If the SDE of an Ito process is identically zero, are the drift and volatility both identically zero?

Question

If the Ito process $\mathrm{d}X_{t}=\mu(X_{t})\mathrm{d}t+\sigma (X_{t})\mathrm{d}W\equiv0$, it seems that $\mu(X_{t})=0$ and $\sigma(X_{t}% )=0$. Is it true? If yes, how to prove? We know that $\mathrm{d}X_{t}% =\mu(X_{t})\mathrm{d}t+\sigma(X_{t})\mathrm{d}W\equiv0$ exactly means given time $s$ with $X_{s}\neq0$ $$ \forall u>s\qquad0\equiv X_{u}-X_{s}=\int_{s}^{u}\mu(X_{t})\,\mathrm{d}% t+\int_{s}^{u}\sigma(X_{t})\,\mathrm{d}W_{t}% $$ then $\mu(X_{t})\equiv0$ and $\sigma(X_{t})\equiv0$ for $s<t<u$.

Answer 1

It is true whether $X_{s}=0$ or not

Remark: If $\int_{s}^{x}f(t)\,\mathrm{d}t=0$ for any $x>s$, then $f(x)=\frac{\mathrm{d}}{\mathrm{d}x}\left( \int_{s}^{x}f(t)\,\mathrm{d}% t\right) =\frac{\mathrm{d}}{\mathrm{d}x}(0)=0$

For $X_{t}=X_{s}$ is constant for any $t>s$, the quadratic variation is zero $$ 0=[X_{t}]=\int_{s}^{u}\sigma^{2}(X_{t})\,\mathrm{d}t $$ which means $\sigma(X_{t})=0$. And for any $u>s$ $$ \int_{s}^{u}\mu(X_{t})\,\mathrm{d}t=(X_{u}-X_{s})-\int_{s}^{u}\sigma (X_{t})\,\mathrm{d}W_{t}=0-0=0 $$ thus $\mu(X_{t})=0$.

Answer 2

No, $X_t-X_0=0$ does, in general, not imply $\sigma=\mu=0$. Consider for instance the SDE

$$dX_t = X_t \, dt + X_t \, dW_t, \qquad X_0 := 0.$$

You can easily check that $X_t := 0$ solves the SDE, but the coefficients $\sigma(x) = \mu(x)=x$ are not identically zero.