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Let $X_t, Y_t$ be stochastic processes with almost sure continuous paths defined on some probability space. Define $$v(t):=E(|X_t-Y_t|^2)$$ for $0\leq t\leq T$. Clearly $v(t)$ is continuous on $[0,T]$. Now suppose $v(t)=0$ for all $t\in [0,T]$. From this can we conclude that $$P\{\omega:|X_t(\omega)-Y_t(\omega)|=0,\forall t\in [0,T]\}=1$$

I am wondering how this can be proved. Any help is appreciated!

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    $\begingroup$ Shouldn't the equality be $P=1$, not $P=0$? $\endgroup$ Sep 14, 2019 at 17:41
  • $\begingroup$ @herbsteinberg fixed it!. Thank you for noticing! $\endgroup$
    – Heisenberg
    Sep 14, 2019 at 17:50

1 Answer 1

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Hints:

  1. Let $f:[0,T] \to \mathbb{R}$ be a continuous function. Show that $f$ is not identical zero, if and only if, there exist some $k \in \mathbb{N}$ and $q \in \mathbb{Q} \cap [0,T]$ such that $$|f(q)| > \frac{1}{k}.$$
  2. Deduce from Step 1 and the a.s. continuity of the sample paths that $$\mathbb{P}\left(\exists t \in [0,T]: |X_t-Y_t| \neq 0 \right) \leq \sum_{q \in \mathbb{Q} \cap [0,T]} \sum_{k \in \mathbb{N}} \mathbb{P}(|X_q-Y_q|>1/k). \tag{1}$$
  3. Use the fact that $\mathbb{E}(|X_q-Y_q|^2)=0$ to conclude that the right-hand side (and hence, the left-hand side) of $(1)$ equals zero.
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  • $\begingroup$ Thank you. Makes a lot of sense. Since $E|X_q-Y_q|^2=0$ and if $A=\{|X_q-Y_q|^2=0,\forall q\}=\{|X_q-Y_q|=0,\forall q\}$ then $P(A)=0$ which proves the r.h.s of (1). Is this correct? $\endgroup$
    – Heisenberg
    Sep 14, 2019 at 20:35
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    $\begingroup$ @Heisenberg Probably you meant to write $P(A)=1$...? It suffices to note that $\mathbb{E}(U)=0$ implies $\mathbb{P}(U \neq 0)=0$ for any non-negative random variable $U$. Hence, $\mathbb{E}(|X_q-Y_q|^2)=0$ implies $\mathbb{P}(|X_q-Y_q| \neq 0)=0$. Alternatively, you can apply Markov's inequality: $$\mathbb{P}(|X_q-Y_q| > 1/k) \leq k^2 \mathbb{E}(|X_q-Y_q|^2)=0.$$ $\endgroup$
    – saz
    Sep 14, 2019 at 20:40
  • $\begingroup$ Thank you! this helped a lot $\endgroup$
    – Heisenberg
    Sep 14, 2019 at 20:43

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