Let $(X_t)_{t\geq 0}$ be the zero-mean Ornstein-Uhlenbeck process such that $X_0 = 0$ almost surely, i.e. $$X_t = \sigma e^{-\alpha t}\int_0^t e^{\alpha s}\,dB_s \quad \qquad (\triangle)$$ On the other hand, $(X_t)$ is the unique process that satisfies the SDE $$dX_t = \alpha X_t\,dt + \sigma\,dB_t \quad X_0 = 0 \qquad (\square)$$

Since the SDE $(\square)$ satisfies the growth and the Lipschitz conditions, we know that the strong solution to this SDE, $(X_t)$, exists, is unique and continuous.

From the latter $[X,X]_t = [\int \alpha X_s\,ds + \int \sigma\,dB_s,\int \alpha X_s\,ds + \int \sigma\,dB_s]_t = [\int \sigma\,dB_s, \int \sigma\,dB_s]_t = \sigma^2 t$

Here I really needed continuity to make sure that the contribution of $\int \alpha X_s\,ds$ to the quadratic variation is $0$ since then this integral is of bounded variation and is continuous.

Anyway, my question is how would one compute $[X,X]_t$ by just using $(\triangle)$? I know that for semimartingales $M,N$ $$[G\cdot M,H\cdot N]_t = \int_{(0,t]}G_sH_s\,d[M,N]_t$$ I applied this result to my case as follows \begin{align}[X,X]_t =& [\sigma e^{-\alpha t}\int_0^t e^{\alpha s}\,dB_s,\sigma e^{-\alpha t}\int_0^t e^{\alpha s}\,dB_s]\\ =& \sigma^2 [\int_0^t e^{-\alpha (t-s)}\,dB_s, \int_0^t e^{-\alpha (t-s)}\,dB_s]\\=& \sigma^2 \int_0^t e^{-2\alpha (t-s)}ds \neq \sigma^2 \end{align} Clearly, I am doing something wrong in the last step. I feel like the $e^{-\alpha t}$ term in $X_t$ should be handled in a different way but I couldn't wrap my head around this.


You cannot apply the formula

$$[G \bullet M]_t = \int_0^t G_s^2 \, d[M]_s \tag{1}$$

because the Ornstein-Uhlenbeck process $X$ is not of the form

$$X_t = (G \bullet B)_t,$$

but of the form $$X_t = (G_t \bullet B)_t$$ and -as your calculation show- we cannot expect that $(1)$ extends to this larger class of processes. The reason is, roughly, that $dt$-terms need a different compensation than $dB_t$-terms - and if you shift the multiplicative $dt$-term under the stochastic integral, then you pretend that it behaves, in some sense, like a $dB_t$-term ... but it doesn't.

The proper way is the following:

  1. Define $$Y_t := \int_0^t e^{\alpha s} \, dB_s.$$ Calculate $[Y]_t$ (that you can do using $(1)$.)
  2. Apply Itô's formula to find the stochastic differential $$d(X_t^2) = \sigma^2 d(e^{-2\alpha t} Y_t^2).$$
  3. The $dt$-term of the stochastic differential $d(X_t^2)$, obtained in step 2, equals the quadratic variation $[X]_t$.
  • $\begingroup$ The third point is not clear. If $X_t=\int_0^tB_sds$, what are the $[X]_t$ and the $dt$-term of stochastic differential $d(X^2_t)$? are they equal? $\endgroup$ – JGWang Sep 11 '17 at 7:59
  • $\begingroup$ @JGWang Sorry, but I don't understand your question. You are considering a totally different process $X_t$ than the OP.... so how is your question related to what I wrote in my answer? $\endgroup$ – saz Sep 11 '17 at 8:10
  • $\begingroup$ Thank you for your reply. My question is what is the definition of $[X]_t$ and for OU-process in $(\Delta)$, is $[X]_t=\sigma^2t$ right? $\endgroup$ – JGWang Sep 12 '17 at 0:59
  • $\begingroup$ @JGWang Well, the OU-process is a semimartingale and therefore we can use the definition of the quadratic variation for semimartingales: $$[X]_t = X_t^2 -X_0^2 - 2 \int_0^t X_s \, dX_s.$$ $\endgroup$ – saz Sep 12 '17 at 4:57
  • $\begingroup$ Thank you for your clear reply. $\endgroup$ – JGWang Sep 13 '17 at 1:07

In Formula $$[G\cdot M,H\cdot N]_t = \int_{(0,t]}G_sH_s\,d[M,N]_t,\tag{1}$$ $G$ and $H$ are functions of $s$ only. However, in your case, $G=H=\sigma e^{-\alpha(t-s)}$, which are not functions of $s$ only, and then Formula$(1)$ is not applicable.


You can use it but first you need to use integration by parts to the product of $$X_t = \sigma e^{-\alpha t}\int_0^t e^{\alpha s}\,dB_s \quad \qquad (\triangle)$$

Doing this you will get

$$\sigma e^{-\alpha t}\int_0^t e^{\alpha s}\,dB_s = \sigma ( B_t - B_0 + -a \int_0^t \int_0^s e^{au} dB_u e^{-as} ds) $$

Since the second integral is continuous in t you get that the quadratic variation of $(\triangle)$ is $\sigma [B,B]_t = \sigma t$.


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