From what I know so far, the process $W = \{W(t): t\geq 0\}$ is a Wiener process provided.
(a) $W$ is time-homogeneous, i.e., $\{W(t+h)-W(h)\}$ has the same distribution as $W(t)$.
(b) $W$ has independent increments, i.e., the increments $W(t_i) – W(t_i^*)$, $i \geq 1$ are independent whenever the intervals $(t_i,t_i^*]$ are disjoint. I found another definition that tell us that for any $t_0 \leq t_1 \leq \cdots \leq t_n$, the increments
$$W(t_1)-W(t_0), \ldots, W(t_n)-W(t_{n-1}),$$
are independent.
(c) $W(t)$ has normal distribution mean zero and variance $\sigma^2 t$ for some constant $\sigma^2$.
Are the following Wiener Process:
$$(1) -W(t), \quad (2)\sqrt{t}W(1), \quad (3)W(2t)-W(t).$$
I was given a hint that only (a) defines a Wiener process, but (b) and (c) doesn't have independent normally distributed increments.
(1) Set $K = \{-W(t): t \geq 0\}$, then we have that $K$ is time-homogeneous. For instance,
$$K(t+h) – k(h) = -W(t+h) – (-W(h)) = – (W(t+h) – W(h)) = – W(t) = k(t),$$
so they have the same distribution. Now, $E(K(t)) = -E(W(t)) = 0$ and
$$Var(K(t)) = Var(-W(t)) = Var(W(t)) = \sigma^2t,$$
as required. We proceed to show that $K$ has independent increment. We have that since $W(t)$ is a Wiener, for any disjoint interval $(t_i, t_i^*]$ ($i \geq 1$), the increments $W(t_i) – W(t_i^*)$ are independent. Thus,
$$K(t_i) – K(t_i^*) = -W(t_i) – (-W(t_i^*)) = -(W(t_i) – W(t_i^*)),$$
so the increments $K(t_i) – K(t_i^*)$ are independent as required.
Do I need to show that $K$ is normally distributed in order for $K$ to be a Weiner or this three conditions are sufficient? We have
$$F_K(x) = P(-W(t) \leq x) = P(W(t) \geq -x) = 1 – F_W(-x),$$
where $F_W$ represents the distribution function of $W$.
Now, for part(2) and (3) I am not understanding how to show that the increments are not independent.
(2) the only argument I can think is the following: let $0 \leq t_1 < t_2 \leq t_3 < t_4$, then for $\mathcal{O} = \{\sqrt{t}W(1): t \geq 0\},$ we have
$$\mathcal{O}(t_2) – \mathcal{O}(t_1) = W(1)(\sqrt{t_2} – \sqrt{t_1}),$$
$$\mathcal{O}(t_4) – \mathcal{O}(t_3) = W(1)(\sqrt{t_4} – \sqrt{t_3}),$$
and it follows that
$$[\mathcal{O}(t_4) – \mathcal{O}(t_3)](\sqrt{t_2} – \sqrt{t_1})= W(1)[(\sqrt{t_2} – \sqrt{t_1})](\sqrt{t_4} – \sqrt{t_3}),$$
which gives
$$[\mathcal{O}(t_4) – \mathcal{O}(t_3)](\sqrt{t_2} – \sqrt{t_1}) = [\mathcal{O}(t_2) – \mathcal{O}(t_1)](\sqrt{t_4} – \sqrt{t_3}).$$
Therefore, $\mathcal{O} = \{\sqrt{t}W(1): t \geq 0\}$ is not a Wiener.
(3)Set $X = \{W(2t) – W(t): t \geq 0\}$ and consider $0 \leq t_1 < t_2 \leq t_3 < t_4$. Then,
$$X(t_2) – X(t_2) = W(2t_2) – W(t_2) – (W(2t_1) – W(t_1)) = W(2t_2)-W(2t_1) – (W(t_2) -W(t_1)),$$
$$X(t_4) – X(t_3) = W(2t_4) – W(t_4) – (W(2t_3) – W(t_3)) = W(2t_4)-W(2t_3) – (W(t_4) -W(t_3)).$$
What I know is that $W$ is a Wiener, which means that $W(t_4) -W(t_3)$ and $W(t_2) -W(t_1)$ are independent. I am not sure how to proceed from the previous argument.
Best Answer
Using your defintion of Wiener process :
b)The independence of the increments implies that $cov(W_t,W_s)=\sigma s$ if $s<t$
However, $cov(\sqrt{t}W_1,\sqrt{s}W_1)=\sqrt{ts}\sigma$
c) same thing, let $X_t=W(2t)-W(t)$, if it was a Wiener process, $cov(X_{2t}-X_{t},X_{t})=0$
We have $X_{2t}-X_{t}=W_{4t}-W_{t}$, and finally $cov(X_{2t}-X_{t},X_{t})=2t\sigma$