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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 $\Delta ... 15 From what I remember, there is no real relation between Markov and Martingale, and my intuition was confirmed by this post. Basically, it says that you can say neither of the following: If A is Markov, then A is a martingale. If A is a martingale, then A is Markov. further down the post, you can find two counter examples:$dX_t = a dt + \sigma dW_t$is ... 14 A martingale is a random process$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$. A martingale is a random walk, but not every random walk is a martingale. A Brownian random walk is a martingale if it does not have drift. Also, a ... 12 I will defer to others answering the parts of your question concerning the relationship between Markov processes and martingales (@SRKX has already given a good explanation of the relationship) and concerning statistical testing. Broadly, however, it is not possible to "prove" either assumption, but only to fail to reject them. A Non-Random Walk Down Wall ... 6 Roughly speaking, we can express the difference between a Markov process and a martingale as follows: A Markov process is one for which conditioning its future value on its history is the same as conditioning its future value on its present value, so that$E(h(X_t)\,|\,X_u,\,u\leq s)=E(h(X_t)\,|\,X_s)$, for any appropriate function$h$; A martingale is a ... 6 Often one will find the argument that a random walk of price changes would be a proof of the efficient market hypothesis, but this is (IMO) a logical fallacy: Only because the EMH does imply random walks in the price changes, the finding of random walks does not imply automagically that the EMH is true. 5 A martingale can be viewed as a fair game (a game in which there is no arbitrage strategy) A (centered) random walk is a martingale (think of it as the total Gain of the fair game) If EFH is in order, then you can think that all information is in the current price, I think this more comparable to Markov Property than to Martingale property. Hope that ... 5 In general, if you have a process that you can write under the form$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$. Then if the resulting SDE has a null drift (that's where Black Scholes PDE comes from), and you get a only local martingale. For it to be a proper martingale you can look at ... 4 Suppose that there are multiple martingale measures$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 ... 2 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 ... 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 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 ... 1 Perhaps not the most encouraging answer, but: I would think that it is contingent upon the specific implementation, magnitude, regularity, and transiency of arbitrage available as well as the volatility estimate time-scale. In a very simple case, the existence of arbitrage opportunities would likely result in larger fraction of informed traders (relative to ... 1 If you look at it from a mathematical point of view - presence of arbitrage should not matter for volatility estimates. Absence of arbitrage can be associated with the existence of an equivalent martingale measure for the bank account numeraire. (first fundamental theorem of asset pricing) Let's assume the real world process is something like ... 1 I strongly recommend not assesing risk using the risk neutral measure. Doesn't this already sound like a contradiction (risk and risk-neutral)? The risk neutral measure is there to derive prices (for derivatives e.g.) that fit to the prices of related contracts and traded assets. With "fit" I mean not allowing for arbitrage. For example if I calculate the ... 1 In reality, you needn't bring exotics into consideration to think about this issue. Consider the case of a shop that has fundamental analysts but also trades options on those equities. The fact that the fundamental analysts trade stocks means they think those prices are somehow "wrong". So of course it seems from their point of view that the options ... 1 In the equilibrium models you can assume that there exists so called Alpha, i.e. an opportunity that can be exploited. Most of the buy side models (i.e. asset allocation, portfolio construction) are based on this idea. As a theoretical model, you can consider CAPM with heterogeneous beliefs: Hedge funds claim to generate the “Alpha”, i.e., excess ... 1 Let (\Omega,\mathcal{F},\mathbb{F},\mathbb{\mu}) be a filtered probability space. Market efficiency implies that the stock price process is Markov with \mathbb{E}[f(X_t)|\mathbb{F}_s] = g(X_s) for 0 \leq s \leq t where f and g are Borel measurable functions. It additionally implies that the discounted stock price process is a martingale w.r.t. ... 1 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 ...