# Modifying the covariance of Brownian motion, what Gaussian process do we get?

If $$(B_t)$$ is a Brownian motion, then $$Cov(B_t,B_s)=min(t,s)$$.

Take a Gaussian process $$(X_t)$$ with mean $$0$$ and covariance $$Cov(X_t,X_s)=f(min(t,s))$$ for a given function $$f$$ such that the covariance is still positive definite. Is $$X$$ related to Brownian motion ?

For example, let's take an easy function: $$f(x)=x+1$$, then clearly $$X_t=B_{t+1}$$, or more generally, if $$f$$ is monotone increasing, then $$X_t=B_{f(t)}= \int_0^t \sqrt{f'(s)} \, dB_s$$ (where the last equality is valid under integrability/differentiability conditions).

But what happens when $$f$$ is not monotone ? For example, $$Cov(X_t,X_s)=min(t,s)(1-min(t,s))$$. This is a covariance function on $$[0,1]$$. Can we describe $$X$$ using Brownian motion on $$[0,1]$$?

Idea: decompose $$f$$ on intervals on which it is increasing and decreasing. Use the above for the increasing parts. But what happens when $$f$$ is decreasing ? For our example that would be $$f(x)=x(1-x)$$ on $$[0.5,1]$$ ?

• Oops, you're right. Feb 25, 2021 at 21:44
• Is that decreasing f a covariance function? Doesn't the argument here (math.stackexchange.com/questions/266222/…) show that it is not the case? (Deterministic at t=1 but covariance of 0 and 1 are 1?
– E-A
Mar 10, 2021 at 17:49
• @E-A Corrected the typo, thanks Mar 10, 2021 at 21:33
• The function $f$ must be monotone and even it must be increasing by definition. Effectively, from the definition $$B_{f(t)} = \int_0^t\sqrt{f'(s)}dB_s$$ we must have $f'(s) \ge 0$ or $f(t)$ is an increasing function.  So, if you really want to construct a process with $f$ not monotone. You should define completely all definitions relating the Brownian motion.
– NN2
Mar 10, 2021 at 22:01
• @W.Volante correcting the typo does not help since the same problem still holds; deterministic at t=1 but Cov(X_1, X_{1/2}) = 1/4. I don't think that is a valid covariance function. Indeed, as is said in the comment above, monotonicity may be necessary.
– E-A
Mar 10, 2021 at 22:32

The function $$f(x)=x(1-x)$$ applied to $$\min(x,y)$$ is not a covariance function since it is not positive definite on $$[0.5,1]$$. To see this, just calculate the determinant of the "covariance" matrix $$C$$ of the points $$v=0.6$$ and $$w=0.8$$. You find $$C(v,v)=0.6*0.4=0.24$$, $$C(w,w)=0.8*0.2=0.16$$ and $$C(v,w)=f(0.6)=0.24.$$ Now $$\det C= 0.24 * 0.16 - (0.24)^2<0.$$
This example gives already a good idea of what is going on or better going wrong with non-monotonic $$f$$. In fact I claim:
If $$f$$ is such that there exist points $$v,w$$ with $$v< w$$ and $$f(v)> f(w)$$ then $$C(x,y)=f(\min(x,y))$$ is not positive definite.
The first necessary condition is $$C(x,x)=f(x)>0$$ for all $$x$$ in the domain of $$f$$. We assume this holds and calculate $$C(v,v)C(w,w) - C(v,w)^2=f(v)f(w) - f(v)^2 = f(v)( f(w) - f(v))<0$$ by assumption on $$f$$.