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18

From what I remember, there is no real relation between Markov and Martingale, and my intuition was confirmed by this post. Basically, it says that you can say neither of the following: If A is Markov, then A is a martingale. If A is a martingale, then A is Markov. further down the post, you can find two counter examples: $dX_t = a dt + \sigma dW_t$ is ...


15

I will defer to others answering the parts of your question concerning the relationship between Markov processes and martingales (@SRKX has already given a good explanation of the relationship) and concerning statistical testing. Broadly, however, it is not possible to "prove" either assumption, but only to fail to reject them. A Non-Random Walk Down Wall ...


11

I think to understand the martingale/local martingale distinction, it helps to bring in a third class of processes, the uniformly integrable martingale. I would argue that the local martingale and the non-uniformly integrable (true) martingale are actually fairly similar. The key property that a uniformly integrable martingale has is the so-called closure ...


11

We assume that, under the probability measure $Q$, \begin{align*} dS_t &= S_t\big(r_t dt + \sigma dW_s(t)\big),\\ dr_t &= -k\, r_t dt + \alpha dW_r(t),\tag{1} \end{align*} where $d\langle W_s(t), W_r(t)\rangle_t = \rho dt$. From $(1)$, for $s\ge t$, \begin{align*} r_s = e^{-k(s-t)}r_t + \alpha\int_t^s e^{-k(s-u)} dW_r(u). \end{align*} Then, for $T\ge ...


10

In the integral $$\int_0^t S_u dW^{*}_u \, ,$$ $dW^{*}_u \equiv W^{*}_{u+du} - W^{*}_u$ is independent from the integrand $S_u$. So, $\mathbb{E}\left[ \int_0^t S_u dW^{*}_u\middle\vert \mathcal{F}_0\right] = \int_0^t \mathbb{E}\left[S_u \middle\vert \mathcal{F}_0\right]\mathbb{E}\left[dW^{*}_u\middle\vert \mathcal{F}_0\right] = 0$, since $\mathbb{E}\...


9

In general, if you have a process that you can write under the form $F(B_t,t)$ where $F$ is $\mathcal{C}^{2,1}$ then Itô's lemma gives you the drift term and diffusion term of $dF$. Then if the resulting SDE has a null drift (that's where Black Scholes PDE comes from), and you get a only local martingale. For it to be a proper martingale you can look at ...


9

Roughly speaking, we can express the difference between a Markov process and a martingale as follows: A Markov process is one for which conditioning its future value on its history is the same as conditioning its future value on its present value, so that $E(h(X_t)\,|\,X_u,\,u\leq s)=E(h(X_t)\,|\,X_s)$, for any appropriate function $h$; A martingale is a ...


9

Whether it's called volatility pumping, rebalancing premium, or Shannon's Demon it would just be a form of replicating a short gamma option strategy (eg. selling straddles). Intuitively, you are systematically selling at higher levels and buying at lower levels. The payoff for continuously rebalancing an equity/cash portfolio without friction when the ...


6

Measure theory helps us overcome some of the drawbacks of constructing measures (measure of probability when ranged at $[0,1]$). Classic probability theory is effective for probability models whose sample space $\Omega$ is a finite or countable set ($P: 2^{\Omega} \to [0,1]$). But for uncountable sets, such as $\mathbb{R}$ (which is uncountable) the ...


5

The martingale representation theorem says that for any martingale $M$, there exists a unique stochastic process $\nu_t$ such that \begin{align*} M_t = \mathbb{E}(M_0) + \int_0^t\nu_sdW_s. \end{align*} See, for example, this book. Since \begin{align*} C_t &= \Bbb{E} \left( \int_t^T f(r_s, \theta_s, W_s) ds \mid \mathcal{F}_t \right)\\ &= -\int_0^t ...


5

The Snell envelope is the smallest super-martingale that is greater than $X$. Since $\tau \le N$, it is obvious that $A_N^{\tau} = A_{N\wedge \tau} = A_{\tau}$. For part (b), note that, from the Doob decomposition, $M$ is a martingale, $A$ is increasing, $M_0=Z_0$, and $A_0=0$. If $Z^{\tau}= \{Z_n^{\tau}\}_{n=1}^N$ is also a martingale, then \begin{align*}...


4

Suppose that there are multiple martingale measures $Q_1$ and $Q_2$ that attain the minimal variance. Then the convex combination $Q_* := \frac{1}{2}Q_1 + \frac{1}{2}Q_2$ is also a martingale measure. Due to the strict convexity of $f(x) = x^2$, it can be shown that $$ E_P \left[\frac{dQ_*}{dP}^2 \right] < \frac{1}{2} E_P \left[ \frac{dQ_1}{dP}^2 \right]...


4

Let's consider a random process $X$. If $X$ is an adapted process, then we know, without any uncertainty, what its value is at the present time. This idea is formalized with measure theory. For $X$ to be a martingale, it needs to have the following property: at any given time, our best estimate of the value at some point in the future (i.e. forecast), is ...


4

Most of the time, when you have a simple SDE without a drift, it's a martingale because the Wiener process itself is a martingale. In your example, you have a constant with the Wiener process, therefore the whole process must also be a martingale because the expectation is clearly X(t). However, we can't conclude a driftless SDE is always a martingale. ...


4

Based on Ito's isometry, \begin{align*} E_t (r^2_{t+1}) &= E_t \bigg(\int_t^{t+1} \sigma_s dW_s \int_t^{t+1} \sigma_s dW_s\bigg)\\ &= E_t \bigg(\int_t^{t+1} \sigma_{\tau}^2 \,d\tau\bigg) \\ &= E_t\bigg(\int_0^1 \sigma_{\tau+t}^2 \,d\tau\bigg) \\ &=\int_0^1 E_t\big(\sigma_{\tau+t}^2\big) \,d\tau. \end{align*} The identity \begin{align*} E_t (r^...


4

Generally Kurtosis measures the degree to which a distribution is more or less peaked than a normal distribution. Positive kurtosis indicates a relatively peaked distribution. Negative kurtosis indicates a relatively flat distribution. In time series we can encounter high kurtosis which is caused by "fat tails" (higher frequencies of outcomes) at the ...


4

Perhaps an answer coming from a different angle and giving you some perspective: The typical approach taken by statistics is top-down: Just looking at the data and finding patterns and stylized facts (like excess volatility, volatility clustering, fat tails, no autocorrelation in returns but significant autocorrelation in absolute returns etc.) The problem ...


4

Under a Black-Scholes framework, the dynamics of the stock price under the risk-neutral measure $\mathbb{Q}$ are given by ... $$ S_t = r S_tdt +\sigma S_tdW^{\mathbb{Q}}_t $$ ... and those of the risk-free bond by: $$ \begin{align} dB_t = rB_tdt \end{align} $$ Let us define the derivative value as $V(t,S_t)$, which only depends on the time $t$ and the ...


4

We decree that $D_t$ has a certain process which makes it a martingale. In particular, we let $$ D_t = \mathbb{E} ( D_T \, | \, \mathcal{F}_t) $$ This is trivially a martingale by the tower law. Since the discounted stock price and discounted bond price are martingales, we have made everything a martingale and that ensures that there is no arbitrage in the ...


4

We assume a Black-Scholes world except the dynamics of the stock price, namely: No arbitrage opportunities. No dividend payments from the stock. Existence of a riskless asset yielding the risk free rate $-$ which here we assume non-constant, $(r_t)_{t \geq 0}$. Possibility to borrow and lend infinitely at the risk-free rate. Possibility to buy and sell ...


4

You may find the following paper worthwhile. It addresses most of the above points (and many more) in a systematic way: Dubikovsky, Vladislav and Susinno, Gabriele, Demystifying Rebalancing Premium and Extending Portfolio Theory in the Process (May 20, 2015). Available at SSRN: https://ssrn.com/abstract=2927791 or http://dx.doi.org/10.2139/ssrn.2927791 ...


4

Assume that: $$ S_0^1(1+r)\leq a,b $$ Arbitrage for a portfolio $V_t$ is defined as: $$V_0\leq0, \quad P(V_1\geq0)=1, \quad P(V_1>0)>0$$ Consider borrowing at rate $r$ to buy the risky asset such that $V_0=0$. Then, assuming $a\not= b$: $$\begin{align} \min_{\omega}V_1(\omega)=a-S_0^1(1+r)\geq 0 \\ \max_{\omega}V_1(\omega)=b-S_0^1(1+r)> 0 \end{...


3

For Itô Processes $dX(t) = \mu(t) \mathrm{d}t + \sigma(t) \mathrm{d}W(t)$ you have the result that (under appropriate assumptions which ensure that the local martingale is a martingale, e.g. $E( (\int \sigma(t)^2 \mathrm{d}t )^{1/2} ) < \infty$, etc.): $X$ is a martingale $\Leftrightarrow$ $\mu(t) = 0$. So in order to check if a process $X$ is a ...


3

You have been given good answers above. Basically, a stochastic process ${X_t}$ is a Markov process if $P(\{X_{t} \leq x\} | \mathcal{F}_{s}) = P(\{X_{t} \leq x\} | X_{s})$, for $s \leq t$. Here $\mathcal{F}_{s}$ is a $\sigma$-algebra, a special collection of subsets of the underlying sample space $\Omega$, containing all information about the process $\{X_t\...


3

Martingale and Markov process are both stochastic processes where the sequences of random variables are not entirely independent, and their differences are: In martingale, the expectation of the next value IS the present value, so this property is sometimes called 'fair game'. In Markov process, the expectation of the next value only DEPENDS ON the present ...


3

I aim to give a careful mathematical treatment to this answer, whilst following the fantastic book "Basic Stochastic Processes" by Brzezniak and Zastawniak. The reason I am putting this answer on is twofold: first, to compliment @ William S. Wong's answer by adding greater mathematical intricacy for other users of the website, and secondly to confirm that ...


3

By definition of the $T$-forward measure $P_T$, the process $\Big\{\frac{P(t,S)}{P(t,T)} \mid t\geq 0\Big\}$ is a martingale under the measure $P_T$, without assuming any specific models of the short rate $r_t$. That is, this martingale property is model independent. However, as a good exercise, you can also do the following: Given the CIR interest rate ...


3

I think there are a few conflating ideas here. With respect to the sum of logs idea, I think you're thinking about infinitely divisible distributions (https://en.wikipedia.org/wiki/Infinite_divisibility_(probability)). These ideas are indeed used to build more complicated models (i.e. Levy processes) for asset returns. With regards to the Efficient Market ...


3

At what scale do you see kurtosis? Daily data? Single stocks or indices? Let us not look at single stock data, because you always find crazy stocks whose price process breaks all rules. Talking about daily data of indices: they could be thought of the sum of hourly returns or other returns of high frequency (minute returns, milliseconds ...). What are the ...


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