# If $E(|X_t-Y_t|^2)=0$, then $P\{\omega:|X_t(\omega)-Y_t(\omega)|=0,\forall t\in [0,T]\}=1$?

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!

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

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.
• 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? Sep 14, 2019 at 20:35
• @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: $$\mathbb{P}(|X_q-Y_q| > 1/k) \leq k^2 \mathbb{E}(|X_q-Y_q|^2)=0.$$