \frac{1}{1-t}\, dW_{t} \\ &= -\frac{X_{t}}{1-t}\, dt + dW_{t} \end{align*}

and so there is indeed a typo.

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If you want to solve

$$dX_{t} = \frac{X_{t}}{1-t}\, dt + dW_{t},$$

then (as in ODEs) use integrating factor

$$\mu(t) = e^{-\int_{0}^{t} \frac{1}{1-s}\, ds } = 1-t$$

to solve SDE

\begin{align*} d \left (X_{t}(1-t) \right) = (1-t)\, dX_{t} - X_{t}\, dt =: (1-t)\, dW_{t} \end{align*}

for solution

\begin{align*} X_{t} = \frac{1}{1-t}X_{0} + \frac{1}{1-t} \int_{0}^{t} (1-s)\, dW_{s}. \end{align*}

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Note of caution: you should not apply It^o's lemma to solve the SDE. This only works in the case that it admits a strong solution (cf. Oksendal, Chapter 5).