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Questions tagged [radon-nikodym]

For questions involving the notion of the Radon-Nikodym derivative or the Radon-Nikodym theorem. Use this tag along with (probability-theory) or (measure-theory).

1 vote
1 answer
70 views

I need to prove a property about the linear operator defined by the following theorem: Let $(X,\mathcal{B},\mu)$ be a finite measure space and $\mathcal{A}\subset \mathcal{B}$ a sub $\sigma-$algebra. ...
Daniel Aricatt's user avatar
10 votes
0 answers
171 views

The Fourier Transform $\int_{\mathbb R^d}e^{-2\pi is\cdot x}\mathrm{d}\mu$ is an injective linear map from the space of bounded-variation Borel measures to the space of continuous functions; so in ...
Lavender's user avatar
  • 1,664
2 votes
1 answer
94 views

Let $(\Omega, (\mathcal{F}_t)_{0\leq t\leq 1}, P)$ be some filtered probability space where the filtration satisfiy the usual assumptions. This means $\mathcal{F}_0$ contains all null sets of $\...
user1294729's user avatar
  • 2,045
1 vote
0 answers
48 views

I am struggling with a measure theory problem that seems simple, but my intuition tells me I am making a mistake somewhere. I would appreciate your help. Here, all measures are Levy measures. Let's ...
Gregório Vieira's user avatar
3 votes
1 answer
80 views

I know that a pdf (probability density function) of a random variable $X$ is the Radon-Nikodym derivative of the distribution of $X$ (which is a pushforward measure of the underlying probability ...
Jan Zapal's user avatar
  • 3,719
0 votes
1 answer
76 views

I am trying to solve the following exercise: Let $m$ denote the Lebesgue measure , and let $\lambda$ be a real valued positive measure. Argue that there is a set $A$ of full Lebesgue measure (i.e. $m(...
notimportant's user avatar
0 votes
0 answers
84 views

Let $(X,\mathcal{F})$ be a measurable space. $\nu,\mu$ are two $\sigma$-finite measure on $\Omega$ such that $\mu\ll \nu$. $P:X\times \mathcal{F}$ is a Markov kernel. One can shows $\mu P\ll \nu P$ as ...
mounta's user avatar
  • 93
1 vote
0 answers
50 views

Let $(\mathcal{X}, \mathcal{F})$ and $(\mathcal{Y}, \mathcal{G})$ be measure spaces. Let $\mu \ll \nu$ measures on $\mathcal{X}$ and $K:\mathcal{X}\times\mathcal{G}\rightarrow[0,1]$ a Markov kernel. ...
guest1's user avatar
  • 956
0 votes
1 answer
132 views

Let $(\mathcal{X}, \mathcal{F})$ and $(\mathcal{Y}, \mathcal{G})$ be measure spaces and let $K:\mathcal{X}\times\mathcal{G}\rightarrow [0,1]$ be a markov kernel connecting these two spaces. Let $\mu$ ...
guest1's user avatar
  • 956
1 vote
1 answer
135 views

Let $(\mathcal{X}, \mathcal{F})$ and $(\mathcal{Y}, \mathcal{G})$ be measure spaces and $K:\mathcal{X}\times\mathcal{G}\rightarrow [0,1]$ be a Markov kernel. Let $\mu \ll \nu$ be $\sigma$-finite ...
guest1's user avatar
  • 956
0 votes
1 answer
76 views

Let $S$ be a compact Haursdorff space. Let $\mu$ be a regular complex Borel measure on S. Let $f$ be a non-negative real function on S. Is it the same to define $\sigma$ by $$d\sigma=fd\mu$$ and by$$\...
user1234's user avatar
1 vote
1 answer
47 views

I have been stuck for a while with the following couple of exercises: (1) Let $(X, \mathcal{M}, \mu )$ be a measure space, $(f_n)_{n \in \mathbb{N}}$ a sequence of functions in $L^{1}(X, \mathcal{M}, \...
user594756's user avatar
2 votes
0 answers
242 views

The following definition is from Kalton's book on Banach space theory. I have a question regarding the definition; why is the space specifically $L^1([0,1])$. I know there is another definition ...
user124910's user avatar
  • 3,389
0 votes
0 answers
83 views

The Borel-Kolmogorov Paradox shows trying to condition on a measure zero event like $Y = y$ can give you different answers if you describe the event $Y = y$ using different random variables. So not ...
Zoe Allen's user avatar
  • 8,062
4 votes
1 answer
101 views

The Problem: Let $\mu$ and $\nu$ be measures on a measurable space $(X, \mathcal{F})$ such that $\mu$ is $\sigma$-finite, $\nu$ is nonzero (i.e., $\nu(X) > 0$) and finite. Assume also that $\nu \ll ...
Jamie Kamsky's user avatar

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