What is the distribution of the subtract of two random variables? Definition) A stochastic process $\{X(t), t \geqslant 0\}$ is said to be Brownian motion process with drift coefficient $\mu$ and variance parameter $\sigma^2$, if it satisfies that


*

*$X(0)=0$.

*$\{X(t), t \geqslant 0\}$ has independent and stationary increments.

*For every $t \gt 0$, $X(t) \sim N(\mu t, \sigma^2 t)$.



Let $\{X(t), t \geqslant 0\}$ be a Brownian motion process with drift coefficient $0$ and variation parameter $1$. Then, for every $t$, $X(t)$ is normally distributed with mean $0$ and variation $t$.
According to the texts, for $s \lt t$, $X(t)-X(s)$ is normally distributed with mean $0$ and variation $t-s$.
Can someone explain why it is?

My trial:
First, Since $X(t)$ and $X(s)$ are normally distributed, linear combination of them is also normally distributed although they are dependent of each.
By using the fact that $E[X(t)]=0$ and $Var[X(t)]=E[X^2(t)]=t$,
\begin{align}
E\left[X(t)-X(s)\right] &=  E\left[X(t)\right] - E\left[X(s)\right] = 0 \\
Var\left[X(t)-X(s)\right] &= E\left[ \left\{X(t)-X(s)\right\}^2 \right] \\
&= E\left[ X^2(t) - 2X(t)X(s) + X^2(s) \right] \\
&= E\left[ X^2(t)\right] + E\left[X^2(s) \right] - 2E\left[X(t)X(s)\right] \\
&= t+s - 2E\left[X(t)X(s)\right] \\
&= t+s - 2\iint_{\Bbb{R}^2} xy f_{X(t), X(s)}(x, y) dtds
\end{align}
Since $X(t)$ is not independent of $X(s)$, I cannot write $f_{X(t), X(s)}(x, y)$ as
$$f_{X(t)}(x) f_{X(s)}(y)=\frac{1}{2\pi\sqrt{ts}}\exp{\left\{ -\frac{x^2}{2t} -\frac{y^2}{2s} \right\}}$$
Please let me know why
$$X(t)-X(s) \sim N(0, t-s)$$
 A: To get $E[X(s)E(t)]$, you could do
\begin{align*}
E[X(s)E(t)]& = E[X(s)\{X(t)-X(s)+X(s)\}]\\
&=E[X(s)\{X(t)-X(s)\}]+E[\{X(s)\}^2]\\
&=E[X(s)]E[X(t)-X(s)]+\text{Var}(X(s))+\{E[X(s)]\}^2\tag 1\\
&=0\cdot 0+ s+0^2\\
&=s
\end{align*}
where $(1)$ is true since $X(s)$ and $X(t)-X(s)$ are independent.
A: Hint: The Moment Generation Function for the sum of independent random variables has an interesting property.
$$\begin{align}\mathsf M_X(z) :=& ~ \mathsf E(\mathsf e^{zX})\\[2ex] \mathsf M_{X+Y}(z) =&~ \mathsf E(\mathsf e^{z(X+Y)}) \\[1ex] \mathop{=}^{\small X\perp Y}&~ \mathsf E(\mathsf e^{zX})~\mathsf E(\mathsf e^{zY}) \\[1ex]=&~ \mathsf M_X(z)~\mathsf M_Y(z)\end{align}$$
Further, the moment generating function of a normally distributed random variable, $X\sim\mathcal N(\mu_{\lower{0.5ex}X}, \sigma_{\lower{0.5ex}X}^2)$, is known: $$\mathsf M_X(z) = \mathsf e^{(z\mu_{\lower{0.25ex}X}+z^2\sigma_{\lower{0.25ex}X}^2/2)}$$

Now $X(s)$ and $X(t)$ are not independent, however, $X(s)$ is independent of $[X(t)-X(s)]$ due to property (2) "$\{X(t): t\geqslant 0\}$ has independent stationary increments".   They are values from disjoint intervals.
Also $X(s)\sim \mathcal N(s\mu, s\sigma^2)$ and $X(t)\sim \mathcal N(t\mu, t\sigma^2)$ due to property (3).  
The moment generating function of a sum of independent random variables is the product of their moment generating functions, so:
$$\mathsf M_{X(t)}(z) = \mathsf M_{[X(t)-X(s)]}(z)~\mathsf M_{X(s)}(z)$$
Find and identify $\mathsf M_{[X(t)-X(s)]}(z)$ .
