Showing that a Bessel-squared process is time homogeneous I'm trying to show that a squared Bessel process is time homogeneous.
So far, I've shown that $Y_t=f(X_t)$ is a time-homogeneous Markov process with transition semigroup $(\nu_t)_{t\geq 0}$ w.r.t. filtration $(\mathcal{F}_t)_{t\geq 0}$ if $\mu_t\phi = \phi\nu_t$ for all $t\geq 0$ and $\phi$ is the kernel defined by $\phi(x,\cdot)= \delta_{f(x)}$.

Suppose $(B_t)_{t\geq 0}$ is a $d$-dimensional standard Brownian motion. Why is $(|B_t|^2)_{t\geq 0}$ is a time-homogeneous Markov process?
I see that we can just set $f$ above to be defined as $f(x) = |x|^2$. But how do I show that exists $\nu_t$ such that $\mu_t \phi = \phi \nu_t$. If I can show that then I'm done.
 A: You seem to be working towards a proof based purely on Markov theory. However, in this particular case, a simpler and more natural proof can be obtained through stochastic calculus, if you are willing to take a few theorems for granted. We proceed as follows.
As you write yourself, let $B$ denote a $d$-dimensional standard Brownian motion. Let $\|\cdot\|_2$ denote the Euclidean norm on $\mathbb{R}^d$. The Bessel process $X$ of order $d$ is defined to be $X_t = \|B\|_2$. We seek to prove that $X^2$ is a time-homogeneous Markov process. For convenience, let $Z=X^2$, we thus need to prove that $Z$ is a time-homogeneous Markov process.
Let $f(x) = \|x\|^2_2$. Ito's formula states that
\begin{align}
Z_t = f(B_t) &= f(B_0)+ \sum_{i=1}^n \int_0^t \frac{\partial f}{\partial x_i}(B_s)dB^i_s
             + \frac{1}{2}\sum_{i=1}^n \sum_{j=1}^n
               \int_0^t \frac{\partial^2 f}{\partial x_i\partial x_j}(B_s)
               d[B^i,B^j]_s \\
    &= 2 \sum_{i=1}^n \int_0^t B^i_s dB^i_s + \sum_{i=1}^n \int_0^t d[B^i,B^i]_s\\
    &= 2 \sum_{i=1}^n \int_0^t B^i_s dB^i_s + nt.
\end{align}
Now define
\begin{align}
  \beta_t = \sum_{i=1}^n \int_0^t \frac{B^i_s}{\sqrt{Z_s}}d B^i_s,
\end{align}
we then have
\begin{align}
  \sum_{i=1}^n\int_0^t B^i_s dB^i_s
  &= \sum_{i=1}^n \int_0^t \sqrt{Z_s} \frac{B^i_s}{\sqrt{Z_s}} dB^i_s
   = \int_0^t \sqrt{Z_s} d \beta_s,
\end{align}
which yields
\begin{align}
  Z_t &= 2 \int_0^t \sqrt{Z_s} d\beta_s + nt.
\end{align}
Now note that
\begin{align}
  [\beta]_t &= \sum_{i=1}^n \int_0^t \left(\frac{B^i_s}{\sqrt{Z_s}}\right)^2 d[B^i]_s
  =\sum_{i=1}^n \int_0^t \frac{(B^i_s)^2}{Z_s} ds
  = t,
\end{align}
so by Levy's characterisation of Brownian motion, $\beta$ is a one-dimensional Brownian motion. Thus, we conclude that $Z$ satisfies the SDE
\begin{align}
  d Z_t &= 2\sqrt{Z_t}d\beta_t + n dt,
\end{align}
where $\beta$ is a one-dimensional standard Brownian motion. By standard results on the Markov properties of SDEs, the solution to this SDE is a time-homogeneous Markov process (and its generator can be inferred directly from the coefficients of the SDE).
You can find details on stochastic calculus, Markov processes and SDEs in, for example, "Diffusions, Markov processes and martingales" by L. C. G. Rogers and D. Williams, or "Brownian motion and stochastic calculus" by I. Karatzas and S. E. Shreve.
