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16m
comment Compute the stationary distribution of a Markov Chain on an infinite state space
Since there are absorbing states, the stationary distributions are trivial: precisely those distributions that assign total probability 1 to the absorbing states. Is this really what you want to ask?
31m
comment It is true that $\int_{0}^{\infty}\mathbb{P}(x<m \ \cap Y \leq k-x) f_{X}(x)dx= \int_{0}^{m}\mathbb{P}( Y \leq k-x) f_{X}(x)dx$?
@Edin_91: Yes, that's right. The independence of $X,Y$ is not needed.
1h
comment It is true that $\int_{0}^{\infty}\mathbb{P}(x<m \ \cap Y \leq k-x) f_{X}(x)dx= \int_{0}^{m}\mathbb{P}( Y \leq k-x) f_{X}(x)dx$?
@Edin_91: It seems so clear that I'm not really sure what to say, but let's start with this. Do you understand why $\mathbb{P}(3 > 4) = 0$? Note that $\mathbb{P}$ is a map whose inputs are subsets of the sample space $\Omega$, so this is shorthand for $\mathbb{P}(\{\omega \in \Omega : 3 > 4\})$. Now, no matter what $\omega$ is, it is not true that $3>4$, so $\{\omega \in \Omega : 3 > 4\} = \emptyset$. Thus $\mathbb{P}(3 > 4) = \mathbb{P}(\emptyset) = 0$.
1h
answered It is true that $\int_{0}^{\infty}\mathbb{P}(x<m \ \cap Y \leq k-x) f_{X}(x)dx= \int_{0}^{m}\mathbb{P}( Y \leq k-x) f_{X}(x)dx$?
6h
comment No drift brownian motion problem
I'm also confused by your definition of the processes $G_1, G_2$. Is $G_1$ supposed to satisfy the SDE $dG_1 = \sigma_1 G_1 dW$ for $W$ a Brownian motion? If so, then $G_1$ is not itself a Brownian motion. Why does $W$ have a subscript $g$? And are both processes $G_1, G_2$ supposed to be driven by the same Brownian motion $W$?
6h
comment No drift brownian motion problem
Ok, so if we let $\tau_1 = \inf\{t : G_1(t) \ge P_1\}$ and $\tau_2 = \inf\{t : G_2(t) \ge P_2\}$, your question is "What is $\mathbb{P}(\tau_1 < \tau_2)$?" Is that right?
6h
comment zero drift brownian motions and barriers problem
Please don't repost closed questions. Your previous question was closed as unclear; you can edit it to make necessary clarifications, and then use the "flag" button with choice "in need of moderator intervention" to request that it be reopened.
6h
comment No drift brownian motion problem
Can you clarify your last sentence? I don't understand the role of $\tau$. And when you say "hit the barrier faster" do you just mean "earlier"?
1d
comment measure equality on two sigma algebras also holds on the combined sigma algebra?
So in other words, it's possible to have two measures $P,Q$ on $\mathcal{F} \vee \mathcal{G}$ which agree on $\mathcal{F} \cup \mathcal{G}$ yet are not equal. Would that resolve your question?
1d
comment measure equality on two sigma algebras also holds on the combined sigma algebra?
Well, what makes you think it is true? I do not know a counterexample to existence of an extension off the top of my head, but uniqueness definitely can fail. Consider a sample space with two coin flips, $\mathcal{F}$ generated by the first flip and $\mathcal{G}$ by the second. Defining a measure on $\mathcal{F} \cup \mathcal{G}$, you can ensure that each coin has probability $1/2$ of coming up heads, but there are infinitely many extensions to $\mathcal{F} \vee \mathcal{G}$, depending on how correlated you want the coins to be.
1d
answered Show that an operator is closable
1d
comment measure equality on two sigma algebras also holds on the combined sigma algebra?
Anyway, what do you mean by saying $P=Q$ on $\mathcal{F} \vee \mathcal{G}$? They are only defined on $\mathcal{F} \cup \mathcal{G}$, and it's not clear that either one has a countably additive extension to $\mathcal{F} \vee \mathcal{G}$.
1d
comment measure equality on two sigma algebras also holds on the combined sigma algebra?
$\mathcal{F} \cup \mathcal{G}$ is not stable under intersections, in general.
2d
comment A measure theory question-1
en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
2d
comment Can a compact set of $\mathbb{R}$ have some properties and not being convex
Oh, did you mean to write $P,R \in int(C)$?
2d
comment Invertible , bounded linear operator on a Hilbert space
It depends on what exactly is meant by "invertible". It has an inverse which is not defined on all of $H$. If "invertible" means "bijective" then indeed your operator is not bijective, and in fact every bijective bounded linear operator on a Hilbert space does have a bounded inverse.
2d
comment Can a compact set of $\mathbb{R}$ have some properties and not being convex
I'm worried by the WLOG.
2d
comment Invertible , bounded linear operator on a Hilbert space
If by "invertible" you mean "bijective" then this is a well known theorem.
2d
comment Invertible , bounded linear operator on a Hilbert space
And the second display should have $\|(a_i)_{i \in \mathbb{N}}\|^2$.
2d
comment Invertible , bounded linear operator on a Hilbert space
Nitpick: you don't really mean $H = \mathbb{R}^{\mathbb{N}}$ which is the set of all sequences, not just those which are square-summable. The usual notation for the space of square-summable sequences is $\ell^2(\mathbb{N})$.