Could someone help me to figure out solutions of following problems?:

Let $X = (X_t)_{t \geq 0}$ be a Gaussian, zero-mean stochastic process starting from $0$, i.e. $X_0 = 0$. Moreover, assume that the process has stationary increments, meaning that for all $t_1 \geq s_1, t_2 \geq s_2, ... , t_n \geq s_n$, the distribution of the vector $(X_{t_1} - X_{s_1} , ... ,X_{t_n} - X_{s_n})$ only depends on the time points through the differences $t_1 - s_1, ... , t_n - s_n$.


Show that for all $s, t \geq 0$

$$EX_sX_t =1/2 (v(s) + v(t) - v(|t - s|))$$

where the function $v$ is given by $v(t) = EX^2_t$


In addition to stationarity of the increments we now assume that $X$ is $H$-self similar for some parameter $H > 0$. Recall that this means that for every $a > 0$, the process $(X_{at})_t$ has the same finite dimensional distributions as $(a^HX_t)_t$.

Show that the variance function $v(t) = EX^2_t$ must be of the form $v(t) = Ct^{2H}$ for some constant $C>0$.


In view of the $(a),(b)$ we now assume that $X$ is a zero-mean Gaussian process with covariance function $EX_sX_t = 1/2(s^{2H} + t^{2H} - |t - s|^{2H})$ for some $H > 0$.

Show that we must have $H \leq 1$. (Hint: you may use that by Cauchy-Schwarz, the (semi-)metric $d(s, t) = \sqrt{ E(X_s - X_t)^2}$ on $[0, 1)$ satisfies the triangle inequality).


Show that for $H = 1$, we have $X_t = tZ$ a.s., for a standard normal random variable $Z$ not depending on $t$.


Show that for every value of the parameter $H \in (0, 1]$, the process $X$ has a continuous modification.

For now I have some ideas about $(a)$ and $(d)$ but have no clue about $(b), (c), (e)$

For $(a)$ my reasoning would be as follows:

$$EX_sX_t = E[(X_s-X_t)+X_t][(X_t - X_s)+X_s]$$ $$=-E(X_s-X_t)^2+E(X_s-X_t)X_s+EX_t(X_t-X_s)+EX_tX_s$$ $$=-E(X_s-X_t)^2+EX_s^2-EX_tX_s+EX_t^2-EX_tX_s+EX_tX_s$$

so we get

$$EX_sX_t=1/2 (v(s) + v(t) - E(X_s-X_t)^2)$$

but how to show that $E(X_s-X_t)^2 = v(|t - s|)$?

I would say that if $s \leq t$ then $E(X_s-X_t)^2 = E(X_{t-s}-X_0)^2= E(X_{t-s})^2=v(t - s)$ and by symmetry in the case of $s \geq t$ we get $E(X_s-X_t)^2 = v(|t - s|)$. Is this correct?

For $(d)$ I think in this way:

If $H=1$ then $EX_sX_t = 1/2(s^{2H} + t^{2H} - |t - s|^{2H})$ simplyfies to $1/2(s^2 + t^2 - |t - s|^2)$ and this is equal to $st$

on the other hand

$EtZsZ=stEZ^2=st$ in the case of $Z$ being standard Gaussian. Because we are talking about Gaussian proccess it is described by mean $(=0)$ and covariance (st) we are done. Is this the right way?

As I wrote above I have no reasonable ideas about $(b), (c), (e)$.


1 Answer 1


For $(a)$ it can be convenient to look at the right-hand side. Suppose $t\geq s\geq 0$, then $v(|t-s|)=v(t-s)=E[X_{t-s}^2]=E[(X_t-X_s)^2]$. Therefore $$ \frac{1}{2}\left(v(t)+v(s)-v(|t-s|)\right)=\frac{1}{2}\left(E[X_t^2]+E[X_s^2]-E[(X_t-X_s)^2]\right)\\ =\frac{1}{2}\left(2E[X_tX_s]\right)=E[X_tX_s]. $$ If on the other hand $s>t\geq 0$, then $v(|t-s|)=v(s-t)=E[(X_s-X_t)^2]$ and here the same calculations apply.

Concerning $(b)$ we have that $a^H X_t\sim X_{at}$ for every $t\geq 0$ and $a>0$. In particular $$ a^{2H}E[X_t^2]=E[(a^H X_t)^2]=E[X_{at}^2], $$ and hence $v(at)=a^{2H} v(t)$ holds for every $t\geq 0$ and $a>0$. I am fairly certain, that the only function satisfying this property is the function $t\mapsto Ct^{2H}$ for some constant $C>0$. The positivity of $C$ is due to the fact that $t\mapsto v(t)$ is a non-negative function.

For $(c)$ we may use that the pseudo-metric $d(s,t)=\sqrt{E[(X_s-X_t)^2]}$ satisfies the triangle inequality. Note that $d(s,t)=\sqrt{v(|s-t|)}$ and that $v(t)=t^{2H}$ in this setup. Now the triangle inequality says that $$ d(s,t)\leq d(s,u)+d(u,t) $$ for every $s,t,u\geq 0$. By means of $v$ this is $$ \sqrt{|s-t|^{2H}}\leq \sqrt{|s-u|^{2H}}+\sqrt{|u-t|^{2H}} $$ which implies $$ |s-t|^H\leq |s-u|^H+|u-t|^H $$ for every $s,t,u\geq 0$. For this inequality to hold we must have $H\leq 1$.

For $(d)$ you are right in using that a Gaussian process is uniquely determined by its mean- and covariance function.

  • $\begingroup$ Please, by which argument you affirm that H is less than or equal 1 in the last step. $\endgroup$
    – Zbigniew
    May 10, 2013 at 6:49
  • $\begingroup$ @Jonas Try $(s,u,t)=(0,1,2)$. $\endgroup$
    – Did
    May 14, 2013 at 17:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .