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Alex R.
Alex R.
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I think your formula for $P(N_n=k)$ is false because as you said, paths are dependent on one-another. With that said:

Expectation is linear, regardless of dependence. Let $S_n$ be the index of set of SAWs of length $n$ and $\gamma_{ni}$ denote a self avoiding path of length $n$, $i\in S_n$. You write

$$N_n=\sum_{i\in S_n} 1_{\gamma_{ni}}$$

Now you note that each $\gamma_{ni}$ has probability $p^n$ and $S_n$ has $\sigma_n$ elements in it. Using $E[1_{\gamma_{ni}}]=p^n$, it immediately follows that $E[N_n]=p^n\sigma_n$.

I think your formula for $P(N_n=k)$ is false because as you said, paths are dependent on one-another. With that said:

Expectation is linear, regardless of dependence. Let $S_n$ be the index of set of SAWs of length $n$ and $\gamma_{ni}$ denote a self avoiding path of length $n$, $i\in S_n$. You write

$$N_n=\sum_{i\in S_n} 1_{\gamma_{ni}}$$

Now you note that each $\gamma_{ni}$ has probability $p^n$ and $S_n$ has $\sigma_n$ elements in it. Using $E[1_{\gamma_{ni}}]=p^n$, it immediately follows that $E[N_n]=p^n\sigma_n$.

I think your formula for $P(N_n=k)$ is false because as you said, paths are dependent on one-another. With that said:

Expectation is linear, regardless of dependence. Let $S_n$ be the index of set of SAWs of length $n$ and $\gamma_{ni}$ denote a self avoiding path of length $n$, $i\in S_n$. You write

$$N_n=\sum_{i\in S_n} 1_{\gamma_{ni}}$$

Now you note that each $\gamma_{ni}$ has probability $p^n$ and $S_n$ has $\sigma_n$ elements in it. Using $E[1_{\gamma_{ni}}]=p^n$, it immediately follows that $E[N_n]=p^n\sigma_n$.

Source Link
Alex R.
Alex R.
  • 5k
  • 2
  • 41
  • 66

I think your formula for $P(N_n=k)$ is false because as you said, paths are dependent on one-another. With that said:

Expectation is linear, regardless of dependence. Let $S_n$ be the index of set of SAWs of length $n$ and $\gamma_{ni}$ denote a self avoiding path of length $n$, $i\in S_n$. You write

$$N_n=\sum_{i\in S_n} 1_{\gamma_{ni}}$$

Now you note that each $\gamma_{ni}$ has probability $p^n$ and $S_n$ has $\sigma_n$ elements in it. Using $E[1_{\gamma_{ni}}]=p^n$, it immediately follows that $E[N_n]=p^n\sigma_n$.