Say we have a set of observations $\{X_i : i=0,...,N\}$ from a simple Brownian motion (scaled Winner process) $$dX_t = \sigma dW_t .$$ If we discretise this to
$$X_i-X_{i-1} = \sigma\sqrt{\Delta}Z_i$$
with $Z_i \sim N(0,1)$ then the likelihood function is a product of
$$L_i(\sigma | \{Z_i\}) = \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{Z_i^2}{2}\right)= \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{(X_i-X_{i-1})^2}{2\sigma^2\Delta}\right) .$$
However, if we discretise this to
$$X_i-X_{i-1} = Y_i$$
with $Y_i \sim N(0,\sigma^2\Delta)$ then the likelihood function is a product of
$$L_i(\sigma | \{Y_i\}) = \frac{1}{\sqrt{2\pi\sigma^2\Delta}}\exp\left(-\frac{Y_i^2}{2\sigma^2\Delta}\right) = \frac{1}{\sqrt{2\pi\sigma^2\Delta}}\exp\left(-\frac{(X_i-X_{i-1})^2}{2\sigma^2\Delta}\right) .$$
So there seems to be a difference in the denominator under the square root depending on whether I choose $Z_i$ or $Y_i$, which doesn't seem right to me as it would presumably lead to different estimates of $\sigma$. Where am I going wrong?