Incorrect use of Ito's lemma

Question

Consider a simple Ornstein–Uhlenbeck process $X(t)$:

$$ \mathrm d X(t) = - X(t) \, \mathrm dt + \sqrt{2} \, \mathrm dW(t). \tag{1} $$

If we apply Itô's lemma in its common formulation to get SDE for $X^2(t)$, we obtain

$$ \mathrm d X^2(t) = [- 2X^2(t) + 2] \, \mathrm d t + 2 X(t) \sqrt 2 \, \mathrm dW(t).\tag{2} $$

Note that both equations have the same Wiener process $W(t)$. The fact that they have the same Wiener process seems natural, since $X(t)$ and $X^2(t)$ should be driven by the same source of noise. From these two SDEs we might incorrectly conclude that

$$ \mathrm d X^2(t) - 2 X(t) \, \mathrm d X(t) = 2 \, \mathrm dt. $$

Hence it seems that the quantity $\mathrm d X^2 - 2 X \, \mathrm d X$ is deterministic. However, let us use the Euler–Maruyama discretization scheme ($\xi$ is a normally distributed random number with mean $0$ and variance $1$):

$$ X(t + \Delta t) = X(t) - X(t) \, \Delta t + \sqrt{2} \sqrt{\Delta t} \, \xi(t). $$

From it, we can calculate $\Delta X^2 - 2 X \, \Delta X$ up to $\Delta t$

$$ [X^2(t + \Delta t) - X^2(t)] - 2 X(t) [X(t + \Delta t) - X(t)] = 2 \Delta t \, \xi^2(t) + \ldots, $$

which is a random variable.

Queston: How do I correctly apply Itô's lemma (or something else) to calculate $\mathrm d X^2 - 2 X \, \mathrm d X$ ? Can it be expressed in terms of $dW(t)$? To make it even more clear, I am interested in realization specific identities (strong, not weak sense). How to formulate Ito's lemma so one would avoid paradoxes like that? I used the OU process just as an illustration, I am interested in the case of a general SDE for $X(t)$.

Update: In case you wonder why I insist that $\mathrm dX^2 - 2X \, \mathrm dX \neq 2 \, \mathrm dt $. You can use any program of your choice to simulate $X(t)$ and then calculate $g(t)$ as

$$ g(t) = [X^2(t + \Delta t) - X^2(t)] - 2 X(t) [X(t + \Delta t) - X(t)] $$

for small values of $\Delta t$. You will see that $g(t)$ is random. I attach below a screenshot from Mathematica that does that (I also tried writing my own program in Julia with the same result). I admit that $g(t)$ might not be the correct approximation of $\mathrm dX^2 - 2X \, \mathrm dX$, then please tell me how to do it right.

Answer 1

At the risk of repeating other user's comments:

To add to NN2's answer: $$\tag{A} X_t=X_0\,e^{-t}+\int_0^t\sqrt{2}\,e^{(s-t)}\,dW_s $$ is the correct solution to the SDE $$\tag{B} dX_t=-X_t\,dt+\sqrt{2}\,dW_t\,. $$ By the Ito rule $$\tag{C} d(X_t)^2=2X_t\,dX_t+2\,dt $$ (cf. Lutz Lehmann's answer).

• You wrote "we might incorrectly conclude that" (C) holds. We disagree with this. (C) is correct and holds.

• The Euler-Maruyama discretization does not refute the validity of (C) which holds only in the limit as I will now show with as elementary arguments as possible:

Using (A), (C) can be written in integral form equivalently as \begin{align}\tag{D} X_t^2=\Big(\int_0^t \sqrt{2}e^{s-t}\,dW_s\Big)^2 \end{align} (to simplify notation I assume $X_0=0\,$.) Taking $0=t_0<...<t_n=t$ we get \begin{align}\tag{E} (X_{t_i})^2-(X_{t_{i-1}})^2&=X_{t_i}\Big(X_{t_i}-X_{t_{i-1}}\Big)+X_{t_{i-1}}\Big(X_{t_i}-X_{t_{i-1}}\Big)\\ &=\underbrace{\Big(X_{t_i}-X_{t_{i-1}}\Big)\Big(X_{t_i}-X_{t_{i-1}}\Big)}_{(*)}+\underbrace{2X_{t_{i-1}}\Big(X_{t_i}-X_{t_{i-1}}\Big)}_{(**)}\,. \end{align}

• In (and only in) the limit $n\to\infty$ and $\max\limits_{i=1,...,n}|t_i-t_{i-1}|\to 0$ the sum over $i$ of the terms (**) converges to the Ito integral $2\int_0^tX_s\,dX_s\,.$

• In the same limit the sum of the terms (*) converges to $$\tag{F} \int_0^t2\,ds=2t\,. $$ Proof. From (B) $$ X_t=-\int_0^t X_s\,ds+\sqrt{2}\,W_t=:A_t+M_t $$ where $A_t$ has finite variation and $M_t$ zero variation but finite quadratic variation. Let's be clear that $\langle M\rangle_t$ denotes quadratic variation and not some expected value $\mathbb E[M_t]\,.$ From $\langle M\rangle_t=2\langle W\rangle_t=2t$ and $\langle A,M\rangle_t=0$ (see this answer) the proof of (F) follows.

• Since the sum over $i$ of the LHS of (E) is a telescoping sum we can put it all together to $$ (X_t)^2=2t+\int_0^tX_s\,dX_s\,. $$ Again: this holds in the limit and not in an Euler-Maruyama discretization.

Answer 2

It's wrong to do $$ \mathrm d X = - X \, \mathrm dt + \sqrt{2} \, \mathrm dW(t)\Longrightarrow X(t + \Delta t) \approx X(t) - X(t) \, \Delta t + \sqrt{2} \sqrt{\Delta t} \, \xi(t) \tag{1} $$ The discretization method does not work on the stochastic differential equations. So, $(1)$ is not correct.

If you compute $X_t$ $$X_t =X_0e^{-t}+\int_0^t\sqrt{2}e^{s-t}dW_s \tag{2}$$ you can then apply the Euler–Maruyama discretization scheme.

For information, the formula $dX^2_t-2X_tdX_t = 2dt$ must be correct. You can use $(2)$ to test it.

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From $(2)$, we will prove $$dX_t^2-2X_tdX_t=2dt \tag{3}$$

For the sake of simplicity, we denote $Z_t :=\int_0^t\sqrt{2}e^{s}dW_s$, then $X_t=e^{-t}(X_0+Z_t)$.

We have some results: $$\begin{align} dZ_t &= \sqrt{2}e^tdW_t \tag{4}\\ d(Z_t^2) &= 2Z_tdZ_t + (dZ_t)^2 \stackrel{(4)}{=} 2\sqrt{2}Z_te^tdW_t+2e^{2t}dt\tag{5}\\ X_t^2 &= e^{-2t}(X_0^2+2X_0Z_t+Z_t^2)\tag{6}\\ d(X_t^2) &\stackrel{(6)}{=} e^{-2t}d(X_0^2+2X_0Z_t+Z_t^2)-2e^{-2t}(X_0^2+2X_0Z_t+Z_t^2)dt\\ &=e^{-2t} \left( 2X_0dZ_t+d(Z_t^2) - 2(X_0^2+2X_0Z_t+Z_t^2)dt \right)\\ &\stackrel{(4,5)}{=}e^{-2t} \left( \color{red}{2X_0\sqrt{2}e^tdW_t+2\sqrt{2}Z_te^tdW_t}+2e^{2t}dt - 2(X_0^2+2X_0Z_t+Z_t^2)dt \right)\\ &=e^{-2t} \left( \color{red}{2\sqrt{2}e^{2t}X_tdW_t}+2e^{2t}dt - 2(X_0^2+2X_0Z_t+Z_t^2)dt \right)\\ &= 2\sqrt{2}X_tdW_t+2dt - 2(X_0^2+2X_0Z_t+Z_t^2)e^{-2t}dt \\ &= 2\sqrt{2}X_tdW_t+2dt - 2X_t^2dt \tag{7}\\ X_tdX_t &= \sqrt{2}X_tdW_t -X_t^2 dt \tag{8} \end{align}$$

Finally, from $(7)(8)$, we can prove $(3)$ $$\begin{align} \color{red}{dX_t^2- 2 X_tdX_t} &= (2\sqrt{2}X_tdW_t+2dt - 2X_t^2dt) - 2(\sqrt{2}X_tdW_t -X_t^2 dt) = \color{red}{2dt} \end{align}$$

Remark: it's quite time-consuming! Luckily I can reach the end of the proof.

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Again, we cannot apply the discretization scheme to SDEs because that is the source of errors. Take for example a well-know SDE $$\frac{dS_t}{S_t}=\sigma dW_t \tag{9}$$ where the solution is $$S_t = S_s\cdot \exp\left(-\frac{1}{2}\sigma^2 (t-s) + \sigma (W_t-W_s) \right) \tag{10}$$

With a discretization time step $t_n = \Delta t \cdot n$, from $(10)$, we have $$\begin{align} S_{t_n} &= S_{t_{n-1}}\exp\left(-\frac{1}{2}\sigma^2 \Delta t + \sigma \sqrt{\Delta t} \mathcal{N}(0,1) \right) \\ \text{or}\hspace{0.5cm} S_{t_n} &\approx S_{t_{n-1}} \left(1 \color{red}{-\frac{1}{2}\sigma^2 \Delta t} + \sigma \sqrt{\Delta t} \mathcal{N}(0,1) \right) \end{align}$$

If we use $(9)$, the red term is missing $$S_{t_n} \approx S_{t_{n-1}}\left(1 + \sigma \sqrt{\Delta t} \mathcal{N}(0,1) \right)$$

Answer 3

By the Ito formula in question, you get in general for the squared process $$ d(X^2)=2X\,dx+d\langle X\rangle_t $$ where with $dX=a\,dt+b\,dW_t$ one gets the quadratic variation as $d⟨X⟩_t=b^2\,dt$, so here $=2\,dt$.

Thus the computed identity is completely normal and expected.