Calculating joint density function of Brownian motion I read in my book today regarding the calculation of the joint density function of a brownian motion process and it went as follows:
If we define $X(t)$ as a Brownian motion process with mean $0$ and variance $t$, to obtain the joint density function of $X(t_1), ... , X(t_n)$ for $t_1 < \cdots < t_n$, note that the set of equalities, 
$$
X(t_1) = x_1, X(t_2) = x_2, ... , X(t_n) = x_n
$$
is equivalent to 
$$
X(t_1) = x_1, X(t_2)-X(t_1) = x_2-x_1, ... , X(t_n)-X(t_{n-1}) = x_n-x_{n-1}.
$$
then, the joint density of $X(t_1) = x_1, X(t_2) = x_2, ... , X(t_n)$ is 
$$
f(x_1,...,x_n) = f_{t_1}(x_1) \cdot f_{t_2 - t_1}(x_2-x_1)\cdots f_{t_n - t_{n-1}}(x_n - x_{n-1}). 
$$
I am confused why we are allowed to directly input the second set of equalities into the density function. Wouldn't it yield a different result as we are indirectly calculating each $x_i$? Would anyone be able to help me? Thank you!
 A: The other answer doesn't really give the full story. It's the transformation theorem. I just had an answer involving it here. 
Call 
$$X(t_1) = Y_1,~X(t_2) = Y_2 + Y_1, ...,~X(t_n) = Y_n + \cdots + Y_1.$$ 
The inverse transformations are $$Y_1 = X(t_1),~Y_2 = X(t_2)-X(t_1), ..., ~Y_n = X(t_n) - X(t_{n-1}).$$ 
We know the joint density of all the $Y$s is the product of mean zero gaussians with variances $\sigma^2 t_1$, $\sigma^2 (t_2-t_1),\ldots, \sigma^2(t_n-t_{n-1})$. We're using the independent and stationary increments properties here. We will call this product of Gaussians $f_{Y_1,\ldots,Y_n}(y_1,\ldots,y_n) = f_{Y_1}(y_1)\cdots f_{Y_n}(y_n)$. Looking at the matrix of partial derivatives, its determinant is $1$. 
So this is why you just plug stuff in. The density of $X(t_1),\ldots, X(t_n)$ is 
\begin{align*}
g_{X(t_1),\ldots, X(t_n)}(x_1,\ldots,x_n)&=f_{Y_1,\ldots,Y_n}(y_1[x(t_1),\ldots,x(t_n)],\ldots,y_n[x(t_1),\ldots,x(t_n)])|1|\\
&= f_{Y_1,\ldots,Y_n}(x(t_1),x(t_2)-x(t_1),\ldots,x(t_n)-x(t_{n-1})) \\
&= f_{Y_1}(x(t_1))f_{Y_2}(x(t_2)-x(t_1))\cdots f_{Y_n}(x(t_n)-x(t_{n-1})). 
\end{align*}
A: The fact that the two set of equalities are equivalent can be seen by substituting the second set in the first one and conversely. This means that the two sets of events described are identical. Therefore calculating the probability of one is equivalent to calculating the probability of the second. 
Also observe that the second set of equalities consists of all independent events by the memorylessness of $X(t)$ so the probability of the set factors as the product of the probabilities of the single equalities. 
Now let $f_v(x)$ be the density function of a normal random variable of mean $0$ and variance $v$. We know that  $X(t)$ is a stationary process with variance $t$; so for each $j=1, ... , n$,   $X(t_j)-X(t_{j-1})$ is distributed like $X(t_j-t_{j-1})$ which is what you wanted (remember $X(0)=0$; also, to make this argument more rigorous one should extend the above to inequalities and use cdfs) 
