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Here's an example: The probability mass function for a binomial random variable is given by $$\mathrm{Pr}(X=i)=f(i; n, p) = \begin{pmatrix} n \\ i\end{pmatrix}p^i(1-p)^{n-i}.$$ Expand the generating function $G(z) = [(1-p)+pz]^n$ using the binomial theorem: $$G(z) = \sum_{k=0}^n \begin{pmatrix} n \\ k\end{pmatrix} (pz)^k(1-p)^{n-k}.$$ Taking the $i$th ...

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Rather than looking at the player, I prefer to explain the paradox from the host's standpoint, as this only involves one step. As the player gets one door, the host gets two. There are 3 possibilities with the same probability: donkey-donkey => leaves a donkey after a door is open car-donkey => leaves the car donkey-car => leaves the car So in two cases ...

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You have this matrix formulas (Here $Y$ is a random (column) vector, $A$ is known constant matrix): $$\text{E} AY = A \text{E}Y$$ $$\text{Var} AY = A \text{Var}(Y) A^{T}.$$ The first one gives your result 1). For the second one, calculate $$\text{Var} U^T (X-m) = U^T \text{Var}(X-m) U = U^T \Sigma U = U^T U D U^T U = D$$ (since \$U U^T = ...

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This looks relevant. Kais Hamza and Fima C. Klebaner, “A Family of Non-Gaussian Martingales with Gaussian Marginals,” Journal of Applied Mathematics and Stochastic Analysis, vol. 2007, Article ID 92723, 19 pages, 2007. doi:10.1155/2007/92723 http://www.hindawi.com/journals/ijsa/2007/092723/abs/

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