# A continuous analogy of an application of the Markov inequality together with the Borel-Cantelli lemma that implies a.s. convergence

It is well known, due to the Markov inequality and the Borel-Cantelli lemma, that for a sequence of random variables $$(X_n)_{n\in \mathbb{N}}$$ that converge to $$0$$ in probability, from $$\sum E(|X_n|) < + \infty$$ one can deduce that the convergence also happens almost surely.

Question: I am wondering if the same analogy can be made with a stochastic process $$(Y_t)_{t \in \mathbb R^+}$$ (possibly with continuous paths) converging to $$0$$ in probability, for example under the assumptions:

1) $$\sum E(|Y_{n}|) < +\infty$$, and

2) $$\mathbb{R}^+\ni t \mapsto E(|Y_t|)$$ is decreasing.

If not, is there any other assumption under which we could draw such a conclusion (i.e. that $$Y_t \to 0$$ a.s.)?

In the related question Is there a continuous version of the Borel-Cantelli lemma? the answer provided is not sufficient to conclude an a.s. convergence.

• Do you really mean a sum in 1) instead of an integral?
– mag
Apr 29, 2020 at 12:31
• What if $Y_t$ is zero except at points of a Poisson process of rate $\lambda$, when $Y_t=1$. Then $E[|Y_t|]=0$ for all $t$ (since an arrival at time $t$ happens with prob 0) but certainly $Y_t$ does not converge to 0 almost surely because $Y_t=1$ at arbitrarily large points in time. Apr 30, 2020 at 11:12
• @mag Yes I do, but assumption 2) implies the convergence of the integral as well. May 1, 2020 at 21:30

I was asking myself the same question, but under weaker assumptions and I think I finally found an answer. Unfortunately there is a counterexample satisfying even the strong assumptions above. It was motivated by the answer of VF1 for the question regarding the continuous version of the Borel-Cantelli lemma. Unfortunately his answer didn't satisfy the convergence in probability to $$0$$, so I took another function and modified the process. Please check if the couterexample is correct.

For the sake of simplicity we consider only the idex set $$[3,\infty[$$ instead of $$\mathbb{R}_+$$, but the example can be shifted of course.

Consider the test function (i.e. element of $$C^\infty_c(\mathbb{R})$$) $$\phi_b(x)=\begin{cases}\exp(b^2/(x^2-b^2),\qquad&x\in]-b,b[\\ 0,\qquad&x\notin]-b,b[ \end{cases}$$ for $$b>0$$ and the probability space $$(\Omega, \sigma, P)=([0,1[,\mathcal{B}([0,1[),\lambda)$$, whereas $$\lambda$$ is the Lebesgue measure. Note: $$\forall b>0:\phi_b(0)=e^{-1}$$ and $$\forall n\in\mathbb{N}\forall \omega\in[0,1[:$$ $$\phi_{1/n^2}(t-\omega-n)=\begin{cases}\exp(n^{-4}/((t-\omega-n)^2-n^{-4}),\qquad&t\in]n+\omega-n^{-2},n+\omega+n^{-2}[\\ 0,\qquad&t\notin]n+\omega-n^{-2},n+\omega+n^{-2}[. \end{cases}$$ We define the process $$Y_t(\omega):=\sum_{n=1}^\infty \phi_{1/n^2}(t-\omega-n).$$ Note, that $$Y$$ is continuous and for fixed $$\omega$$ there is for $$t\geq3$$ always at most one of the functions inside the sum unequal to $$0$$. Obviously $$Y$$ is not converging to $$0$$ a.s., since $$\forall k\in\mathbb{N}:Y_{\omega+k}(\omega)=e^{-1}$$. But it holds for all $$\epsilon>0$$ \begin{align} P(Y_t>\epsilon)\leq P(Y_t>0)=& P\left(\omega\in[0,1[: \sum_{n=1}^\infty \phi_{1/n^2}(t-\omega-n)>0\right)\\ \leq& 3P\left(\omega\in[0,1[: |t-\lfloor t\rfloor-\omega|\leq\frac{1}{(\lfloor t\rfloor-1)^2} \right)\\ \leq&3\lambda\left([t-\lfloor t\rfloor-\frac{1}{(\lfloor t\rfloor-1)^2},t-\lfloor t\rfloor+\frac{1}{(\lfloor t\rfloor-1)^2}]\right)\\ =&\frac{6}{(\lfloor t\rfloor-1)^2}, \end{align} so it is converging to $$0$$ in probability. Here in the second inequality we estimate with $$\lfloor t\rfloor-1$$ instead of $$\lfloor t\rfloor$$ or $$\lfloor t\rfloor+1$$, since for big $$\omega$$ and small $$t-\lfloor t\rfloor$$ the $$t$$ can be "in the bump belonging to the previous integer". Analogous for small $$\omega$$ and big $$t-\lfloor t\rfloor$$ with the bump of the following integer. The three possible cases cause the factor 3. Furthermore $$\int_3^\infty P(Y_t>\epsilon)\mathrm{d}t\leq\int_3^\infty \frac{6}{(\lfloor t\rfloor-1)^2}\mathrm{d}t=6\sum_{n=2}^{\infty}\frac{1}{n^2}<\infty$$ the continuous analogy of the Borel-Cantelli condition holds and $$\int_3^\infty E(Y_t)\mathrm{d}t=\int_3^\infty E(Y_t\cdot{1}_{Y_t>0})\mathrm{d}t\leq e^{-1}\int_3^\infty P(Y_t>0)\mathrm{d}t\leq e^{-1}\int_3^\infty \frac{6}{(\lfloor t\rfloor-1)^2}\mathrm{d}t<\infty$$ condition 1) is satisfied. Also condition 2) seems to me beeing satisfied, but it is a lot of calculation \begin{align} \frac{\mathrm{d}}{\mathrm{d}t} E(Y_t)=\frac{\mathrm{d}}{\mathrm{d}t}\int_0^1 \sum_{n=1}^\infty \phi_{1/n^2}(t-\omega-n) \mathrm{d}\lambda(\omega)=&\sum_{n=1}^\infty \int_0^1 \frac{\mathrm{d}}{\mathrm{d}t}\phi_{1/n^2}(t-\omega-n) \mathrm{d}\lambda(\omega)\\ =&\sum_{n=1}^\infty \int_0^1 -\frac{\mathrm{d}}{\mathrm{d}\omega}\phi_{1/n^2}(t-\omega-n) \mathrm{d}\lambda(\omega)\\ =&\sum_{n=1}^\infty \phi_{1/n^2}(t-0-n)-\phi_{1/n^2}(t-1-n)\\ =&\sum_{n=1}^\infty \phi_{1/n^2}(t-n)-\phi_{1/n^2}(t-(n+1)). \end{align} To see that this is less or equal to zero we consider \begin{align} &\sum_{n=1}^\infty \phi_{1/n^2}(t-n)-\phi_{1/n^2}(t-(n+1))\\ &=\phi_{1}(t-1)-\phi_{1}(t-2)+\phi_{1/4}(t-2)-\phi_{1/4}(t-3)+\phi_{1/9}(t-3)-...\\ &=\underbrace{\phi_{1}(t-1)}_{=0,\text{ für } t\geq2}+\underbrace{\big(\phi_{1/4}(t-2)-\phi_{1}(t-2)\big)}_{\leq0}+\underbrace{\big(\phi_{1/9}(t-3)-\phi_{1/4}(t-3)\big)}_{\leq0}+... \end{align} Since we consider only $$t\geq3$$, $$\frac{\mathrm{d}}{\mathrm{d}t} E(Y_t)\leq0$$ holds and condition 2) is satisfied.

In summary, we found a nonnegative process $$Y$$ with a.s. $$C^\infty([3,\infty[)$$ paths and

1. $$\int_3^\infty P(|Y_t|>\epsilon)\mathrm{d}t<\infty$$,
2. $$\int_3^\infty E(|Y_t|)\mathrm{d}t<\infty$$,
3. $$[3,\infty[\ni t \mapsto E(|Y_t|)$$ is decreasing,
4. $$Y$$ is converging to $$0$$ in probability,
5. $$Y$$ is not converging to $$0$$ a.s.

Furthermore, from 2. and 3. we can conclude

1. $$Y$$ is converging to $$0$$ in $$\mathcal{L}^1(P)$$.