Sheet 11

Exercise 1 (Canonical Continuous Process).

Suppose $X={(X_t)}_{t\ge 0}$ is a continuous stochastic process on some probability space $(\mathrm{\Omega },𝒜,\mathbb{P})$ and ${\mathbb{P}}^X$ its law on $C([0,\mathrm{\infty })$. Define

• $\tilde{\mathrm{\Omega }}:=C([0,\mathrm{\infty }))$, $\widetilde{𝒜}:=ℬ(C([0,\mathrm{\infty })))$ and $\tilde{\mathbb{P}}:={\mathbb{P}}^{X,c}$

• $\tilde{X}:=\text{id}:C([0,\mathrm{\infty }))\to C([0,\mathrm{\infty }))$ with ${\tilde{X}}_t:={\pi }_t(\tilde{X})$.

Show that ${({\tilde{X}}_t)}_{t\ge 0}$ is a continuous stochastic process on $(\tilde{\mathrm{\Omega }},\widetilde{𝒜},\tilde{\mathbb{P}})$, which has the same finite dimensional marginals (same law) as $X$.

Solution.

Due to $\tilde{X}(\omega )\in C([0,\mathrm{\infty }))$ for all $\omega$, we have

${\tilde{X}}_s(\omega )$ $={\pi }_s(\tilde{X}(\omega ))=\tilde{X}(\omega )(s)$

$\to \tilde{X}(\omega )(t)={\tilde{X}}_t(\omega )\mathit{\hspace{1em}}s\to t,$

for all $\omega \in \tilde{\mathrm{\Omega }}$, so ${({\tilde{X}}_t)}_{t\ge 0}$ is a continuous stochastic process. Now the law of $\tilde{X}$ is

${\mathbb{P}}^{\tilde{X}}(B)=\tilde{\mathbb{P}}(\tilde{X}\in B)\stackrel{\tilde{X}=\text{id}}{=}\tilde{\mathbb{P}}(B)={\mathbb{P}}^{X,c}(B).$

And by definition ${\mathbb{P}}_{t_1,\mathrm{\dots },t_n}^{X,c}:={\mathbb{P}}_{t_1,\mathrm{\dots },t_n}^X$, therefore we have that the finite dimensional marginals are the same

${\mathbb{P}}_{t_1,\mathrm{\dots },t_n}^{\tilde{X}}$ $={\mathbb{P}}_{t_1,\mathrm{\dots },t_n}^X.\mathit{∎}$

Exercise 2 (Modification vs. Indistinguishable).

1. Let $X,Y$ be stochastic processes. Show the implications (i)(a)$\Rightarrow$(i)(b)$\Rightarrow$(i)(c) for

   1. $X$ and $Y$ are indistinguishable

   2. $X$ is a modification of $Y$

   3. $X$ and $Y$ have the same finite dimensional distributions

   Solution.

   (i)(a)$\Rightarrow$(i)(b) We have for all $t\in E$

   $$\{X_t\ne Y_t\}\subseteq \{\exists s\in E:X_s\ne Y_s\}\subseteq N$$

   where $N$ is a measurable zero set as $X$ and $Y$ are indistinguishable. Therefore $\{X_t\ne Y_t\}$ is a zero set and we have $\mathbb{P}(X_t=Y_t)=1$. So by definition $X$ is a modification of $Y$.

   (i)(b)$\Rightarrow$(i)(c) For every $t$ we have $X_t=Y_t$ almost surely. So for a finite set $\{t_1,\mathrm{\dots },t_n\}$ we can just take the union of zero sets to get $(X_{t_1},\mathrm{\dots }{X}_{t_n})=(Y_{t_1},\mathrm{\dots }{Y}_{t_n})$ almost surely. But this implies that the distributions are also equal and by Prop. 8.1.14 we only need to check the finite dimensional distributions are equal (i.e. ${\mathbb{P}}_{t_1,\mathrm{\dots },t_n}^X={\mathbb{P}}_{t_1,\mathrm{\dots },t_n}^Y$). ∎

2. If $X$ is a modification of $Y$ and both $X$ and $Y$ are continuous on a Polish space $(E,d)$, then $X$ and $Y$ are indistinguishable.

   Solution.

   Let $Q$ be the countable dense subset of $E$, then as $X$ is a modification of $Y$ we have

   $\mathbb{P}(\forall t\in E:X_t=Y_t)$ $\stackrel{X,Y\text{ contin.}}{=}\mathbb{P}(\forall t\in Q:X_t=Y_t)=1-\mathbb{P}\left({\displaystyle \bigcup _{t\in Q}}\{X_t\ne Y_t\}\right)$

   $\ge 1-{\displaystyle \sum _{t\in Q}}\underset{=0}{\underbrace{\mathbb{P}(X_t\ne Y_t)}}.\mathit{∎}$

Exercise 3 (Hölder Continuity).

Prove the following statements

1. If $f$ is locally $\gamma$-Hölder continuous, then $f$ is also locally $\beta$-Hölder continuous for all $\beta \le \gamma$. Why is this wrong for global Hölder continuity?

   Solution.

   Let $U_x$ be the neighborhood of $x$ such that for all $y\in U_x$ we have the $\gamma$-Hölder continuity. Then for all $\beta \le \gamma$ we get

   So using the neighborhood $U_x\cap B_1(x)$ we have local $\beta$-Hölder continuity. Because we can not intersect the neighborhood with a radius one ball in the global case, this is where the proof breaks. And $f(y)={|y|}^{\gamma }$ is already a counterexample of the statement as for $x=0$ we have

   $|f(x)-f(y)|={|x-y|}^{\gamma }>K{|x-y|}^{\beta }\iff {|y|}^{\gamma /\beta }>K\iff |y|\ge K^{\beta /\gamma },$

   so we just have to select $y$ large enough. ∎

2. If $f$ is continuously differentiable, then $f$ is locally $1$-Hölder continuous (Lipschitz continuous).

   Solution.

   Let $ϵ>0$, then for all $y\in B_{ϵ}(x)$ by the mean value theorem, there exists $\xi \in B_{ϵ}(x)$ such that

   $|f(x)-f(y)|$ $=|f^{\prime }(\xi )|\parallel x-y\parallel \le \underset{=:K_x}{\underbrace{\underset{z\in B_{ϵ}(x)}{sup}|f^{\prime }(z)|}}\parallel x-y\parallel .\mathit{∎}$

3. If $f$ is Hölder continuous with index $\gamma >1$, then $f$ is constant.

   Hint.

   Show $f$ is differentiable using the definition of differentiable.

   Solution.

   We have for every $x$, $f^{\prime }(x)=0$ because

   $\underset{y\to x}{lim}{\displaystyle \frac{\parallel f(y)-f(x)-f^{\prime }(x)(x-y)\parallel }{\parallel x-y\parallel }}\le \underset{y\to x}{lim}{K}_x{\parallel x-y\parallel }^{\gamma -1}=0.$

   But $f^{\prime }\equiv 0$ implies $f$ is constant. ∎

4. If $g$ is (locally) $\gamma$-Hölder continuous and $f$ is (locally) $\beta$-Hölder continuous, then $g\circ f$ is (locally) $\gamma \beta$-Hölder continuous.

   Solution.

   Using the respective definitions we get

   $\parallel g\circ f(x)-g\circ f(y)\parallel \le K_g{\parallel f(x)-f(y)\parallel }^{\gamma }\le K_g{(K_f{\parallel x-y\parallel }^{\beta })}^{\gamma }=\underset{=:K}{\underbrace{K_g{K}_f^{\gamma }}}{\parallel x-y\parallel }^{\gamma \beta }.$

   For the local version we need to ensure that $f(y)$ is in an for $g$ appropriate neighborhood of $f(x)$. But due to Hölder continuity of $f$ we can ensure

   $\parallel f(x)-f(y)\parallel \le ϵ$

   by selecting $\parallel x-y\parallel \le {(\frac{ϵ}{K_f})}^{1/\beta }$. ∎

5. For , $f(x)={\parallel x\parallel }^{\beta }$ is Hölder continuous with index $\beta$.

   Hint.

   Show that $g\mathrm{:}{ℝ}_{\mathrm{+}}\mathrm{\to }{x}^{\beta }$ is $\beta$-Hölder continuous and use (iv). Note that $z^{\beta }\mathrm{\ge }z$ for all $z\mathrm{\in }\mathrm{[}\mathrm{0}\mathrm{,}\mathrm{1}\mathrm{]}$. Deduce for all $a\mathrm{,}b\mathrm{\ge }\mathrm{0}$

   ${\left({\displaystyle \frac{a}{a+b}}\right)}^{\beta }+{\left({\displaystyle \frac{b}{a+b}}\right)}^{\beta }\ge 1⟹a^{\beta }+b^{\beta }\ge {(a+b)}^{\beta }$

   which is kind of a triangle inequality. Now you just need to recall how you show Lipschitz continuity for norms.

   Solution.

   We use $f=g\circ \parallel \cdot \parallel$ with $g:{ℝ}_{+}\to x^{\beta }$. The norm is 1-Hölder continuous because of

   $\parallel x\parallel \le \parallel x-y\parallel +\parallel y\parallel ⟹\parallel x\parallel -\parallel y\parallel \le \parallel x-y\parallel$ (1)

   since by symmetry we get Lipschitz continuity

   $\left|\parallel x\parallel -\parallel y\parallel \right|\le \parallel x-y\parallel .$

   So if $g$ is $\beta$-Hölder continuous, we are done by (iv). For we have $z^{\beta }\ge z$ for all $z\in [0,1]$. Using this fact, we get for all $a,b\ge 0$

   $\underset{\ge \frac{a}{a+b}}{\underbrace{{\left({\displaystyle \frac{a}{a+b}}\right)}^{\beta }}}+\underset{\ge \frac{b}{a+b}}{\underbrace{{\left({\displaystyle \frac{b}{a+b}}\right)}^{\beta }}}\ge 1⟹a^{\beta }+b^{\beta }\ge {(a+b)}^{\beta }.$

   This is a type of triangle inequality so we can essentially do (1) again, except we need to keep $a,b\ge 0$. Assuming $x,y\in {ℝ}_{+}$ and without loss of generality $x\ge y$, we define $b=x-y$ and $a=y$ to get

   $x^{\beta }={(a+b)}^{\beta }\le y^{\beta }+{(x-y)}^{\beta }$

   and with $x\ge y$ and monotonicity of $g$ we have

   $|g(x)-g(y)|=x^{\beta }-y^{\beta }\le {(x-y)}^{\beta }={|x-y|}^{\beta }.$

   Therefore $g$ is $\beta$-Hölder continuous. ∎

Exercise 4 (Fractional Gaussian is Hölder).

1. Prove that there exists a centred Gaussian process ${(X_t)}_{t\in ℝ}$ with $\text{Var}(X_t-X_s)={|t-s|}^{2H}$ and $X_0=0$ for all $H\in [0,1]$. You can use that

   $k(s,t)=\frac{1}{2}({|t|}^{2H}+{|s|}^{2H}-{|t-s|}^{2H})$

   is positive semi-definite without proof. Prove that the law of this process is uniquely determined. (This process is called fractional Brownian motion and the BM is a special case with $H=\frac{1}{2}$).

   Solution.

   By Theorem 8.1.18 there exists a centred Gaussian process $X_t$ with covariance $k$. And we have

   $\text{Var}(X_t-X_s)$ $=𝔼[{(X_t-X_s)}^2]=𝔼[X_t^2]-2𝔼[X_t{X}_s]+𝔼[X_s^2]$

   $=\underset{={|t|}^{2H}}{\underbrace{k(t,t)}}-2k(t,s)+\underset{={|s|}^{2H}}{\underbrace{k(s,s)}}$

   $={|t-s|}^{2H}$

   As the covariance function is uniquely determined because of

   $\text{Cov}(X_t,X_s)=\frac{1}{2}(\underset{=\text{Var}(X_t-X_0)}{\underbrace{𝔼[X_t^2]}}-\text{Var}(X_t-X_s)+𝔼[X_s^2]).$

   The law is uniquely determined by Proposition 8.1.14. ∎

2. Prove that for all , the fractional Brownian motion has a modification which is $\gamma$-Hölder continuous.

   Hint.

   For $X\mathrm{\sim }𝒩\mathit{}\mathrm{(}\mathrm{0}\mathrm{,}{\sigma }^{\mathrm{2}}\mathrm{)}$ we have

   where $n\mathrm{!!}\mathrm{=}n\mathit{}\mathrm{(}n\mathrm{-}\mathrm{2}\mathrm{)}\mathit{}\mathrm{(}n\mathrm{-}\mathrm{4}\mathrm{)}\mathit{}\mathrm{\dots }\mathit{}\mathrm{3}\mathrm{\cdot }\mathrm{1}$ for odd $n$.

   Solution.

   Because we have for all $n\in \mathbb{N}$

   $𝔼[{|X_t-X_s|}^{2n}]=(2n-1)!!{|t-s|}^{2nH}$

   We get a $\gamma$-Hölder continuous modification for all

   (2)

   by the Kolmogorov-Chentsov Theorem. But as $n$ was arbitrary, we can find an $n$ for all such that Equation (2) is still satisfied. ∎