$$f(t)=e^{-t^2/4}, \ \ \ t \ge 0$$

I want to show that $f$ is in $M^2$

where $M^2$ denotes the class of stochastic processes $f(t),t\ge0$ such that


and there is a sequence of $f_1,f_2,\ldots \in M^2_{\mathrm{step}}$ of random step processes such that


$M^2_{step}$ is the set of random step processes. We call $f(t),t\ge0$ a random step process if there is a finite sequence of numbers $0=t_0<t_1<\ldots<t_n$ and square integrable random variables $\eta_0, \eta_1, \ldots, \eta_{n-1}$ such that

$$f(t)=\sum_{j=0}^{n-1}\eta_j1_{[t_j,t_{j+1})}(t)$$, where $\eta_j$ is $\mathcal F_{t_j}$-measurable for $j=0,1,\ldots,n-1$

The $E\left(\int_0^\infty|f(t)|^2dt\right)<\infty$ part is clear.

I need help for finding such a sequence $f_n$ of random step processes such that



Hints: Set $$f_n(t) := \sum_{k=0}^{n 2^n} \exp \left[ - \frac{1}{4} \exp \left(- \frac{(k+1)}{2^n} \right)^2 \right] 1_{[k2^{-n},(k+1)2^{-n})}(t).$$

  1. Check that $f_n(t) \leq f(t)$ for all $t \geq 0$. (Hint: $f$ is decreasing.)
  2. Check that $f_n(t) \to f(t)$ as $n \to \infty$. (Hint: $f$ is uniformly continuous)
  3. Conclude from the dominated convergence theorem that $\lim_{n \to \infty} \int_0^{\infty} (f(t)-f_n(t))^2 \, dt =0.$
  • $\begingroup$ how does uniform continuity help here? $\endgroup$ – user671288 May 5 at 15:18
  • $\begingroup$ @user671288 Fix $t \in [0,N]$ for some $N \in \mathbb{N}$. Then $$|f(t)-f_n(t) = |f(t)-f(t_n)|, \qquad n \geq N$$ where $t_n$ is the smallest number of the form $k2^{-n}$ which is strictly larger than $t$; in particular $|t_n-t| \leq 2^{-n}$. Uniform continuity of $f$ now gives pointwise convergence. $\endgroup$ – saz May 5 at 16:07

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