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I wonder how to solve the following quadratic optimization problem?

$$\min_{f:[0,1]\to \mathbb{R}} \int_0^1 f(t)^2dt-2\int_0^1 f(t)dB_t$$

where $B(t)$ is standard Brownian motion.

Intuitively, the optimization can be performed pointwise, i.e., for any $t\in[0,1]$, solve

$$\min_{f(t)} f(t)^2dt-2f(t)dB_t,$$

however the solution is $f(t)=dB_t/dt$ which doesn't make sense since $B_t$ is not differentiable.

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  • $\begingroup$ Is f required to be adapted? If not, how do you define the stochastic integral? $\endgroup$
    – ofer zeitouni
    Commented Jun 17, 2022 at 8:18
  • $\begingroup$ Thanks Ofer, we can assume f to be adapted. $\endgroup$
    – Zuofeng Shang
    Commented Jun 18, 2022 at 3:04

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The minimum (which is an infimum) is $-\infty$.

We have that $\limsup_{t\to 0} B_t/\sqrt{t}=\infty$ (this follows in particular from the LIL). This means that there exists $t_M\in (0,1)$ arbitrarily small so that $B_{t_M}>M \sqrt{t_K}$. Let us disregard for a moment that $t_M$ is not a stopping time.

Take now $f(t)=(1/\sqrt{t_M}) 1_{t<t_M}$. Then $\int f(t)^2 dt=1$, while $\int f(t) dB_t= M$. Then $J(f)=\int f(t)^2 dt-2\int f(t) dB_t=-2 M+1\to_{M\to\infty} -\infty$.

Now, in that construction $t_M$ is not a stopping time, so the definition of the stochastic integral requires some care. However, we can modify and simplify the construction as follows. For $K$ the square root of an integer, consider the two functions $f_{1,K}=K1_{t<1/K^2}$ and $f_{2,K}=-f_{1,K}$. The value of $\min(J(f_{1,K}),J(f_{2,K}))=1-|B_{1/K^2} K|$. Then, again by an application of the LIL on the discrete sequence $j$ (using that $t B_{1/t^2}$ is a Brownian motion), one has $$ \inf_{i=1,2, j\in N} J(f_{i,\sqrt{j}})=-\infty,$$ almost surely.

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  • $\begingroup$ Thanks for your excellent example. Motivated by your example, the infinite minimum is probably due to the discontinuity of f. If we require f being sufficiently smooth, shall the minimum exist? For example, if we require f being constant, the minimum is -(B_1-B_0)^2 achieved at f=B_1-B_0. I am not sure if it is possible to find a weakest smooth condition on f such that the minimum finitely exists. $\endgroup$
    – Zuofeng Shang
    Commented Jun 18, 2022 at 3:04
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    $\begingroup$ That's a different question. Note that if $f$ is differentiable, the stochastic integral becomes $B(1) f(1)-\int_0^1 f'(s) B_s ds$. If you put a constraint on (say) the $H^1$ norm, you can optimize, it is a standard optimization problem (but the final f will not be adapted). $\endgroup$
    – ofer zeitouni
    Commented Jun 18, 2022 at 14:27
  • $\begingroup$ When $f \in H^1(0,1)$, using the integration by part proposed by ofer zeitouni in the comment above, you can solve the optimization problem $\omega$ by $\omega$ on the canonical space of the Brownian motion. The Wiener measure becomes an example amongst others. $\endgroup$
    – megaproba
    Commented Jun 19, 2022 at 6:38

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