# Clarifications about SDEs, Differentials & Derivatives

A general SDE look like the following:

$$\mathrm{d}\psi=a\mathop{}\!\mathrm{d}t+b\mathop{}\!\mathrm{d}W,\tag{1}$$

where $\psi:t\mapsto y = \psi(t)$ is the solution, while $a$ and $b$ can be both, constants or any function, right?

Equation $1$ can be rewritten as:

$$\frac{\mathrm{d}\psi}{\mathrm{d}t}=a\mathop{}\!\mathrm{d}t+b\mathop{}\!\mathrm{d}W,$$

or as:

$$\psi(t)-\psi(t_0)=\int a\mathop{}\!\mathrm{d}t+\int b\mathop{}\!\mathrm{d}W.$$

so it's not clear to me why it's common to use $\mathrm{d}\psi$ instead of $\psi'(t)$ to abbreviate. Also, is $\psi(t_0)$ the integration constant of $\int\frac{\mathrm{d}\psi}{\mathrm{d}t}\mathop{}\!\mathrm{d}t$? If so, why it has the minus sing?

The differential of the function $y = \psi(t)$ is:

$$\mathop{}\!\mathrm{d}\psi=\psi'(t)\mathop{}\!\mathrm{d}t=\frac{\mathrm{d}y}{\mathrm{d}t}\mathop{}\!\mathrm{d}t,$$

and since the two $\mathop{}\!\mathrm{d}t$ simplify, the final result would be $\mathrm{d}y$? Then in this case, Ito's lemma would help me find the expression of this $\mathrm{d}y$?

In the case of a system of two SDEs, can I just use Ito's lemma on both the equations and then match the initial conditions with the constants?

• "so it's not clear to me why it's common to use $\mathrm{d}\psi$ instead of $\psi'(t)$ to abbreviate" First thing is that $t\mapsto\psi(t)$ is in general not differentiable, with probability 1.
– Did
Commented Jan 2, 2015 at 16:10
• @Did mmm ok, so why SDEs use $\mathrm{d}\psi$, even if the funciton is not differentiable? I don't get the meaning of that writing. Commented Jan 6, 2015 at 16:04
• The meaning of $\mathrm d\psi_t=a_t\mathrm dW_t+b_t\mathrm dt$ is that $$\psi_t=\psi_0+\int_0^ta_s\mathrm dW_s+\int_0^t\mathrm b_sds,$$ where the first integral on the RHS is a stochastic integral.
– Did
Commented Jan 6, 2015 at 16:59

$$\psi(t)=\int_{0}^{t}b_{s}ds+\int_{0}^{t}a_{s}dW_{s}$$
is not differentiable for nonzero $$a$$ eg. take $$a=1$$
$$\psi(t)=\int_{0}^{t}b_{s}ds+B_{t}.$$