Question

Prove $E^{\mathbb Q}[Y]=E^{\mathbb P}[XY]$ if $E^{\mathbb P}[X]=\mathbb P(X>0)=1$ and $ \mathbb Q(A)=E^{\mathbb Q}[X1_A] $

score 355 · Answer 1 · 2018-03-14 01:36:20

355

Views

Prove $E^{\mathbb Q}[Y]=E^{\mathbb P}[XY]$ if $E^{\mathbb P}[X]=\mathbb P(X>0)=1$ and $ \mathbb Q(A)=E^{\mathbb Q}[X1_A] $

Published on 14 Mar 2018 - 1:36

score 1.9k · Answer 2 · 2018-03-16 18:45:56

1.9k

Views

Computing expectation under a change of measure

Published on 16 Mar 2018 - 18:45

score 120 · Answer 3 · 2018-03-22 06:12:09

120

Views

Show that the Radon-Nikodym density is $\mathcal{G}$ measurable iff $E_P[X|G] = E_Q[X|G]$

Published on 22 Mar 2018 - 6:12

score 457 · Answer 4 · 2018-03-23 17:57:49

457

Views

ergodic measure and absolutely continuous measure

Published on 23 Mar 2018 - 17:57

score 237 · Answer 5 · 2018-03-24 13:35:00

237

Views

What is the Radon-Nikodym density $dP^∗ /d P$ of the unique $P^ ∗ ∼ P$ such that the discounted price $S^*_t := S_t /B_t$ is a $P^∗$-martingale

Published on 24 Mar 2018 - 13:35

score 54 · Answer 6 · 2018-03-29 04:32:25

54

Views

Application of chain rule, is this correct?

Published on 29 Mar 2018 - 4:32

score 305 · Answer 7 · 2018-04-09 12:55:16

305

Views

Radon Nikodym derivative $\frac{\mathrm d(fλ)}{\mathrm d(gλ)}$

Published on 09 Apr 2018 - 12:55

score 180 · Answer 8 · 2018-04-11 15:39:38

180

Views

Radon-Nykodym Derivative process-Property of Conditional Expectation

Published on 11 Apr 2018 - 15:39

score 791 · Answer 9 · 2018-04-20 16:15:10

791

Views

Finding the Radon-Nikodym Derivative

Published on 20 Apr 2018 - 16:15

score 340 · Answer 10 · 2018-04-27 01:37:10

340

Views

The Dual of $L^{\infty}$ and the Radon-Nikodym Theorem

Published on 27 Apr 2018 - 1:37

score 380 · Answer 11 · 2018-05-08 16:10:51

380

Views

Lebesgue-Radon-Nikodym representation

Published on 08 May 2018 - 16:10

score 50 · Answer 12 · 2018-05-11 01:56:39

50

Views

Deriving the posterior distribution given arbitrary measures

Published on 11 May 2018 - 1:56

score 148 · Answer 13 · 2018-05-16 14:02:04

148

Views

Can divergence be thought of as a Radon-Nikodym derivative?

Published on 16 May 2018 - 14:02

score 58 · Answer 14 · 2018-05-20 06:39:57

58

Views

Change of measure as functions of random variable

Published on 20 May 2018 - 6:39

score 337 · Answer 15 · 2018-05-22 22:20:36

337

Views

Radon–Nikodym derivative with another variable is a martingale

Published on 22 May 2018 - 22:20

List Question

Prove $E^{\mathbb Q}[Y]=E^{\mathbb P}[XY]$ if $E^{\mathbb P}[X]=\mathbb P(X>0)=1$ and $ \mathbb Q(A)=E^{\mathbb Q}[X1_A] $

Computing expectation under a change of measure

Show that the Radon-Nikodym density is $\mathcal{G}$ measurable iff $E_P[X|G] = E_Q[X|G]$

ergodic measure and absolutely continuous measure

What is the Radon-Nikodym density $dP^∗ /d P$ of the unique $P^ ∗ ∼ P$ such that the discounted price $S^*_t := S_t /B_t$ is a $P^∗$-martingale

Application of chain rule, is this correct?

Radon Nikodym derivative $\frac{\mathrm d(fλ)}{\mathrm d(gλ)}$

Radon-Nykodym Derivative process-Property of Conditional Expectation

Finding the Radon-Nikodym Derivative

The Dual of $L^{\infty}$ and the Radon-Nikodym Theorem

Lebesgue-Radon-Nikodym representation

Deriving the posterior distribution given arbitrary measures

Can divergence be thought of as a Radon-Nikodym derivative?

Change of measure as functions of random variable

Radon–Nikodym derivative with another variable is a martingale

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