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3

Using $q = 1-p$ we can work out the root as: $$\sqrt{1-4pq} = \sqrt{1-4p(1-p)} = \sqrt{1-4p+4p^2} = \sqrt{(1-2p)^2}$$ Taking the positive root reduces this to $(1-2p)$. This gives for the fraction: $$\frac{1 + \sqrt{1-4pq}}{2p} = \frac{1 + (1-2p)}{2p} = \frac{1-p}{p}$$ This also holds inside the logarithm.

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It's true that in general if $\sum_i f_i g_i = \sum_i f_i h_i$, we do not automatically have $g_i = h_i$. But this sum is special, because all $f_i$ are monomials (i.e. of the form $\alpha^n$). This makes the sum a power series (of the form $\sum_{n=0}^\infty \alpha^n C_n$), and these series have a lot of nice properties such as continuity and ...

1

If you have a vector $X = (X_1,\ldots,X_n)$ of a multivariate normal distribution with covariance matrix $\Sigma$ and $F_i$ is the marginal cumulative distribution function of $X_i$ then $F_i(X_i)$ is uniformly distributed. So what you can do: generate uniforms (e.g. Sobol or Halton) transform to uncorrelated Gaussians transform these Gaussians to ...

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