# 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!

• By definition of the Brownian motion, disjoint time increments are independent. – Clarinetist Jun 5 '15 at 7:32
• sorry, should have clarified it a bit more, I meant what allows us to use the fact that the two sets of equalities are the same? What allows us to use it in the joint density? – user123276 Jun 5 '15 at 7:35
• What I am saying here probably won't help, but the way I see it, we know that disjoint time increments are independent, so it is more convenient to work with the time increments rather than the times themselves. This probably isn't right, but my best guess. – Clarinetist Jun 5 '15 at 7:38
• The sets of equalities are indeed equivalent - solve for $X(t_j)$ for each $j$ in the second system of equalities and you will see that they are identical to the first system. – Clarinetist Jun 5 '15 at 7:40

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*}
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)