Cfrgtkky

Let

• 
  $(Omega,mathcal A)$ and $(E,mathcal E)$ be measurable spaces


• 
  $(Y_n)_{ninmathbb N_0}$ be a $(E,mathcal E)$-valued stochastic process on $(Omega,mathcal A)$
  


• 
  $(N_t)_{tge0}$ be a $mathbb N_0$-valued stochastic process on $(Omega,mathcal A)$
  

Don't know if it matters, but in my application, $Y$ is a Markov chain and $N$ is a Poisson process, independent from $Y$.

Moreover, let $$mathcal F^N_t:=sigma(N_s:0le sle t);;;text{for }tge0$$ and $$mathcal F^Y_n:=sigma(Y_m:minleft{0,ldots,nright});;;text{for }ninmathbb N_0.$$

Now, let $$X_t:=Y_{N_t};;;text{for }tge0$$ and $$mathcal F^X_t:=sigma(X_s:0le sle t);;;text{for }tge0$$

How can we show that $$sigmaleft(left{Acap Bcapleft{N_t=nright}:Ainmathcal F^N_t,ninmathbb N_0text{ and }Binmathcal F^Y_nright}right)=mathcal F_t:=sigma(mathcal F_t^Ncupmathcal F_t^X)$$ for all $tge0$?

Let $tge0$. I even fail to show that if $Ainmathcal F^N_t$, $ninmathbb N_0$ and $Binmathcal F^Y_n$, then $Acap Bcapleft{N_t=nright}inmathcal F_t$. Though, it would be clear, if $Bcapleft{N_t=nright}inmathcal F^X_t$. I guess we somehow need to use that $X_t$ coincindes with $Y_n$ on $left{N_t=nright}$.

Observe that for any $Ain {cal F}^Y_n$ and any $Bin{cal F}^N_t$, up to null events (not sure if completions were considered here), there exist $tilde Ain {cal E}$ and $tilde Bsubset {mathbb N}_0$ such that
$A = {Y_n in tilde {A}}$.
$B ={N_tin tilde{B}}.$

This should help. Indeed,

$$A cap {N_t=n} ={Y_n in tilde A,N_t=n}={X_t in tilde A,N_t=n}$$
and
$$ B cap {N_t=n} ={N_tin tilde Bcap {n}}.$$

Therefore,

$$ A cap B cap {N_t=n} ={underset{in {cal F}^X_t}{underbrace{ X_t in tilde A}},underset{in {cal F}^Y_t}{underbrace{N_tin tilde B cap {n}}}}.$$

As RHS is an intersection of two events in ${cal F}^N_t$ and ${cal F}^X_t$, it is in $sigma({cal F}^N_tcup {cal F}^X_t)$.