Central limit theorem for the Edwards model
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Citations
A specification test for nonlinear nonstationary models
The SDE solved by local times of a brownian excursion or bridge derived from the height profile of a random tree or forest
A survey of one-dimensional random polymers
Some Brownian functionals and their laws
References
Brownian Motion and Stochastic Calculus
Continuous martingales and Brownian motion
Markovian Bridges: Construction, Palm Interpretation, and Splicing
On Edwards' model for long polymer chains
On Edwards' Model for Polymer Chains: II. The Self-Consistent Potential
Related Papers (5)
Frequently Asked Questions (9)
Q2. what is the distribution of y#t t0 under Ph?
In other words, the distribution of Y#t t≥0 under P#h is equal to that of B4t∧T0 t≥0 under Ph, which in turn is equal to that of 2Bt∧T0 t≥0 under Ph/2.
Q3. What is the proof of the first term?
The first term on the r.h.s. of (3.12) vanishes as T → ∞ because of the continuity of a → xa ∈ L2 R+ and a → ρ′ a (see HH, Lemma 22). ✷4. Integrability for the boundary pieces.
Q4. What is the proof of the proof of KS?
Since ya 0 = za 0 = 1 and since za is bounded on R+0 , it suffices to show that ya satisfies the same differential equation as za [see (4.3)].
Q5. how many ts is a probability density?
Since t → 12t−3/2 is a probability density on 1 ∞ , Jensen’s inequality and the boundedness of x2−pa∗ on R+ 0 give∫ ∞1 W 1 p t dt ≤ c ∫ ∞ 1 1 2t −3/2 dt (∫ ∞ 0 h1−pt3p/2wa∗ h t p dh )1/p≤ c (∫ ∞1 dt ∫ ∞ 0 dhh1−pt 3/2 p−1 wa∗ h t p )1/p (4.16)Use (4.10), Jensen’s inequality and the Brownian scaling property to estimate wa∗ h t p ≤ ϕh t p−1ϕh t Eh/2 ( exp ( a∗pt− p ∫ t 0 2Bs ds ) ∣∣∣T0 = t ) ≤ chp−1t− 3/2 p−1 ϕhp1/3 tp2/3×E
Q6. what is the a l a l a?
define α l a ∈ R and y l a ∈ L2 ◦ R+0 byρ l a− α l a = ρ a and y l a = x l a−α l ax l a−α l a ◦L2l ∈ N0 (3.6)Note that α 0 a = 0, y 0 a = xa/ √
Q7. What is the proof of Proposition 2?
Proof of Proposition 2. Recall Lemma 3 and (2.26) to see that the l.h.s. of (2.28) is equal toexp ( −λ √ T ρ′ a ) ∫ ∞ 0 dhf h× Eh ( exp ( − ∫ A−1 T′0( Fa Xs −λ√ T) ds ) g XA−1 T′ ) (3.1) According to (2.29), ρ aλ T = ρ a − λ/ √ T . Since T′ = ∫A−1 T′ 0
Q8. ?
ρ′ a f xa L2 g xa ◦ L2 ∣∣∣∣ + ∞∑ l=1 ∣∣∣∣ (∫ ∞ 0 dhf h xaT h Ê a h ( y l aT xaT YT )) g y l aT ◦L2 ∣∣∣∣ (3.12)With the help of Step 2, the second term on the r.h.s. of (3.12) equals∞∑ l=1 exp α l aT T ∣∣∣∣ (∫ ∞ 0 dhf h xaT h y l aT xaT h ) g y l aT ◦L2 ∣∣∣∣ ≤ exp α 1 aT
Q9. what is the eigenvalue of the process?
Use (2.7) and (2.14) to compute, for f ∈ C2 R+ ,( G̃a ( fxa)) u = f u uxa u ( 2uf′′ u + 2f′ u f u − 2ux′′a u + 2x′a u xa u ) (3.9)Apply this for f = y l a , use (0.5) and the eigenvalue relation a′x l a′ = ρ l a′ x l a′ for a′ l = a 0 and for a′ l = a− α l a l , to obtainG̃a ( y l axa) = α l a y l axa (3.10)Thus, G̃a being the generator of the process