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 $[0.5,1]$ ?

  • $\begingroup$ Oops, you're right. $\endgroup$ Feb 25, 2021 at 21:44
  • $\begingroup$ 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? $\endgroup$
    – E-A
    Mar 10, 2021 at 17:49
  • $\begingroup$ @E-A Corrected the typo, thanks $\endgroup$
    – W. Volante
    Mar 10, 2021 at 21:33
  • $\begingroup$ 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. $\endgroup$
    – NN2
    Mar 10, 2021 at 22:01
  • $\begingroup$ @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. $\endgroup$
    – E-A
    Mar 10, 2021 at 22:32

1 Answer 1


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 $f$.


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