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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

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 ...


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 ...


7

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 ...


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 ...


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

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 ...


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 ...


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 ...


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.


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 ...


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 ...


2

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 ...


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

First I must appreciate the @Richard's help that cause to solved this question. The Dothan model with this dynamic " $dr_t=ar_tdt+\sigma r_tdW_t$ " is easily integrated $r(t)=r(s)exp ( \mu (t-s)+\sigma (W_t-W_s))$ Where $\mu=a-\frac{\sigma^2}{2}$ so We have $E^Q[B_t]=E^Q[exp(\int_0^t r(u)du)]\approx E^Q[e^{e^y}]$ Where $y$ is Gaussian distributed so ...


2

I think that you are a bit confused: the support of the Black-Scholes model is $(0,+\infty)$, that is to say the underlying asset price is non-negative, like a stock. The Vasicek model has an OU process whose support is $(-\infty,+\infty)$, that is to say the underlying can be negative. Therefore all equivalent measures (of which the martingale is one) ...


1

As in the vonjd's answer martingale property makes some sense only if considered with the risk premium and risk-free rate ("stochastic discount factor" they say). Discounted stock price process is assumed to be a martingale in many studies. The root of H02's "evil" is Fama's Efficient Market Hypothesis Survey. It is the most clear and comprehensive survey ...


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

For a stochastic process $\left(X_{t}\right)$ to be adopted to a filtration $\left(\mathcal{F}_{t}\right)_{t\in T}$ the random variable $X_{t}$ must be $\mathcal{F}_{t}$-measurable for each $t\in T$. A stochastic process is a collection of random variables $X_{t}$, indexed by some set $T$. Each random variable is a mapping from a probability space into a ...


1

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.


1

See http://kalx.net/ftapd.pdf for a rigorous proof of the FTAP accessible at the masters level.



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