Estimating the value of $\sigma$ for Brownian motion

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Let $X_t=\sigma W_t$ be a stochastic process, where $W_t$ is the Wiener process and $\sigma$ is an unknown parameter.

I want a formula to estimate the value of $\sigma$ (which could not be found in my book) from $n$ successive measurements $(x_{t_k})_{k=1}^n$. Now, I understand that $(X_{t_k})_{k=1}^n$ is a normally distributed vector in $\mathbb R^n$, but so I could NOT estimate its co-variance matrix because I only have the sample of ONE vector in $\mathbb R^n$, and co-variance could NOT be estimated from ONE sample. That stops me from estimating $\sigma$. How could I estimate $\sigma$?

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Since Wiener increments are independent, so are the $X$ increments.

By definition, $W_t-W_s \sim N(0, (t-s))$ so $V[X_{t_k} - X_{t_{k-1}}] = (t_k - t_{k-1})\sigma^2$.

In other words: $\Delta X$ is a sequence of $n-1$ normal variables with the variance above. Normalizing by the standard deviation gives $n-1$ independent variables with std $\sigma$:

$$ V[Z_k] = V[\frac{X_{t_k} - X_{t_{k-1}}}{\sqrt{t_k - t_{k-1}}}] = \sigma^2. $$

Now, $\sigma$ can be estimated as usual for the iid normal sample $Z$.