I'm reading Itô's lemma from two sources.

  1. Wikipedia page.

In its simplest form, Itô's lemma states the following: for an Itô drift-diffusion process $$ d X_t=\mu_t d t+\sigma_t d B_t $$ and any twice differentiable scalar function $f(t, x)$ of two real variables $t$ and $x$, one has $$ d f\left(t, X_t\right)=\left(\frac{\partial f}{\partial t}+\mu_t \frac{\partial f}{\partial x}+\frac{\sigma_t^2}{2} \frac{\partial^2 f}{\partial x^2}\right) d t+\sigma_t \frac{\partial f}{\partial x} d B_t $$ This immediately implies that $f\left(t, X_t\right)$ is itself an Itô drift-diffusion process. In higher dimensions, if $\mathbf{X}_t=\left(X_t^1, X_t^2, \ldots, X_t^n\right)^T$ is a vector of Itô processes such that $$ d \mathbf{X}_t=\mu_t d t+\mathbf{G}_t d \mathbf{B}_t $$ for a vector $\boldsymbol{\mu}_t$ and matrix $\mathbf{G}_t$, Itô's lemma then states that $$ \begin{aligned} d f\left(t, \mathbf{X}_t\right) & =\frac{\partial f}{\partial t} d t+\left(\nabla_{\mathbf{X}} f\right)^T d \mathbf{X}_t+\frac{1}{2}\left(d \mathbf{X}_t\right)^T\left(H_{\mathbf{X}} f\right) d \mathbf{X}_t \\ & =\left\{\frac{\partial f}{\partial t}+\left(\nabla_{\mathbf{X}} f\right)^T \boldsymbol{\mu}_t+\frac{1}{2} \color{blue}{\operatorname{Tr}\left[\mathbf{G}_t^T\left(H_{\mathbf{X}} f\right) \mathbf{G}_t\right]}\right\} d t+\left(\nabla_{\mathbf{X}} f\right)^T \mathbf{G}_t d \mathbf{B}_t \end{aligned} $$ where $\nabla_{\mathbf{X}} f$ is the gradient of $f$ w.r.t. $X, H_{\mathbf{X}} f$ is the Hessian matrix of $f$ w.r.t. $X$, and Tr is the trace operator.

  1. A lecture note.

Itô formula (Multidimensional). Let $X_t$ solve $d X_t=b(t, \omega) d t+\sigma(t, \omega) d W_t$, where $X_t \in \mathbb{R}^n, \sigma \in \mathbb{R}^{n \times m}$, $W_t \in \mathbb{R}^m$. Let $Y_t=f\left(X_t\right)$, where $f \in C^2\left(\mathbb{R}^n\right)$. Then $$ d Y_t=\nabla f\left(X_t\right) \cdot d X_t+\frac{1}{2}\left(d X_t\right)^T \nabla^2 f\left(X_t\right) d X_t $$ where $\nabla^2 f=\left(\frac{\partial^2 f}{\partial x_i \partial x_j}\right)_{i, j}$ is the Hessian matrix of $f$, and products of increments are evaluated using the rules following (11) plus the additional rule $$ d W_t^i \cdot d W_t^i=d t, \quad d W_t^i \cdot d W_t^j=0 \quad \text { for } i \neq j $$ where $W_t^i$ is the ith component of $W_t$. Therefore $Y_t$ solves the equation $$ d Y_t=\left(b \cdot \nabla f+\frac{1}{2} \color{blue}{\sigma \sigma^T: \nabla^2 f}\right) d t+(\nabla f)^T \sigma d W_t $$ where $A: B=\operatorname{Tr}\left(A^T B\right)=\sum_{i, j} a_{i j} b_{i j}$.

I already verified the formula in the note here. Following notation in the note, $$ \sigma \sigma^T: \nabla^2 f = \operatorname{Tr} ((\sigma \sigma^T)^T (\nabla^2 f)) = \operatorname{Tr} ((\nabla^2 f)^T \sigma \sigma^T) = \operatorname{Tr} ((\nabla^2 f) \sigma \sigma^T). $$

For the Wikipedia page to be consistent with the note, I expect that $$ \operatorname{Tr}\left[\mathbf{G}_t^T\left(H_{\mathbf{X}} f\right) \mathbf{G}_t\right] = \operatorname{Tr}\left[\left(H_{\mathbf{X}} f\right) \mathbf{G}_t \mathbf{G}_t^T\right] \tag{1}. $$

However, it seems (1) is not generally true unless $\mathbf{G}_t$ is symmetric.

Could you elaborate on my confusion?

  • 1
    $\begingroup$ The cyclic property $Tr(ABC)=Tr(BCA)$ doesn't require symmetry, does it? $\endgroup$ May 24, 2023 at 17:15
  • $\begingroup$ @user6247850 You are right! Thank you so much for your elaboration! $\endgroup$
    – Analyst
    May 24, 2023 at 17:32

1 Answer 1


We have from this Wikipedia that

The trace of a square matrix which is the product of two real matrices can be rewritten as the sum of entry-wise products of their elements, i.e. as the sum of all elements of their Hadamard product. Phrased directly, if $\mathbf{A}$ and $\mathbf{B}$ are two $m \times n$ real matrices, then: $$ \operatorname{tr}\left(\mathbf{A}^{\top} \mathbf{B}\right)=\operatorname{tr}\left(\mathbf{A B}^{\boldsymbol{\top}}\right)=\operatorname{tr}\left(\mathbf{B}^{\top} \mathbf{A}\right)=\operatorname{tr}\left(\mathbf{B} \mathbf{A}^{\top}\right)=\sum_{i=1}^m \sum_{j=1}^n a_{i j} b_{i j}. $$

We apply above formula with $\mathbf{A} := \mathbf{G}_t$ and $\mathbf{B} := \left(H_{\mathbf{X}} f\right) \mathbf{G}_t$ and get $$ \operatorname{Tr}\left[\mathbf{G}_t^T\left(H_{\mathbf{X}} f\right) \mathbf{G}_t\right] = \operatorname{Tr}\left[\left(H_{\mathbf{X}} f\right) \mathbf{G}_t \mathbf{G}_t^T\right]. $$


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