Quadratic variation of the Ornstein-Uhlenbeck process

Question

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.

Answer 1

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. Use $[X] = X^2 - X^2_0 - 2X \bullet X$ (a consequence of Itô's formula) in combination with the result of step 2 to get the quadratic variation $[X]_t$.

Answer 2

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.

Answer 3

You can use it, but first you need to use integration by parts in the integral of $$X_t = \sigma e^{-\alpha t}\int_0^t e^{\alpha s}\,dB_s. \tag{$\triangle$}$$ Here, you can regard the integral as a stochastic integral or just as a Wiener integral.

Doing this, you will get

$$\sigma e^{-\alpha t}\int_0^t e^{\alpha s}\,dB_s = \sigma e^{-\alpha t}\left[e^{\alpha t} B_t - \int_0^t B_s \alpha e^{\alpha s}\,ds \right].$$

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

More generally, if $X$ and $Y$ are continuous semimartingales, then integration by parts (for stochastic integrals) says that $$XY - X_0 Y_0 = X \cdot Y + Y \cdot X + [X, Y],$$ whence $$[XY, XY] = X^2 \cdot [Y,Y] + 2(XY) \cdot [Y, X] + Y^2 \cdot [X, X].$$