8 votes

stochastic process: how can probability space be the same?

"What am I doing wrong"? Answering that first, what you're doing wrong is changing your source of randomness to fit the random variable in question. You shouldn't do that. In experiment ...
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7 votes
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Is the sum of two Brownian motions always a martingale, even if the two are possibly correlated?

I interpret the question to mean: If $\{B(t)\}$ and $\{W(t)\}$ are two Brownian motions defined on the same probability space, is $\{B(t)+W(t)\}$ necessarily a Martingale even if $B$ and $W$ are ...
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4 votes
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Book suggestions (Stochastics, Brownian Motion etc.)

I would suggest the following books: J. M. Steele, Stochastic Calculus and Financial Applications (*) R. van Handel, Stochastic Calculus, Filtering, and Stochastic Control (*) J. R. Norris, Markov ...
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4 votes
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If $\frac{dX_{t}}{X_{t}} = dL_{t}$, where $L_{t}$ is a local martingale, then is $X_{t}$ a local martingale?

There is essentially only one process $X$ satisfying $dX_t = X_t dL_t$, namely $X_t := C\exp(L_t - \frac 12 \langle L,L\rangle_t)$ where $C \in \mathbb{R}$ can be arbitrary. More precisely, if $Y$ is ...
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  • 8,662
3 votes
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Ito integral over an indicator function

As the comment says, the question makes little sense for a geometric Brownian motion. So let $P$ be just a general Ito process with $P_t = \int_0^t a_s ds + \int_0^t b_s dW_s$. You want to write the ...
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  • 23.1k
3 votes
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Proof of a martingale not being uniformly integrable

If the martingale $\left(\varphi\left(S_n\right)\right)_{n\geqslant 0}$ was uniformly integrable, then the optional stopping theorem for uniformly integrable martingale would give that $\mathbb E\left[...
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3 votes
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Martingale and independence

Your idea is correct. Is it true that $\mathcal{H}_u=\sigma(\sigma(X_v,0 \leq v \leq u)\cup\sigma(Y_v, 0 \leq v \leq u))$ Of course. Indeed, $\sigma(X_v,0 \leq v \leq u)\subset\sigma(X_v,Y_v,0 \leq ...
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  • 3,429
3 votes
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Is it possible that the first arrival time of a poisson process is infinite?

A homogeneous Poisson process with intensity $\lambda > 0$ almost surely has a finite first arrival time, since $$\lim_{t \to \infty} \Pr[T \le t] = \lim_{t \to \infty} 1 - e^{-\lambda t} = 1.$$ ...
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  • 113k
3 votes

Binomial distribution approximated by normal distribution: How to arrive at this identity?

If $X \sim \text{Binomial}(N, |\beta|^2)$, then $r(t)$ is $\mathbb{E}[e^{ig(N-2X)t}] = e^{igNt} \mathbb{E}[e^{-2igtX}]$. The normal approximation approximates this Binomial distribution with $\mathcal{...
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  • 81.7k
3 votes

Show markov property for a s.p. with independent increment

Let $\sigma(X)\perp \sigma(Y,Z)$, where $X,Y,Z\in L^1$. We want to show that for $f$ measurable and bounded we have $$E[f(X,Y)|Y,Z]=E[f(X,Y)|Y]$$ Define the set of functions $$\mathscr{H}:=\{h:E[h(X,Y)...
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  • 8,626
3 votes
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Distribution of difference of first hitting times of Brownian motion

Yes, assuming that $0<b<a$, the random variables $T_a-T_b$ and $T_{a-b}$ have the same distribution. This follows from the strong Markov property of Brownian motion at time $T_b$. Note that if $...
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2 votes
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Showing that Brownian motion will hit sphere with an exact integer distance

First, since the BM will eventually move arbitrarily far from the origin, we have $\limsup\lvert B_t + z \rvert = \infty$, so if $\lvert{B_0 + z \rvert} < r$ for any integral point $z$ then we are ...
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2 votes

Show markov property for a s.p. with independent increment

After doing some small manipulations about the idea of independent increment, I think I have figured out a method: \begin{align*} \mathbb{P}(X_t \leq x | X_{t_1},\ldots ,X_{t_n}) &= \mathbb{P}(...
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2 votes
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Intuition behind the definition of Markov Transition Kernel, defined in the following way

An intuitive interpretation: Suppose that this describes the motion of a particle the trajectory is given by $X(t)$. If $X(t_0)=x$ then the conditional probability of finding it in some Borel set $A$...
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  • 34.1k
2 votes

Solution of a particular Backward SDE

Suppose for simplicity that $f=0$. With $Z_{s}=c$ we get \begin{equation*} Y_{t}=\xi-\int_{t}^{T} Z_{s}dW_{s}=\xi-c\left(W_{T}-W_{t}\right). \end{equation*} Unless $\xi$ looks something like $a +cW_{T}...
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  • 714
2 votes

Example of natural filtration which is not right-continuous

Let $(W_n)_{n \in \mathbb{Z}_+}$ be a simple symmetric random walk on $\mathbb{Z}_+$ starting at $0$. Interpolate $W$ linearly on each time interval $[n,n+1]$. This defines a continuous and piecewise ...
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2 votes
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Showing that a martingale is constantly zero

Consider the Martingale $M_k:=W(Y_{k \wedge T})$, where $T$ is the first time $Y$ hits $i_0$, and $Y_0=i$. From $E[(M_k-M_0)^2]<\infty$, the Martingale is bounded in $L^2$, so it is uniformly ...
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1 vote

$P(|B(2)|>|B(1)|)$ for Brownian motion

Write $X = B(1)$ and $Y = B(2) - B(1)$, and note that $X$ and $Y$ are i.i.d. standard normal variables. Then \begin{align*} |B(2)| \geq |B(1)| &\quad\iff\quad |X+Y| \geq |X| \\[0.25em] &\quad\...
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1 vote
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Why do we have there exists a constant $c>0$ so that for any $x$, $P(\sup_{t\le T}X_t\ge x)\le P(A^c)+P(X_0\ge cx) $?

Take $c=1$. $$ P(\sup_{t\le T}X_t\ge x)$$ $$= P(A \cap (\sup_{t\le T}X_t\ge x))$$ $$+P(A^{c}\cap (\sup_{t\le T}X_t\ge x))$$ $$\le P(X_0\ge cx)+P(A^c).$$
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1 vote
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martingale, stopping times

It is immediate that $$ \{T\leqslant t\}\in \mathcal F_t $$ for all $t$. Now, since $\sigma+\tau$ and $T$ are stopping times, we have $$ \{(\sigma+\tau)\wedge T\leqslant t\} = \{\sigma+\tau\leqslant t\...
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  • 33.2k
1 vote

Basic application of Ito's Lemma

In the first formula you are using the Ito formula for non-autonomous transformations $y=f(t,x)$ with the increments $$ dY_t = f_t\,dt+f_x\,dX+\frac12f_{xx}\,d\langle X\rangle_t. $$ In the example ...
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1 vote

Negative moment of the number of arrivals in a renewal process

This is not possible, unfortunately. Let us play a little bit: $$ \mathbb{E}\Big[\frac{t}{\max(1,N(t))}\Big] = t\int_0^1 \mathbb{P}(N(t)^{-1}\ge x)dx\\ = t\int_1^\infty \mathbb{P}(N(t)\le u)\frac{du}{...
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  • 23.1k
1 vote
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Proving Zero-One law for tail events using Bluementhal's law.

This follows immediately from Bluementhal's $0-1$ law applied to the BM $(X(s))$ since $\sigma(X(s): 0 \leq s \leq 1/t)=\sigma(B(s): s \geq t)$ which implies $\bigcap_{t\geq 0} \mathcal{G}(t)=\bigcap_{...
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  • 2,226
1 vote

$(a)$ Calculate $P(X_5>0|W_3=1)$ given that $\{W_t\}$ is an Simple Brownian Motion and $\{X_t\}$ a stochastic process

Hint \begin{align*} \mathbb P\{X_5>0\mid W_3=1\}&=\mathbb P\{W_5^2>5\mid W_3=1\}\\ &=\mathbb P\{(W_5-W_3)^2+2(W_5-W_3)W_3+W_3^2>5\mid W_3=1\}\\ &=\mathbb P\{(W_5-W_3)^2+2(W_5-W_3)+...
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  • 51.1k
1 vote
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Does local Hölder-continuity suffice to get an upper bound for Hausdorff dimension - or does it need to be global? ((Mistake in the literature?))

The issue is the definition of local Holder continuity. In general, we say that a property $P$ of metric spaces is satisfied locally in a metric space $M$, if every point in $M$ has an open ...
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1 vote

Quadratic variation of linear combination of two processes

Since either $W = 0$ or $W = 1$, $Z_t := W X_t + (1-W)Y_t$ will either equal $X_t$ or $Y_t$ for all $t$. Hence $[Z]_t = W [X]_t + (1-W)[Y]_t$. This could be verified by checking that $Z_t^2 - [Z]_t$ ...
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  • 8,662
1 vote
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How can we show $\operatorname E\left[Y_\tau;t\ge\tau\mid\tau=s\right]=1_{\{\:t\:\ge\:s\:\}}\operatorname E\left[Y_s\right]$?

You are right. (1) is wrong. Here is a counter-example: let $(Y_t)$ be Brownian motion and $\tau=\inf \{t \geq 0: Y_t=1\}$. Then $Y_{\tau}=1$. If we take $s \leq t$ the LHS of (1) is $1$ and RHS is $...
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  • 2,226
1 vote
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How $\{B(t): t \leq T\}$ is $\mathcal{F}^+-$measurable in Brownian Motion, with $T$ a stopping time

Following up on my comment, we will show that given $A=[B_T \in I]$ for any interval $I$, we have $A \cap [T\le t] \in \mathcal{F}^+(t)$ for all $t$. To take this to a Borel set of $\mathbb{R}$ rather ...
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  • 1,702
1 vote
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Brownian motion and martingale

Yes, this can be proved in the same way. As noted in the comments, when $f$ is harmonic the second term vanishes.
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