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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.

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  • $\begingroup$ Would you be trying to check condition (a) for process (a), condition (b) for process (b) and condition (c) for process (c)? If you are, note that this is utterly absurd. You are aked to show that process (a) satisfies conditions (a)-(b)-(c) (hence is a Wiener process) and that processes (b) and (c) do not. $\endgroup$
    – Did
    Dec 2, 2016 at 11:29
  • $\begingroup$ I will edit the question. Using the hint, I was thinking to show (a),(b),(c) for (1). And only show the (2),(3) doesn't satisfies the time increment condition. $\endgroup$
    – okie
    Dec 3, 2016 at 20:57

1 Answer 1

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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 $cov(X_{2t}-X_{t},X_{t})=2t\sigma$

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  • $\begingroup$ Please see my comment on main for what could actually be the OP's misconception. $\endgroup$
    – Did
    Dec 2, 2016 at 11:31

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