Let \(W_t=(W^{(1)}_t,\dots,W^{(n)}_t) \) be an \(n\)-dimensional Brownian motion with \( W^{(i)}_t\) standard independent 1-dim brownian motions and \(n \geq 2\).

Let
\[X_t = \|W_t\| = \Big(\sum_{i=1}^n  (W^{(i)}_t)^2\Big)^{\frac12}\]
be the norm of  the brownian motions. Even though the absolute value is not differentiable at zero we can still apply Itos formula since Brownian motion never visits the origin if the dimension is greater than zeros.

1. Use Ito’s formula to show that \(X_t\) satisfies the Ito process
   \[  dX_t = \frac{n-1}{2 X_t} dt + \sum_{i=1}^n \frac{W^{(i)}_t }{X_t} dW^{(i)}_t \]
2. Using the Levy-Doob Theorem show that
   \[Z_t =\sum_{i=1}^n  \int_0^t \frac{W^{(i)}_t }{X_t} dW^{(i)}_t \]
   is a standard Brownian Motion.
3. In light of the above discussion argue that \(X_t\) and \(Y_t\) have the same distribution if   \(Y_t\) is defined by
   \[ dY_t = \frac{n-1}{2 Y_t} dt + dB_t\]
   where \(B_t\) is a standard Brownian Motion.

Take a moment to reflect on what has been shown. \(W_t\) is a \(\mathbf R^n\) dimensional Markov Process. However, there is no guarantee that the  one dimensional process \(X_t\) will again be a Markov process, much less a diffusion. The above calculation shows that the distribution of  \(X_{t+h}\) is determined completely by \(X_t\) . In particular, it solves a one dimensional SDE. We were sure that \(X_t\) would be an Ito process but we had no guarantee that it could be written as a single closed SDE. (Namely that the coefficients would be only functions of \(X_t\) and not of the details of the \(W^{(i)}_t\)’s.