Given the Brownian Motion with drift
$$ dX(t) = \mu dt + \sigma dW(t) $$
It is well known that its distribution has the following form
$$ f_t(x) = \frac{1}{\sqrt(2 \pi \sigma^2 t)} e^{-\frac{(x -\mu t)^2}{2 \sigma^2 t}} $$
So following examples online for the Normal distribution, I get the following formulas for the parameters of the BM distribution
$$ \mu = \frac{1}{n} \sum_{i=1}^{n} \frac{x_i}{t_i} \\ \sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i-\mu t_i)^2 $$
Does this look correct? Please let me know if I need to show my work. Thanks
Edit 2 - Let's start this over
Step 1 - Define the Likelihood function:
$$ \mathcal{L}(\mu, \sigma;x)=\prod_{i=1}^{n} Pr(x_i|\mu, \sigma)= \left(2 \pi t_1\right)^{-\frac{1}{2}} \times \left(2 \pi t_2\right)^{-\frac{1}{2}} \times ... \times \left(2 \pi t_n\right)^{-\frac{1}{2}} \left(\sigma^2 \right)^{-\left(\frac{n}{2}\right)}e^{-\frac{1}{2 \sigma^2} \sum_{i=1}^{n} \frac{(x_i -\mu t_i)^2}{t_i}} $$
Step 2 - Take the log of the likelihood function
$$ \mathcal{l}(\mu, \sigma;x) = -\frac{1}{2}\ln(2 \pi t_1) -\frac{1}{2}\ln(2 \pi t_2) - ... -\frac{1}{2}\ln(2 \pi t_n) -\frac{n}{2} \ln(\sigma^2) -\frac{1}{2 \sigma^2} \sum_{i=1}^{n} \frac{(x_i - \mu t_i)}{t_i} $$
So, before we take derivatives - I am not sure how to go from this
$$ -\frac{1}{2}\ln(2 \pi t_1) -\frac{1}{2}\ln(2 \pi t_2) - ... -\frac{1}{2}\ln(2 \pi t_n) $$
to this for the first term
$$ - \frac{1}{2} \sum_{i=1}^{n} \ln(2 \pi (t_i - t_{i-1})) $$
Are we making any assumptions? The algebra does not work out for me.