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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

Using $q = 1-p$ we can work out the root as: $$\sqrt{1-4pq} = \sqrt{1-4p(1-p)} = \sqrt{1-4p+4p^2} = \sqrt{(1-2p)^2}$$ Taking the positive root reduces this to $(1-2p)$. This gives for the fraction: $$\frac{1 + \sqrt{1-4pq}}{2p} = \frac{1 + (1-2p)}{2p} = \frac{1-p}{p}$$ This also holds inside the logarithm.


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 ...


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

If you have a vector $X = (X_1,\ldots,X_n)$ of a multivariate normal distribution with covariance matrix $\Sigma$ and $F_i$ is the marginal cumulative distribution function of $X_i$ then $F_i(X_i)$ is uniformly distributed. So what you can do: generate uniforms (e.g. Sobol or Halton) transform to uncorrelated Gaussians transform these Gaussians to ...


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} ...


1

It's true that in general if $\sum_i f_i g_i = \sum_i f_i h_i$, we do not automatically have $g_i = h_i$. But this sum is special, because all $f_i$ are monomials (i.e. of the form $\alpha^n$). This makes the sum a power series (of the form $\sum_{n=0}^\infty \alpha^n C_n$), and these series have a lot of nice properties such as continuity and ...



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