Show that a Wiener Process $X(t)$ is a normal process? Show that a Wiener Process $X(t)$ is a normal process?
Consider an arbitrary linear combination:
$$\sum \limits_{i=1}^{n} a_iX(t_i) = a_1 X(t_1) + a_2 X(t_2) + \cdots + a_n X(t_n)\tag{1}$$
where $0 < t_1 < \cdots < t_n$ and $a_i$ are real constants, and $X(0)=0$.
Now we write:
$$\begin{aligned}\sum \limits_{i=1}^{n} a_i~X(t_i)&= (a_1 + \cdots + a_n)~[X(t_1) - X(0)] \\&+ (a_2 + \cdots + a_n)~[X(t_2) -X(t_1)] \\&+ \cdots \\&+ (a_{n-1} + a_{n})~[X(t_{n-1}) - (t_{n-2})]\\& + a_n~[X(t_n)-X(t_{n-1})]\end{aligned}\tag{2}$$
How did they factor (1) to obtain (2)?
 A: $X(0)$ is zero so $(a_1+\cdots+a_n)X(0)=0$. The term involving $X(t_1)$ is
$$X(t_1)[(a_1+\cdots+a_n)-(a_2+\cdots+a_n)]=a_1X(t_1)$$
The term involving $X(t_2)$ is
$$X(t_2)[(a_2+\cdots+a_n)-(a_3+\cdots+a_n)]=a_2X(t_2)$$
iterating the term involving $X(t_n)$ for $n>0$ is $a_nX(t_n)$ so these two expressions are equal.
A: You have written something as a linear combination of $X(0), X(t_1), \ldots,X(t_n).$
You want to write it as a linear combination of $$X(t_1)-X(0), \quad X(t_2)-X(t_1), \quad\ldots, \quad X(t_n) - X(t_{n-1})$$ because those are probabilitistically independent.
\begin{align}
& b_1\big( X(t_1)-X(0)\big) + b_2\big( X(t_2) - X( t_1) \big) + \cdots + b_{n-1}\big(X(t_n) - X(t_{n-1}) \big) \\[12pt]
= {} & (-b_1)X(0) + (b_1-b_2) X(t_1) + ( b_2-b_3)X(t_2) + \cdots \\
& \qquad\qquad\qquad\qquad {} \cdots + (b_{n-2}-b_{n-1})X(t_{n-1}) + b_{n-1} X(t_n) \\[12pt]
= {} & a_1X(t_1) + \cdots + a_n X(t_n)
\end{align}
So
\begin{align}
b_1-b_2 & = a_1 \\
b_2-b_3 & = a_2 \\
& \,\,\,\vdots \\
b_{n-2} - b_{n-1} & = a_{n-1} \\
b_{n-1} & = a_n
\end{align}
The very last line gives you $b_{n-1}.$ Then the line before that tells you $b_{n-2} = a_{n-1} + a_n.$
Then the line before that tells you $b_{n-3} = a_{n-2} + a_{n-1} + a_n$
And so on.
A: Show the wiener Process X(t) is a Gaussian process. Note, that $X(0) = 0$
Starting with an arbitrary linear combination random variables from the process:
$$\sum \limits_{i=1}^{n} a_iX(t_i) = a_1 X(t_1) + a_2 X(t_2) + \cdots + a_n X(t_n)\tag{1}$$

$$a_0 X(0) = 0 -\Big(\sum \limits_{i=1}^{n} a_i\Big) X(0)\tag{0}$$
$$a_1 X(t_1) = \Big(\sum \limits_{i=1}^{n} a_i\Big) X(t_1) -  \Big( \sum \limits_{i=2}^{n}a_i \Big) X(t_1)\tag{1}$$
$$a_2 X(t_2) = \Big(\sum \limits_{i=2}^{n} a_i\Big) X(t_2) -  \Big( \sum \limits_{i=3}^{n-1}a_i \Big) X(t_2)\tag{2}$$
$$\cdots$$
$$a_{n-1} X(t_{n-1}) = \Big(\sum \limits_{i=n-1}^{n} a_i\Big) X(t_{n-1}) -  a_{n}  X(t_{n-1})\tag{n}$$
$$a_{n} X(t_{n}) = a_n X(t_n)$$
Adding equations $(0), (1), (2),\cdots, (n)$ together we get:
$$\begin{aligned}\sum \limits_{i=1}^{n} a_iX(t_i) &=\Big(\sum \limits_{i=1}^{n} a_i\Big) X(t_1) -\Big(\sum \limits_{i=1}^{n} a_i\Big) X(0) \\ &+ \Big(\sum \limits_{i=2}^{n} a_i\Big) X(t_2) -  \Big( \sum \limits_{i=2}^{n}a_i \Big) X(t_1) \\ &+\cdots \\&+ a_n X(t_n) - a_{n}  X(t_{n-1})\end{aligned}$$
Cleaning up:
$$\boxed{\begin{aligned}\sum \limits_{i=1}^{n} a_iX(t_i) &= \Big(\sum \limits_{i=1}^{n} a_i\Big)~ [X(t_1) -X(0)] \\ &+ \Big(\sum \limits_{i=2}^{n} a_i\Big)~[ X(t_2) - X(t_1)] \\ &+\cdots \\ &+ \Big(\sum \limits_{i=n-1}^{n} a_i\Big)~[ X(t_{n-1}) - X(t_{n-2})] \\&+ a_n~ [X(t_n) - X(t_{n-1})]\end{aligned}}$$
According to Definition of a Wiener process:

*

*X(t) has stationary increaments

*The increment X(t) - X(s), where t>s, is normally (aka Gaussian) distributed.

*E[X(t)] = 0

*X(0) = 0

The RHS of boxed equation is a linear combination of Gaussian random variables according to Wiener definition.  Since the increments are Gaussian and multiplied by a constant.
In a previously problem, we proved that any linear combination of Gaussian random variables is also Gaussian, this includes Gaussian wiener increments.   Thus, LHS of equation is also Gaussian since RHS is Gaussin. $\sum \limits_{i=1}^{n} a_iX(t_i)$ is Gaussian.
This means that $X(t_i)$ is Gaussian, and $X(t_i)$ is the Wiener process.
