11

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 ...


7

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 ...


6

They are different concepts, and the relation between them can be described as a conditional: "if EMH holds (all available information about future price movements is already priced into the market), then future price movements will follow a purely random walk as new and unpredictable information emerges"


6

The way I understand it is: In equation 2 $x_{j, t + 1}$ is defined as the change in of $p_j$ over the period $t$ to $t + 1$. The formula says that the expectation of the change is zero which is the same as saying that the expectation of the original variable at $t+1$ is equal to its current value.


5

Historically the RWT (Random Walk Theory) came first, as empirical observations by for example M.F.M. Osborne (1959) and others in the 1960s. The EMH came about as a result of theoretical work by Samuelson in 1965 ("Proof that properly discounted prices...") and E.Fama (1969) as a general empirical/theoretical hypothesis that guided the field for many ...


5

I think the main difference even in this little example is the gain-loss asymmetry which is a known stylized fact: When you look at the big bump both time series posses your artificial one is perfectly symmetric whereas the real one takes longer for going up and then crashes in a relatively shorter time frame. This is a known phenomenon in real financial ...


5

If you allow for exchanges between the people in the queue, then there is no problem to be solved. So the implicit assumption must be that people in the queue only deal with the counter.


4

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 ...


4

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 ...


4

I don't think I was clear in my comment so I'm putting it in an answer to have more space. The variance of a brownian motion, z, is $t$. (i.e: $E(z^{2}) = t$ ). Notice that $R_{i}$ really equals $\sqrt{\frac{t}{n}} \times \epsilon$ where $\epsilon \sim N(0,1)$. I think they leave the $\epsilon$ out because the variance is 1 but showing the consistency is ...


4

So there are several issues with your posting that you will need to resolve. The first one is your concept of randomness and distinguishing between a random event and a non-random event. To understand the problem, I think I should tell you a story. You go home for a family reunion and see a tree you used to climb as a small child. You see the branch you ...


4

$X_j$ can be either 1 or -1 with 50% probability each. So this step is just applying the expectation to both possible cases. See definition of the Expectation... \begin{align} {\mathbb E}\bigl[ X \bigr] = \sum_i i \cdot p(x = i) \end{align} It's the sum over all possibilities of the probability of getting that value (both ${\frac 1 2}$ in your case) ...


3

This looks to me like a range accrual. Let $t_1, \ldots, t_n$, where $0 < t_1 < \cdots < t_n$ be business days that are being considered. We compute \begin{align*} E\left(\sum_{i=1}^n \pmb{1}_{b_1 < S_{t_i} < b_2} \right) &=\sum_{i=1}^n E\left(\pmb{1}_{b_1 < S_{t_i} < b_2} \right)\\ &=\sum_{i=1}^n \left[E\big(\pmb{1}_{S_{t_i} >...


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.


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

In the Ljung-Box test, the null hypothesis is: $H_0$: The data are independently distributed So, your p-values of 0 indeed indicate that you should reject the null hypothesis, but it means that your data is not independently distributed, and in particular that there is some significant autocorrelation in the process. This is obviously the case, because ...


3

It makes no difference. Starting with a capital of 1, let $X_i$ be the multiplying factor for the $i$th day, so $X_i\in\{1+r,1-r\}$ with each possibility having probability 1/2. The expected capital after one day is $$\mathbb E(X_1)=\frac12((1+r)+(1-r))=1.$$ After $n$ days, your capital is $X_1X_2\cdots X_n$, and $$\mathbb E(X_1\cdots X_n)=\mathbb E(X_1)\...


3

For the two-dimensional case, the Cholesky decomposition of the covariance matrix \begin{equation} \Sigma = \left( \begin{array}{c c} \sigma_1^2 & \rho \sigma_1 \sigma_2\\ \rho \sigma_1 \sigma_2 & \sigma_2^2 \end{array} \right) \end{equation} is given by \begin{equation} B = \left( \begin{array}{c c} \sigma_1 & 0\\ \rho \sigma_2 & \sigma_2 ...


3

The correlation matrix refers to the correlations between the asset returns. In fact, it can be seen as follows. Each asset follows a geometric Brownian motion, i.e., $$ \frac{{\rm d}S_t^i}{S_t^i}=\mu_i{\rm d}t+\sigma_i{\rm d}W_t^i, $$ where the correlation between $W_t^i$ and $W_t^j$ is supposed to be $$ \text{Corr}\left(W_t^i,W_t^j\right)=\rho_{ij}. $$ ...


3

I guess the concept you're looking for are martingales. These are stochastic processes which remain on their current level (in expectation!). Ignoring some technical conditions, a stochastic process $(X_t)$ is called a martingale if for all time points $t\geq s$, $$\mathbb{E}[X_t|\mathcal{F}_s]=X_s.$$ Here, $(\mathcal{F}_s)$ refers to a filtration, the ...


3

Echoing some of the comments to the OP above, the only real difference between random walks and Brownian motions is a question of time frequency. IE a Brownian motion is just an aggregation of a (binary) random walk with higher frequency. Given both will always be at best an approximation of reality, asking for which is "better" becomes a bit of a ...


2

The answer can be found here under 1.3) Random Walk Hitting Probabilities (when events have equal probability of $\frac{1}{2}$ each). \begin{equation} p(a) = \frac{b}{a+b} \end{equation} $p(a)$ would be the probability of take-profit hit first. To look at probability of stop-loss being hit first, just take 1 minus the above, resulting with $a$ on the top (...


2

There are several reasons: Expected payoff = 0 I don't know why you selected 0.5 ATR and 2 ATR away from the market but let's go with it for a while. This means that you want to gain 2x while risking only 0.5x. For now let's assume that FX log-returns are normal. To bring it to a higher level, we can use a piece of Black Scholes formula, namely the ...


2

For both time-series, just plot the log returns. You will see that one is not a Random-Walk .. the S&P500 since you will get values that far beyond the normal distribution. Just watch this video by Benoit Mandelbrot (starting at 11min:54sec). Looking at both graphs, your eyes can fool you making you believe that both are generated by Random Walks...


2

The first the solution to: $$dS_t = S_t\left[\mu dt +\sigma dW_t\right]$$ The second is the solution to: $$ dS_t = S_t\left[\left(\mu -\frac{\sigma^2}{2}\right)dt + \sigma dW_t\right]$$ The difference is that the first one is a martingale when $\mu$ is equal to zero while the second one is not: $$ \mathbb{E}[S_0 exp(\sigma W_t)]= S_0exp(-\frac{\sigma^2}{2}t)...


2

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.


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 ...


2

Firstly, this is very normal fare for Stochastics which studies properties of such processes. So if you're not finding anything, look up any stochastics textbook. One thing that is not clear from your explanation is what $X[N]$ is when Slowmo reached the boundary early. Does $X$ continue to evolve, or does it stop? It makes a difference because if he can ...


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