Hot answers tagged martingales
36
Suppose, for simplicity, that the probability of winning one round of this game is $\frac{1}{2}$, and the probability of losing is also $\frac{1}{2}$. (Roulette in real life is not such a game, unfortunately.) Let $X_0$ be the initial wealth of the player, and write $X_t$ for the wealth of the player at time $t$. Assuming that the outcome of each round of ...
16
This betting strategy is very smart if you have access to infinite wealth or can go into infinite debt. In reality however, you will eventually lose all or most of your money.
Say your friend had $k$ gold at the beginning. I assume that this simple roulette has a probability of both win and loss equal to $0.5$.
First, let's see how many times you need to ...
9
The product of two independent martingales is a martingale--or rather it is or it is not, depending on the precise formulation of the hypothesis! When it is, one says that the martingales are orthogonal. This is explained, for example, by Alexander Cherny in the chapter Some Particular Problems of Martingale Theory of the Shiryaev Festschrift.
And yes, the ...
8
Here you are :
From Protter's book "Stochastic Integration and Differential Equations" Second Edition (page 73 and 74)
First :
Let $M$ be a local martingale. Then $M$ is a martingale with
$E(M_t^2) < \infty, \forall t > 0$, if and only if $E([M,M]_t) < \infty, \forall t > 0$. If $E([M,M]_t) < \infty$, then $E(M_t^2) = E([M,M]_t)$.
Second ...
7
Yes, $\bar X_n = S_n / n$ converges to $\mu$ in $\mathscr{L}_1$. It suffices to show that $\bar X_n$ is uniformly integrable since we already know $\bar X \to \mu$ in probability. This follows very quickly from the fact that the $X_i$ are trivially uniformly integrable (just fix $\epsilon > 0$ and choose $\delta > 0$ so that if $P(A) < \delta$ then ...
7
If $(S_n)$ is a simple symmetric random walk on the integers, then $S_n^3-3nS_n$ is a martingale. Applying the optional stopping theorem for martingales, we get
$$ \mathbb{E}_x(S_T^3-3TS_T)=\mathbb{E}_x(S_0^3-0)=x^3.$$ Here $-a\leq x\leq a$ is the starting point of the random walk and $T$ is the hitting time of the boundary $\{-a,a\}$.
Standard results ...
6
If $M$ is a martingale in the filtration $(F_n)$, $\|M\|$ is a sub-martingale in $(F_n)$. Assume that $\|M\|$ is a martingale in another filtration $(G_n)$, then $\mathrm E(\|M_n\|)$ is constant. Any submartingale with constant expectation is a martingale hence $\|M\|$ is a martingale in $(F_n)$.
Finally, it seems that a martingale $M$ is such that there ...
6
The answer is YES if one replaces $\gt$ by $\geqslant$ and if one assumes that $(X_t)$ has finite fourth moments. This is based on three simple properties of conditional expectation, namely:
(C) Convexity. (T) Tower property. (M) Taking out what is measurable.
(C) means that if $\varphi$ is convex and if $Z$ and $\varphi(Z)$ are integrable, ...
5
Set $S_n:=\sum_{i=1}^n X_i$ and $S_0=0$. $S$ is a martingale (wrt the natural filtration), so $S^2$ is a sub-martingale.
Using the Doob's decomposition we can write $S^2=M+A$ where M is a martingale and A is predictable (increasing) process, both null at 0.
It turns out that $A_n=\sum_{i=1}^n \sigma_i^2$.
Define the stopping time $T_\alpha$ as ...
5
The $\sigma^2t$ term is deterministic hence one can move it out of the conditional expectation. The $B(t)$ term is $(B(t)-B(s))+B(s)$. The $B(s)$ term is $\mathscr F_s$-measurable hence one can move it out of the conditional expectation. The $B(t)-B(s)$ term is independent of $\mathscr F_s$ hence one can integrate it. To summarize,
$$
\mathbb E(\mathrm ...
4
Since $(X(t))_t$ is a martingale in the filtration $(\mathcal F_t)_t$, one knows that $\mathrm E(X(u)\mid\mathcal F_s)$ is $X(u)$ is $u\leqslant s$ and $X(s)$ if $u\geqslant s$. Hence, for every $s\leqslant t$,
$$
\mathrm E(g(t)\mid\mathcal F_s)=g(s)+\int_s^tf(u)\mathrm E\left(X(u)\mid F_s\right)\mathrm du=g(s)+X(s)\cdot \int_s^tf(u)\mathrm du.
$$
The only ...
4
I think the easiest way is through strategy 2). Let $X$ be a transient irreducible Markov Chain. Consider an arbitrary $x_{0}$ and define:
$$\tau = \inf\{n \geq 0: X_{n} = x_{0}\}$$
Define
$$f(x) = P(\tau < \infty|X_{0}=x)$$
By construction, $f(x) \in [0,1]$ and $f(x_{0}) = 1$. Since $X$ is transient, there exists $y$ such that $f(y) < 1$. ...
4
A careful look at the definition of $E [X_{n+1} |X_0 , X_1 , . . . , X_n ]$ on the left hand side of the equation shows that conditional expectation is defined only almost surely. This is true for any conditional expectation $E[X | {\cal F}]$, whether conditioned on a $\sigma$-algebra or a set of random variables. This has nothing to do with martingales, per ...
4
This is because conditional expectations can only be defined up to almost sure events.
Recall that $Z=E[X|Y]$ is defined as any random variable $Z$ such that (1) $Z$ is $\sigma(Y)$-measurable, and (2) $E[Z\varphi(Y)]=E[X\varphi(Y)]$ for every bounded measurable function $\varphi$. Hence, if $Z$ fulfills (1) and (2) and $P(Z'=Z)=1$, then $Z'$ ...
4
Q1: In general it is not true that $X^d_t$ is equal to $\sum_{s\le t}\Delta X_s$. In order for this equality to make sense, you need assume something like that the sum of the jumps of $X$ is absolutely convergent (Hypothesis A), although that still does not imply that the identity holds. The term $X^d$ is the purely discontinuous part of the semimartingale ...
4
We note $M^*_t = C_\alpha^{-1/2}\sup_{0 \leq s \leq t}|M^\alpha_s|$ so that the hypothesis writes $E\left[(M^*_t)^2\right] \leq t^{1-\alpha}$.
Fix $\epsilon > 0$. Chebyshev's inequality shows that, for all $n \geq 1$,
$$
P\left(\frac{M_n^*}{n^\beta} \geq \epsilon\right) \leq \epsilon^{-2}\frac{1}{n^{2\beta-1+\alpha}}.
$$
Suppose that $\beta > 1 - ...
4
Assuming your process is one dimensional, by representation theorem there exist a Brownian motion $B$ and a (predictible) process $Y_s$ verifying $\forall t>0$, $\int_0^tY_s^2ds<+\infty$ almost surely such that :
$X_t=E[X_0]+\int_0^tY_sdB_s$ for all $t>0$
Then $[X]_t=\int_0^tY_s^2ds$ which is almost surely finite by representation theorem.
Note ...
4
Yes. Let $\alpha$ be a random variable with uniform distribution on $[0,1]$, we set
\begin{align*}
X_t=\mathbb{1}_{\alpha+\mathbb{Q}}(t).
\end{align*}
$X$ is clearly discontinuous everywhere. However, it is a martingale with respect to
the filtration defined by ${\cal F}_t=\sigma(\alpha)$ for all $t\geq 0$. Indeed, for all $0\leq s\leq t$,
\begin{align*}
...
3
Let $(B_t)_{t \geq 0}$ a Brownian motion. $(B_t)_{t \geq 0}$ is a martingale with respect to the natural filtration $$\mathcal{F}_t := \sigma(B_s; s \leq t)$$ but for $\mathcal{G}_t := \mathcal{F}_{2t}$ the process $(B_t,\mathcal{G}_t)_{t \geq 0}$ isn't a martingale since for example $$\mathbb{E}(B_t \mid \mathcal{G}_{\frac{t}{2}}) = \mathbb{E}(B_t \mid ...
3
This may not be exactly what you're looking for, but here's a proof which at least uses the specific form of the process (being the exponential local martingale of an integral of a square-integrable deterministic process with respect to a Brownian motion):
First off, in order to make things fit better with the standard framework, I'll assume that $\phi$ is ...
3
First, the question doesn't even make sense if $\alpha$ is not an integer, because $W^\alpha$ is undefined as a real valued process when W goes negative. So, in that case, I'll assume that W is started from a positive value and stopped when it hits zero.
There's two methods: Use Ito's formula for $X=W^\alpha$,
$$
dX = \alpha W^{\alpha-1}\,dW + ...
3
As Did suggested you can apply Doob's inequality. If you want to prove it on your own, define a stopping time $T$ by $$T := \min\{0 \leq k \leq n; Y_k \geq 1\}$$
Note that this implies $$\mathbb{P} \left[ \inf_{0 \leq k \leq n} S_k \leq 0 \right] = \mathbb{P} \left[ \max_{0 \leq k \leq n} Y_k \geq 1 \right] = \mathbb{P}[T<\infty] = ...
3
Hint: Let $M_n=f(x+S_n)$. Then $(M_n)$ is a martingale. To wit, $x+S_{n+1}=x+S_n+X_{n+1}$ where $x+S_n$ is measurable with respect to $\mathcal F_n$ and $X_{n+1}$ is independent of $\mathcal F_n$. By the standard properties of conditional expectation (integrate that which is independent, leave out that which is measurable),
$$
\mathbb E(M_{n+1}\mid\mathcal ...
3
Just check the definition. Let $S_n:=\sum_{k=1}^nX_k$.
We have that $S_n$ is integrable since so are $X_j$.
Let $n$ an integer. We have by linearity of conditional expectation that
$$E[S_{n+1}\mid\mathcal F_n]=E[S_n+X_{n+1}\mid\mathcal F_n]=E[S_n\mid\mathcal F_n]+
E[X_{n+1}\mid\mathcal F_n].$$
Since $S_n$ is $\mathcal F_n$ measurable, we have ...
3
I'll try to answer your second question.
Here we have a filtration $\mathcal{F}_n$ on a probability space $(\Omega, \mathcal{F},\mathbb{P})$. Letting $H = L^2(\mathcal{F},\mathbb{P})$, we can see the sigma-algebra filtration as giving rise to a filtration of subspaces $H_n = L^2(\Omega,\mathcal{F}_n, \mathbb{P})$ (all this means is that $H_1 \subset H_2 ...
3
There seems to be two questions in your post. A martingale which is not a sum of independent increments? Try $X_n=a^{S_n}$, where $(S_n)$ is a $\pm1$ random walk on the integer line with probability $p\ne\frac12$ for the step $+1$ and $1-p$ for the step $-1$, for a suitable value of $a$ which I will let you discover. A good place to learn the basics of ...
3
Since you're curious as to whether or not you're going in the right direction, allow me to address your approach first before giving you my solution (which follows a different approach).
I actually liked what you were trying to do and attempted to rescue it for a while. Then, I realized it might be a little too weak. I don't have a counterexample at hand ...
3
Without loss of generality, we can assume $\pi_{n+1} \backslash \pi_n = \{s\}$ for some $s \in I$ (otherwise we add the missing partitions), i.e. $\pi_{n+1} = \pi_n \cup \{s\}$. Let $$t_j := \sup\{t \in \pi_n; t<s\}$$ Then $$\pi_{n} B- \pi_{n+1} B = (B_{t_{j+1}}-B_{t_j})^2 - (B_{t_{j+1}}-B_s)^2+(B_s-B_{t_j})^2 = 2 (B_{t_{j+1}}-B_s) \cdot (B_s-B_{t_j})$$ ...
3
In general it's neither martingale nor supermartingale. Consider for example $$X_n := \prod_{j=1}^n \xi_j$$ where $(\xi_j)_j$ are independent identically distributed random variables such that $\mathbb{E}\xi_1 = 1$, $\mathbb{E}|\xi_1|^3 < \infty$. Then $(X_n)_n$ is a martingale with respect to the filtration $\mathcal{F}_n := \sigma(\xi_1,\ldots,\xi_n)$ ...
3
The condition $\mathbb E|X_n|\lt n$ is odd. What is required for $(X_n)$ to be a martingale is, in particular, that each $X_n$ is integrable (if only to be able to consider its conditional expectation), but nothing is required about the growth of $\mathbb E|X_n|$.
Consider for example a sequence $(Z_n)$ of i.i.d. centered $\pm1$ Bernoulli random variables, ...
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