# Tag Info

## Hot answers tagged random-walk

7

I will assume a white noise is a process $(\varepsilon_t)$ with zero mean, no autocorrelation and constant variance $\sigma^2 > 0$ while a random walk is a process $(x_t)$ defined by $$x_{t+1} = x_t + \varepsilon_{t+1}$$ where $\varepsilon$ is a white noise. 1) No since $Var(x_{t+1}) = Var(x_t) + Var(\varepsilon_{t+1})$ is stricly increasing while ...

3

Not sure about the correctness of the first approach, but second approach uses $1 /\sqrt k$ to scale the variance of the total sum by $k$. So the difference of two processes (say $W_t$ and $W_{t+\Delta t}$) generated by the random walk would have a variation of $\Delta t$, which satisfies one of conditions needed to get a Wiener's process.

2

Regarding the relationship between white noise and a random walk, I would put it this way: a random walk is integrated white noise. [And vice versa we get a white noise when we differentiate/difference a random walk]. Or to put it in quant finance terms: white noise is like the daily changes in the S&P in points, a random walk is the S&P daily level ...

2

As I mentioned above, I am not sure what the variable $r$ is. If we ignore that, or assume the questioner wanted to say its the risk free interest rate, then it has no effect on the number of paths. Then it is clear that after 50 steps going from \$1024 to \$2500 requires a net of 4 up movements with the given $x=y^{-1}=1.25$. Thus the number of steps ...

1

Two hints : The number of paths never going up to $3125$ when starting from $1024$ and stepping up by a multiplicative factor of $5/4$ and down by a multiplicative factor $4/5$ is the same as the number of paths starting from $0$ and and stepping up by an additive factor $+1$ and stepping down by an additive factor of $-1$ and never going up to $5$ Let ...

1

For positive integer $n$, \begin{align*} \{T=n\} &= \Big(\cap_{k=1}^{n-1} \{X_k \ne b\}\Big) \cap \{X_n = b\} \in \mathscr{F}_n. \end{align*} That is, $T$ is a stopping time.

1

$\because M_n$ is a martingale and $T \wedge k$ is bounded, by Doob's optional stopping theorem, we have $$E[M_{T \wedge k}] = E[M_0] = 0$$ $$\to E[T \wedge k] = \frac{1}{p-q} E[X_{T \wedge k}]$$ By monotone convergence theorem, we have $$E[T] = \lim_{k \to \infty} E[T \wedge k]$$ Finally, by definition of $T$, we have $$X_{T \wedge k} \le ... 1 Let b=1, p=1/3, q=2/3. It is not hard to show that in this case T is finite with probability exactly 1/2. Consequently, E[T] = \infty and your claim does not hold in general. The claim would hold if p\geq q (in which case you could, for example, address it by first showing that it holds for b=1 and proceeding inductively from there on). 1 EMH: An asset always trades at its fair value. That is, all information is continuously being priced in. RWH: The asset price is not predictable and follows a random walk. So RWH is a hypothesis which is consistent with EMH. If every piece of information is being priced in continuously, and you cannot predict what information will become available, then ... 1 First, for Ito processes and Brownian motion. Ito process is a continuous-time trajectory with random evolution, so non-smooth and very kinky - also has a fractal look: no matter how much you'd zoom in, it will look similar. Ito process consists in fact of two parts: the drift part (deterministic evolution) and the diffusion part (where all the kinkiness and ... 1 Assuming no math at all: Using an Ito process we can describe the return of a stock with two components: an average level (the "drift") plus some uncertainty (the "volatility"). This uncertainty is represented by a Brownian Motion. As written in Wikipedia, A random walk is a mathematical formalization of a path that consists of a succession of ... 1 If you go up u times and down d times, your random walk ends up at u-d at time u+d. Since there are {u+d \choose u} ways to distribute the u up moves among the u+d moves$$ P(M_{u+d} = u-d) = {u+d \choose u} \frac{1}{2^{u+d}} $$Setting n = u+d, x = u-d = 2u-n, so u = n+x/2, this is equivalent to$$ P(M_n = x) = {n \choose (n+x)/2} ...

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