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## Hot answers tagged martingale

16

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 ...

16

Samuelson suggested in 1965 that the stock prices follow a martingale (see P. Samuelson “Proof That Properly Anticipated Prices Fluctuate Randomly”). Assume there is a security with a random payoff $X_T$ at date $T$. Let $..., P_{t–1}, P_t, P_{t+1},...$ be the time series of prices of a security with this payoff. Finally, define the price change $\Delta ... 15 A martingale is a random process$X(t)$which has the following properties:$ E[X(T)|\mathcal{F}_t] = X(t) $for$T > t$and$ E[|X(T)|] < \infty $where$\mathcal{F}_t$is the filtration at time$t$. A martingale is a random walk, but not every random walk is a martingale. A Brownian random walk is a martingale if it does not have drift. Also, a ... 13 I will defer to others answering the parts of your question concerning the relationship between Markov processes and martingales (@SRKX has already given a good explanation of the relationship) and concerning statistical testing. Broadly, however, it is not possible to "prove" either assumption, but only to fail to reject them. A Non-Random Walk Down Wall ... 8 Roughly speaking, we can express the difference between a Markov process and a martingale as follows: A Markov process is one for which conditioning its future value on its history is the same as conditioning its future value on its present value, so that$E(h(X_t)\,|\,X_u,\,u\leq s)=E(h(X_t)\,|\,X_s)$, for any appropriate function$h$; A martingale is a ... 6 Often one will find the argument that a random walk of price changes would be a proof of the efficient market hypothesis, but this is (IMO) a logical fallacy: Only because the EMH does imply random walks in the price changes, the finding of random walks does not imply automagically that the EMH is true. 6 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}\...

5

A martingale can be viewed as a fair game (a game in which there is no arbitrage strategy) A (centered) random walk is a martingale (think of it as the total Gain of the fair game) If EFH is in order, then you can think that all information is in the current price, I think this more comparable to Markov Property than to Martingale property. Hope that ...

5

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 ...

5

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 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 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 ... 4 We consider the case where the Novikov condition is satisfied, that is, \begin{align*} E\left[\exp\left(\frac{1}{2}\int_0^T \theta^2_s ds \right)\right] < \infty. \end{align*} Then \{L_t \mid t \ge 0\} is a (\mathscr{F}_t, \mathbb{P})-martingale. On \mathscr{F}_T, we define the probability measure Q by \begin{align*} \frac{dQ}{dP}\big|_{\mathscr{F}... 3 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 ... 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 This is very standard financial theory. The answer to the question is given in the first chapter of Duffie's Dynamic Asset Pricing Theory and in Cochrane's Asset Pricing. The latter is more elementary, but you have to read more to get to the answer. 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 ... 3 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 ... 3 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 ... 2 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 ... 2 Similar to the answer aleady given. We can use a measure Q such that E_Q[A_n] = 0. Let's reformulate the sequence as X_0 =x and X_{n+1} = X_n + A_{n+1}. First, beause expectation is linear: E_Q[X_{n+1}|F_n] = E_Q[X_n|F_n] + E_Q[A_{n+1}|F_n]. $$Now assume that \{F_n\}_{n=0}^\infty is the filtration that represents the information of (X_n)_{n=... 2 Just to give you two examples. Note that dX_t =a \; dt + dW_t is Markov but is not a martingale. dX_t=(\int_0^t X_s ds) \; dW_t is a martingale but is not Markov. 2 A random process that is adapted to a filtration is measurable (ie X_t is F_t-measurable) but not necessarily a martingale. X_t is a martingale if E(X_t | F_s) = X_s for s < t. 2 To give you another perspective: Let us assume that the world had only one risky/noisy asset S(t) and let us further assume that at time T our process cann only have n states - namely (S_1, \dots, S_n) and that the interest rate was flat and given by r Now let's say we have a payoff funtion f(x): \mathbb{R}\to\mathbb{R}. Working under the risk ... 2 This tackles the second part of your question: In a world were interest rates are always zero (for simplicity sake), if the discount price process is a martingale, we have: E[X_T | F_t] = X_t In an arbitrage free world, every price process is a martingale in the risk-neutral measure. Having martingale price processes means that if we build a hedged ... 2 You have two processes, X_t:=\log{\frac{f}{g}} and Y_t=\frac{f}{g}. Note, I use \log for the natural logarithm. Hence we have Y_t=\exp{(X_t)}. Therefore, applying Itô:$$dY_t=\exp{(X_t)}dX_t + \frac{1}{2}\exp{(X_t)}d\langle X,X\rangle_t$$Using the dynamics of X_t, we get$$dY_t=\frac{f}{g}[-\frac{(\sigma_f-\sigma_g)^2}{2}dt+(\sigma_f-\sigma_g)...

2

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 ...

2

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 ...

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