Maximum Likelihood Estimation of Brownian Motion Drift I'm looking at times series of stock movements over 10 minute windows, and am trying to measure the "trend" of these movements. 
Method A is to simply calculate $\Delta P$, the difference between the final value and the initial value.  
Method B is to use MLE to calculate the maximum likelihood value of the drift coefficient using the Brownian Motion model $dP(t)=\mu P(t) dt + \sigma P(t)  dB(t)$.
My intuition is that, if I am able to calculate the estimate $\hat{\mu}$, it will be a better measure of the trend since the Brownian motion is able to take into account volatility. However, another part of me thinks that $\hat{\mu}$ may actually just be a linear combo of  $\Delta P$ (i.e. $\hat{\mu}$ is completely determined by the first and last values of the time series). 
Any thoughts or semi-rigorous explanations would greatly appreciated! And let me know if anything else needs clarifying. 
 A: If $\mu$ and $\sigma$ are constant,  $dP = \mu\; dt + \sigma\; dW(t)$ is equivalent to $P(t) = P(0) + \mu t + \sigma W(t)$ where $W$ is Brownian motion with $W(0)=0$.  Then $\Delta P = P(T) - P(0) = \mu T + \sigma W(T)$ has a normal distribution with mean $\mu T$ and standard deviation $\sigma \sqrt{T}$.  The density at a given $\Delta P$ is maximized with respect to $\mu$ when $\mu = \Delta P/T$. 
Now I think what you're really doing is looking at not just the initial and final values, but some in between as well, say $P(t_j) = P_j$, $j= 0 .. N$, where $0 = t_0  < \ldots  < t_N = T$.
Let $\Delta_j t = t_j - t_{j-1}$ and $\Delta_j P = P(t_j) - P(t_{j-1})$.
  By the fact that Brownian motion has independent increments, the joint distribution of $\Delta_1 P, \ldots, \Delta_N P$ is independent normal with means $\mu \Delta_1 t, \ldots, \mu \Delta_N t$ and variances $\sigma^2 \Delta_1 t, \ldots \sigma^2 \Delta_N t$.
The density of this is proportional to 
$$\exp\left( - \sum_j \dfrac{(\Delta_j P - \mu \Delta_j t)^2}{2\sigma^2 \Delta_j t}\right) = \exp\left((const) + \frac{\mu}{\sigma^2} \sum_j \Delta_j P   - \frac{\mu^2}{2\sigma^2} \sum_{j} \Delta_j t \right) = (const) \exp((2\mu \Delta P - \mu^2 T)/\sigma^2)$$
which is maximized when $\mu = \Delta P/T$.
Thus the two approaches have the same result.  
