## change of measure on continuous random walk

some background on polymer modeling: the discrete polymer model is the change of measure on the random walk space

$\frac{dP_{h,N}}{dP} = e^{hR_N}\delta_0(X_N)Z^{-1}_{h,N}$ for $R_N := \sum_{k=1}^N \delta_0(X_k)$ the number of visits to zero and the typical partition function $Z_{h,N}=E[ e^{hR_N}\delta_0(X_N)]$. Then we know that for the free energy

$\chi(h) := \lim_{N\to\infty} \frac{1}{N} \ln Z_{h,N}$

that $\chi(h) =0$ for $h\le h_c$ and $\chi(h) >0$ for $h>h_c$

where $h_c = -\ln(1-r)$ for $r$ the escape probability of the random walk.

So if our random walk has first return distribution $K(n)=P(X_n=0, X_k \ne 0, k <n)$

then $1-r=\sum K(n)$, so if we redefine $K'(n) = K(n)e^{h_c}$, we have $\sum K'(n)=1$

$Z_{h+h_c,N} = E[e^{(h+h_c)R_N}\delta_0(X_N)]$ $= \sum_{n=1}^N e^{n(h+h_c)} \sum_{j_1+\cdots j_n=N} \prod_{i=1}^n K(j_i)$ $= \sum_{n=1}^N e^{nh} \sum_{j_1+\cdots j_n=N} \prod_{i=1}^n K'(j_i)=E'[e^{hR_N}\delta_0(X_N)]=Z'_{h,N}$

and thus, we can just assume that $h_c=0$ and our walk starts off recurrent.

im working with a continuous polymer model with pinning at the origin

so $\frac{dP_{h,t}}{dP} = e^{hR_t} \delta_0(X_t)Z^{-1}_{h,t}$ for $R_t:= \int_0^t \delta_0(X_s)ds$ the time spent at the origin. and $Z_{h,t} = E[e^{hR_t}\delta_0(X_t)]$ a typical partition.

with free energy, $\chi(h) := \lim_{t\to\infty} \frac{1}{t} \ln Z_{h,t}$

has $\chi(h) >0$ for all $h>r$, and $\chi(h) = 0$ for $h\le r$

I want to be able to say something similar, that i can let $K'(n)=\frac{K(n)}{1-r}$, and then i should have

$Z_{h+r,t} =E[e^{(h+r)R_t}\delta_0(X_t)]= E[e^{hR_t}\delta_0(X_t) \cdot e^{rR_t}]=E_t'[e^{hR_t}]$

but i know that the radon nikodym derivative must then be

$\frac{dP_t'}{dP} = \frac{e^{rR_t}}{E[e^{rR_t}]}$`,

but this suggests that i will be changing more than just the skeletal part of the random walk, since $R_t$ shows up. Is there some way to only change the skeleton and still be able to induce some kind of shift of this critical value to zero?

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