エクセンダールSDEもんだい3.14
A solution of Exercise 3.14 in the Øksendal SDE book is given in this article. It seems this exercise to be complex and require careful observation.
Fig. 1 Extention of the domain of 
(
)
Fig. 2 Shape of 
Let be the filtration generated by the Brownian motion
.
We see obviously that is
-measurable if it is a random variable as in the assertion. Then we show the converse.
Suppose is
-measurable. We now follow the hint: a)-c).
a) Define for
. Then we see that
a.s. If, for every
, we have a sequence
of bounded random variables which converges to
in
, taking an appropriate
yields that
.
Then let . We obtain
\begin{align*}
P(|h-g_n|>\varepsilon)
&\leq P(|h-h_n|>\varepsilon/2) + \frac{2E[|h_n-g_n|]}{\varepsilon} \\
&\leq P(|h-h_n|>\varepsilon/2) + \frac{2}{n\varepsilon}
\end{align*}
for . Since
tends to
as
in probability, so does
in probability. Thus
must have an a.s. convergent subsequence with limit
.
Then it suffices to construct a sequence of as in the assertion which converges to any bounded in
.
b) We now show that
\begin{align*}
\mathcal{F}_t = \bigvee_{n=1}^\infty \mathcal{H}_n =: \mathcal{H} .
\end{align*}
For , take a sequence
from
converging to
. Since
and every
is
-measurable,
is also
-measurable. Thus, we see that
. And then we get the inverse inclusion by definitions of
and
.
By Corollary C.9, it follows that a.s. and in
.
c) By the Doob-Dynkin lemma (Lemma 2.1.2), we have
\begin{align*}
h_n = G_n(B_{t_1},\dots,B_{t_k}),
\end{align*}
where for
. Since
are bounded, we see that
are also bounded.
Let be the set of all continuous functions
, which are univariate functions and whose support are compact when regarding as ones
.
We now consider the following:
Claim Any bounded function admits a sequence
in the subabgebra generated by
such that
\begin{align*}
& \sup_{j\geq 1,~ x\in\mathbb{R}^k}|F_j(x)|<\infty, \quad \lim_{j\to\infty} F_j \to G, \quad \text{a.e.}
\end{align*}
We prepare Lemma 1 and 2 before showing Claim.
Lemma 1 There exists a sequence in
*1 such that
for some
independent of
, and
a.e.
Proof. Let for
. Then every
is in
for
under the Lebesgue measure and tends to
a.e. as
.
For , by density of
in
, one can take an
-convergent sequence
of
with limit
. Moreover, we may assume that
\begin{align*}
\sup_{x\in\mathbb{R}^k}|F^{(N)}_j(x)|\leq K:=\sup_{x\in\mathbb{R}^k}|G(x)|+1 \quad \text{for $j,N\in\mathbb{N}$}.
\end{align*}
Indeed, defining as
\begin{align*}
\tilde{F}^{(N)}_j(x) &=
\begin{cases}
K & ,\text{if $F^{(N)}_j(x)>K$}, \\
-K & ,\text{if $F^{(N)}_j(x)<-K$}, \\
F^{(N)}_j(x) & ,\text{otherwise},
\end{cases}
\end{align*}
leads that and
\begin{align*}
\int_{\mathbb{R}^k} |\tilde{F}^{(N)}_j(x)-G_N(x)|dx
&\leq \int_{\mathbb{R}^k} |F^{(N)}_j(x)-G_N(x)|dx,
\end{align*}
and then it follows that in
. Then the same argument as in a) yields the assertion. □
Lemma 2 For , let
be a
dimensional compact rectangle containing the support of
and let
be the set of all continuous functions
,
. Then each
admits a sequence
of the subalgebra generated by
such that
. Furthermore, we can extend the domain of each
to whole
satisfying
.
Proof. We refer to Miyajima [1], Theorem 7.65, for the Stone-Wierstrass theorem:
Theorem Let
be a compact Hausdorff space and
be a subset of
. Then, the subabgebra generated by
(1) For any, there exists an
such that
.
(2) For any different, there exists an
.
The theorem is stated in other books or media as well, e.g. [2-4].
It is clear that each satisfies (1) and (2) in the theorem. Thus, we see that the former holds.
Next, for the latter. We extend each of the domain of by putting the same values on
, where
, as ones on the boundary of
along each axis (Fig. 1).
And then, rewrite as the product of the extended
above and
, where each
is defined as in Fig. 2. Then we see that each of the extended
is in the subalgebra generated by
, which is defined before Claim, and
. □
Proof of Claim. By Lemma 1 and 2, we may assume that and that
, by renumbering if necessary. Then we have
\begin{align*}
|\tilde{F}^{(j)}_j(x)-G(x)|
&\leq |F_j(x)-G(x)| + \frac{1}{j} \\
& \to 0, \quad \text{as $j\to\infty$}
\end{align*}
almost everywhere. □
We are now back to showing the exercise. By Claim, there exists, for , a sequence
in the subabgebra generated by
converging to
a.e. By the dominated convergence theorem, we see that
as
. Then as repeated above, we assume that
\begin{align*}
E[|G_n(B_{t_1},\dots,B_{t_k}) - F^{(n)}_n(B_{t_1},\dots,B_{t_k})|]\leq \frac{1}{n}.
\end{align*}
Then, letting yields
\begin{align*}
E[|h-F^{(n)}_n(B_{t_1},\dots,B_{t_k})|]
&\leq E[|h-h_n|] + \frac{1}{n} \\
&\quad \to 0.
\end{align*}
As mentioned in a), we constructed a desired convergent sequence. We complete the proof.
References
[1] Miyajima, S., Functional Analysis, Yokohama Publishers, 2007. In Japanese.
[2] Rudin, W., Functional Analysis, McGraw-Hill, 1991.
[3] Yosida, K., Functional Analysis, Springer, 1980.
[4] Wikipedia, Stone–Weierstrass theorem
Hmm, too heavy exercise...x(
*1:It denotes the space of all continuous functions with compact support.
2月27日のしゅくだい
問題4.6
(b)を示せば、とすることで(a)が得られる。よって(a)のみを示す。
中点""で
のユークリッド内積を表すとし、
とおく。
\begin{align*}
& \partial_t f(t,x) = cf(t,x), \\
& \nabla_x f(t,x) = f(t,x)\alpha, \\
& \partial_{x_ix_j}f(t,x) = f(t,x)\alpha_i\alpha_j
\end{align*}
となる。
であり、伊藤の公式より、
\begin{align*}
dX_t
&= \partial_tf(t,B_t)dt + \nabla_xf(t,B_t)\cdot dB_t + \frac{1}{2}\sum_{i,j=1}^n \partial_{x_ix_j}f(t,B_t)dB^i_tdB^j_t \\
&= c X_t dt + X_t \alpha\cdot dB_t + \frac{1}{2}|\alpha|^2X_tdt \\
&= \left( c+\frac{1}{2}|\alpha|^2 \right)X_tdt + X_t \alpha\cdot dB_t
\end{align*}
を得る。
Lévy processのジャンプはいつまでも起き続ける
The following might be nontrivial so I will write down a proof.
Claim
Assume that the filtration satisfies the usual conditions. Let
be a Lévy process (with càdlàg sample paths) adapted to
and
be a Borel set in
such that
. Define the random times
\begin{align*}
T^0_\Lambda &= 0, \\
T^{n}_\Lambda &= \inf\{t>T^{n-1}_\Lambda;\Delta X_t\in\Lambda\}, \quad n=1,2,\dots
\end{align*}
, where . Then the followings hold:
(i) a.s. for
(ii) Every is an
-stopping time.
(iii) a.s.
Proof.
(i) Since ,
holds. Then
implies
.
By right continuity of , there is a
such that
\begin{align*}
\left|X_t-X_{T^{n-1}_\Lambda}\right| \leq \frac{d}{6}
\end{align*}
for all .
For any , taking a sufficiently small
satisfying
yields, by càdlàgness of
,
\begin{align*}
\left| X_{t-{\delta_2}}-X_{T^{n-1}_\Lambda} \right|,|X_{t-{\delta_2}}-X_{t-}| \leq \frac{d}{6}.
\end{align*}
Thus, it holds that
\begin{align*}
\left|\Delta X_t\right|
&\leq |X_t-X_{T^{n-1}_\Lambda}| + |X_{t-{\delta_2}}-X_{T^{n-1}_\Lambda}| + |X_{t-{\delta_2}}-X_{t-}| \\
&\leq \frac{d}{6}+\frac{d}{6}+\frac{d}{6} = \frac{d}{2} < d.
\end{align*}
Then we obtain .
(ii) We use the result from Karatzas-Shreve; Proposition 2.26.
It holds for , since
is
-adapted,
\begin{align*}
\{ T^n_{\Lambda} \leq t \}
&= \bigcup_{k_1,\dots,k_n\in\mathbb{N}} \bigcap_{j=1}^n \left( \{T_{k_j}\leq t\} \cap \{ \Delta X_{T_{k_j}}\in\Lambda \} \right) \\
&\in \mathcal{F}_t,
\end{align*}
where, by the proposition above, is the sequence of stopping times exhausting the jumps of
. Thus we see all
are stopping times.
(iii) Assume that there is a such that
for all
. Then,
is increasing and bounded. Thus,
exists.
By càdlàgness of , it follows
. Since
is left continuous, we get
.
Hence, as
. However,
. Those are contradiction.
2月13日のしゅくだい4
しゅくだいです。正直、イェンセンの不等式から明らかです。
、
を
の部分
-加法族のとき、
\begin{align*}
E[(E[X|\mathcal{G}])^2] \leq E[X^2] \tag{1}
\end{align*}
しょうめい：
は凸関数なので、イェンセンの不等式より、
\begin{align*}
(E[X|\mathcal{H}])^2 \leq E[X^2|\mathcal{H}]
\end{align*}
である。
よって、両辺期待値をとれば、(1)を得る。□
補題6.1.1を見よともあります。
ヒルベルト空間において、
可測な確率変数のなす部分空間を
とする。
補題6.1.1より、]は
への直交射影
なので、
\begin{align*}
E[(E[X|\mathcal{G}])^2]
&= \Vert P_{\mathcal{N}}X \Vert_{L^2}^2 \\
&\leq \Vert P_{\mathcal{N}} \Vert^2 \Vert X \Vert_{L^2}^2 \\
&\leq \Vert X \Vert_{L^2}^2 = E[X^2]
\end{align*}
となる。
2月13日のしゅくだい3
しゅくだいときます。
(i) ならば、
と
は独立
(ii) は定常
(iii) すべてのに対し、
]
を満たす実数値確率過程で連続なものは
なるものを除いて存在しない。
しょうめい：
Step 1) に対し、
とする。
このとき、各に対して、
は連続であるので、これと
との合成である
は(i),(ii)を満たす。
(ヒントの通り計算する。)
\begin{align*}
E[(W^N_t - W^N_s)^2]
&= E[(W^N_t)^2] - 2E[W^N_tW^N_s] + E[(W^N_s)^2]\\
&= 2E[(W^N_0)^2 - (E[W^N_0])^2] \\
&= 2E[(W^N_u - E[W^N_0])^2]
\end{align*}
が任意ので成り立つ。
今、が連続過程であると仮定する。
すると、も連続であり、したがって
とすると、優収束定理(有界収束定理)より
\begin{align*}
E[(W^N_u - E[W^N_0])^2] = 0
\end{align*}
を得る。
よって、
\begin{align*}
W^N_u = E[W^N_0] \tag{1}
\end{align*}
である。
よって、は定数過程である。
Step 2) 任意のに対し(1)が成り立つことより
なので、
\begin{align*}
&\{W_0>N\} = \{W_u>N\} \\
&\{W_0<-N\} = \{W_u<-N\} \\
&\{-N\leq W_0\leq N\} = \{-N\leq W_0\leq N\}
\end{align*}
が分かる。
このとき、任意のに対して、次の①～③のいずれか1つのみが成り立つ：
①
②
③
∵) ①については、とおくと、すでに述べたように
で、(i)より
と
は独立であること、そして(ii)より、
\begin{align*}
P(A_0)
&= P(A_0\cap A_u) \\
&= P(A_0)P(A_u) \\
&= P(A_0)^2
\end{align*}
となり、を得られる。②、③も同様。①～③の事象は互いに素なので、いずれか1つのみが成り立つ。
Step 3) 任意のに対して、
が成り立つことを示す。
あるで①が成り立つとすると、
となり、
に反する。
②も同様。
よって、任意のに対し、
である。
Step 3)より、特にとしておくと、
\begin{align*}
W_u
&= W^N_u \\
&= E[W^N_0] \\
&= E[W_0] = 0
\end{align*}
となり、を得る。この
は(i)～(iii)を満たす。□
2月13日のしゅくだい2
しゅくだいときます。
\begin{align*}
N_t = B_t^3-3tB_t
\end{align*}
がマルチンゲールである。
\begin{align*}
N_t = B_t^3-3tB_t
\end{align*}
がマルチンゲールである。
しょうめい：
をブラウン運動
によって生成されるフィルトレーションとする。
ブラウン運動はガウス分布に従うので、は可積分である。
とする。
\begin{align}
N_t
&= (B_t-B_s + B_s)^3 - 3tB_t \\
&= (B_t-B_s)^3 + 3B_s(B_t-B_s)^2 + 3B_s^2(B_t-B_s) + B_s^3 - 3tB_t
\end{align}
である。
第1項の条件付き期待値は、
\begin{align*}
E[(B_t-B_s)^3|\mathcal{F}_s]
&= E[(B_t-B_s)^3] \\
&= \frac{1}{\sqrt{2\pi (t-s)}} \int x^3 e^{-\frac{x^2}{2(t-s)}} dx \\
&= 0,
\end{align*}
第2項は
\begin{align*}
E[B_s(B_t-B_s)^2|\mathcal{F}_s]
&= B_sE[(B_t-B_s)^2|\mathcal{F_s}] \\
&= B_sE[(B_t-B_s)^2] \\
&= (t-s)B_s,
\end{align*}
第3項は
\begin{align*}
E[B_s^2(B_t-B_s)|\mathcal{F}_s]
&= B_s^2 E[B_t-B_s] \\
&= 0
\end{align*}
である。
よって、がマルチンゲールになることと合わせると
\begin{align*}
E[N_t|\mathcal{F}_s]
&= 3(t-s)B_s + B_s^3 -3tB_s \\
&= B_s^3 - 3sB_s \\
&= N_s
\end{align*}
となる。□
