Let (left(B_{t} ight)_{t geqslant 0}) be a (mathrm{BM}^{1}) and consider the two-dimensional process (X_{t}:=left(t, B_{t} ight), t

Let \(\left(B_{t}\right)_{t \geqslant 0}\) be a \(\mathrm{BM}^{1}\) and consider the two-dimensional process \(X_{t}:=\left(t, B_{t}\right), t \geqslant 0\).

a) Show that \(\left(X_{t}\right)_{t \geqslant 0}\) is a Feller process.

b) Determine its transition semigroup, resolvent and generator.

c) State and prove Theorem 7.30 for this process and compare the result with Theorem 5.6

Data From 5.6 Theorem 

Data From 7.30 Theorem 

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5.6 Theorem. Let (Bt, Ft)to be a d-dimensional Brownian motion and assume that the function f = 1,2 ((0, co) x Rd, R) ne([0, co) x Rd, R) satisfies (5.5). Then t M{ := f(t, B;) f(0, Bo) | Lf(r, B,) dr, t0, 0 is an F+ martingale. Here Lf(t, x) = f(t, x) + 4xf(t, x). (5.6) Proof. Observe that M - M = f(t, B) - f(s, B) - Lf(r, B,) dr for 0 < s < t. We have to show E (M-M | Fs) = 0. Using Bt - Bs 11 Fs we see from Lemma A.3 that E(M-M's | Fs) = E (f(t, B) f(s, Bs) - | Lf(r, Br) dr|Fs) - = E (f(t, (Bt Bs) + Bs) f(S, Bs) - Lf(r, - (B - Bs) + Bs) dr | Fs) Mdr|5) 2. B - B) + 2) dr) | A E (f(t, (B - Bs) + 2) f(s, 2) - Lf(r, (B - Bs) + 2) dr) Set (t, x)=s,z(t, x) = f(t + s, x + z), s > 0, and observe that Bt - Bs - Bt-s. Since fee1,2 ((0,00) x Rd), the shifted function satisfies (5.5) on [0, 7] Rd where we can take C = c(t) =y and the constant y depends only on s, T and z. We have to show that for all z R t-s -LO E (MM) = E(t-s, Bt-s) - (0, Bo) - | L(r, Br) dr = 0. dr) Ex. 5.10 Ex. 18.8