# Law of one Price and Cointegration relationship

I have a question on the relationship between the law of one price and cointegration of (financial) time series.

• Suppose there is an unobserved "efficient" price $$v_t$$ that follows a standard Brownian motion such that $$v_t = v_0 + \int\limits_{0}^t \sigma dW_s$$.
• Two venues trade the asset and quote prices ($$x_t$$ and $$y_t$$) which reflect the efficient price and some stationary noise component: $$x_t = v_t + u_t \text{ and } y_t = v_t + \tilde u_t$$
• Then, both $$x_t$$ and $$y_t$$ exhibit a unit root but $$z_t := x_t - y_t$$ resembles a valid cointegration relationship because $$z_t = u_t + \tilde u_t$$ which is stationary.
• Suppose, there is no noise at one market such that $$x_t = v_t$$.
• The other market simply mirrors the prices but with a possibly random lag such that $$y_t = x_{t-\tau} = v_{t-\tau}$$
• Then, under which circumstances is $$z_t = x_t - y_t = \int\limits_{t-\tau}^t \sigma dW_s \sim N(0, \sigma^2\tau)$$ again a stationary variable and resembles a reasonable cointegration relationship? Is it necessary that $$\tau$$ is stationary or are there other sufficient conditions?
• It's interesting but there's no stochastic process in your second example so I don't think one can define cointegration in that case. This is because $y_t$ is deterministic. – mark leeds Oct 20 '19 at 14:01
• Thanks @markleeds, excellent point. I tried to edit my question a bit, such that you can basically think of $y_t$ being a jump process with random latency. In other words, $y_t$ does not need to be deterministic at all..does this change something from your perspective? – muffin1974 Oct 21 '19 at 1:59