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Is there something like a "self-avoiding Markov chain" on a continuous space?
@0xbadf00d I am not a specialist on this topic. According to Kostya_I, $SLE_\kappa$ is space filling as soon as $\kappa \ge 8$. This may be more visible for larger values of $\kappa$. You can find other simulations there math.arizona.edu/~tgk/sle
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Expectation of inverse of complex Gaussian variables
The modulus of $h$ is an explonential random variable.
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Rigorous estimates on roots of function
I modified my post to answer the right question.
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Rigorous estimates on roots of function
I corrected to answer the right question.
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Rigorous estimates on roots of function
Warning about the inadequate solution.
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On choosing the correct square root of $g^{4n}$ modulo primes
What is $n$? Any integer? An integer in $[0,(p-3)/4]$? What do you mean by the correct square root?
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Probability of multivariant gaussian random variables in different areas
The assumptions look incomplete. First, you should assume that the vector $(X_1,\ldots,X_4)$ is gaussian and not only its components $X_i$. Nest do you assume taht the $X_i$ is centered? If yes, write it explicitly.
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Is there something like a "self-avoiding Markov chain" on a continuous space?
To get a conformal bijection between the half upper plane and a square, look at en.wikipedia.org/wiki/Schwarz%E2%80%93Christoffel_mapping
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Is there something like a "self-avoiding Markov chain" on a continuous space?
Look at the wikipedia page added in the references.
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Is there something like a "self-avoiding Markov chain" on a continuous space?
I added a reference
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Holomorphic function on $\mathbb C^n$
Verbisty Be careful, $F$ is defined on $\Omega$, not on the whole space $\mathbb{C}^n$. So Liouville theorem does not apply.
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Adding an independent variable does not increase conditional information
Can you recall what the notation $I(U ; V | W)$ means?
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A convergence problem
I relaxed the assumption of independence, see my post.
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A convergence problem
I relaxed assumptions
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A convergence problem
You need an extra assumption. For example, imagine that all the draws give the same result as the first one. Then $x_n(b)$ will be $0$ or $1$.
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