New answers tagged martingales
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Applying Ito's rule to: $Y(t)=f(M(t),⟨M⟩(t))$ (a function of Martingale and its quadratic variation)
First we note that $dM = \sigma_t dB_t$ and $d \langle M \rangle = \sigma_{t}^{2} dt$.
From Ito's rule we then have:
$$
dY_t = \bigg(\frac{\partial f}{\partial \langle M \rangle} \sigma_{t}^{2} + \...
- 1,020
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Prove that $S^0=S^1$ so there is one bank account in the market without arbitrage
Your mistake, ultimately, is that your second equation is meaningless. It only makes sense if nobody can sell/buy after the first period. Instead of that second equation, you should have two ...
- 16.1k
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Show that it is a stopping time
Don't forget the quantifiers on $\tau_1$ and $\tau_2$ in your definition of $\nu$. Taking them into account you'll have
$$
\{\nu=t\}=\cup_{\tau_1=0}^{t-2}\cup_{\tau_2=\tau_1+1}^{t-1}\left[\cap_{n=0}^{\...
- 23.9k
2
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Let $M$ be a continuous martingale, $r >0$, and $\tau := \inf \{t \ge 0 : \langle M \rangle_t \ge r \}$. Then $M_\tau$ is square-integrable
Yes, your proof is correct. One could also use Fatou's lemma for a slightly more direct proof and a tighter bound: After showing $\mathbb{E}[|M_{t \wedge \tau}|^2] \le r$, we have
\begin{align*}
\...
- 10.9k
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Brownian motion: how is DCT for conditional expectation applied on this stopped martingale?
By Vitali's convergence theorem, it suffices to show that $( M_{\tau (s_2) \wedge \tau_n}, n \in \mathbb N)$ is uniformly integrable. By OST, $( M_{t \wedge \tau (s_2) \wedge \tau_n}, t \ge 0)$ is a ...
- 13.7k
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Proof of Doobs decomposition theorem
$\mathbb{E}[Y_{n+1}|\mathcal{F}_n]$ is $\mathcal{F}_n$-measurable and $\mathcal{F}_n \subseteq \mathcal{F}_t$ for $0 \leq n \leq t$. Hence, $\mathbb{E}[Y_{n+1}|\mathcal{F}_n]$ is $\mathcal{F}_t$-...
- 21.2k
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Brownian motion: how is $\mathbb E [\sup_{t \ge 0} |\langle M\rangle_{t \wedge \tau_n}|^2] < \infty$ satisfied?
You need two things:
A continuous local martingale $M$ such that $M_0\in L^2$ is a bounded martingale in $L^2$ iff $\mathbb E[\langle M\rangle_{\infty}]<+\infty$. In that case, $M^2-\langle M\...
- 4,989
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Local martingale: how is $\sup_{s \in[0, t]}\left|M_s\right|$ measurable?
Measurability of the supremum cannot be proved. If the process has right(or left) continuous paths then the supremum is measurable. I think condition (**) is to be interpreted as saying that the ...
- 21.2k
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A proof of Lévy's theorem: how to obtain the independence $X_t-X_s \perp \mathcal{F}_s$?
You can find another proof of Lévy's characterisation in Schilling, Partszch, Boettcher; 9.12 p. 148; the use of characteristic functions is less troublesome, in my opinion, than MGFs. In the ...
- 12.3k
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20-sided dice 100 rounds game
I took a bit of a different approach. I assumed that, for each round, there is an optimal value $x_0$ above which the expected gain is maximized.
Assuming $N$ throws and an $n$ sided die, we obtain
$P(...
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If $B$ is a Brownian motion and $f \in \mathcal C^2(\mathbb R^n)$ harmonic, then $f(B)$ is a continuous martingale
$\mathbb E \bigg [ \int_0^t f_{x_i}^{\prime}\left(\underline{B}_s\right) d B_s^{(i)} \bigg ]$ has no reason to be $0$.
And even if it was, having $\mathbb E [f\left(\underline{B}_t\right)] - f\left(0\...
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Branching Process; understanding theorem proof
I think @clark argument is correct but there are still a few details missing. Following his argument, by the Markov property,
$$
\begin{align*}
\mathbb{P}(Z _n = k \text{ for all } n \geq N)
&...
- 332
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Is Brownian motion a semimartingale?
We know that the quadratic variation of a Brownian motion $\left(B_t\right)_{t \geq 0}$ is equal to $t$. Thus $[B]_t = t$. Now if we add to the brownian motion another continuous process and of finite ...
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Independence of random variable to sigma-algebra generated by previous ones
It is clear that if $\mathcal F$ is a $\sigma$-algebra for which $X_1,\dots,X_n$ are measurable, then $S_1,\dots,S_n$ are also $\mathcal F$-measurable. Therefore $$\sigma(S_1,\dots,S_n)\subseteq \...
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Prove that $Z_{n} = X_{n} 1_{ (T>n)} + Y_{n} 1_{ (T \leq n)}$ is a supermartingale.
It seems that I have to associate terms in a good way and play with the set $(T\geq n+1) = (T =n +1) \cup (T> n+1) = (T \leq n)^{c}$ to make indicators $ 1_{(T> n+1) }= 1_{(T \geq n+1) }$ $= 1_{...
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Is M defined as below a martingale?
Which Bessel function do you have in mind? (There are several, and your notation $B_1$ is non-standard.)
A more appropriate condition for testing whether you have a martingale might be Kazamaki's: If ...
- 23.9k
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Application of Cesaro lemma
You just need the particular case of arithmetic mean of Cesàro's lemma, which yields that $\frac{1}{n}\sum_{k=0}^{n-1}M_k$ converges almost surely as $n\to+\infty$ to the same limit as that of $M_n$. ...
- 4,989
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Given a local martingale $M$ the running supremum $N_t=\sup_{0\leq s\leq t}|M_s|$ is locally integrable
Well, $M$ may (and often will) have a jump at $S_n$, so it is quite possible that $|M_{S_n}|>n$. So actually
$$
N_{S_n} \le |M_{S_n}| \vee \sup_{s<S_n} |M_s| \le |M_{S_n}| \vee n.
$$
- 24.5k
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Example of a martingale with non-independent increments and fixed variance
Here's a construction of many examples.
Let $S_n$ be a simple symmetric random walk: $S_0=0$ and
$$
S_n=\sum_{k=1}^n \xi_k,
$$
where the $\xi_k$ are iid with $P[\xi_k=1]=P[\xi_k=-1]=1/2$. Let $\...
- 23.9k
3
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Example of a martingale with non-independent increments and fixed variance
Let $Z_i \sim N(0,1)$ be i.i.d. random variables and define $M_n := Z_0 \sum_{i=1}^n Z_i$. Then $\Delta_n = Z_0 Z_n$, so the increments are identically distributed and hence have constant variance, ...
- 10.9k
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