Quadratic Variation of an Ito Process

Setup

Suppose

$$ dZ_t=X_t\,dt+Y_t\,dB_t. $$

Only the Brownian part contributes to quadratic variation, so

$$ d\langle Z\rangle_t=Y_t^2\,dt. $$

Using differential notation,

$$ (dZ_t)^2=(X_t\,dt+Y_t\,dB_t)^2=Y_t^2\,dt, $$

because

$$ (dt)^2=0, \qquad dt\,dB_t=0, \qquad (dB_t)^2=dt. $$