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The Variance Problem Find $Var(Z)$, where: $$Z = \begin{cases}W + X_1 & \text{if } W \leq c \\ W + X_2 & \text{if } W > c \end{cases}$$ where $\quad W \sim N(\mu_0, \sigma_0^2), \quad X_1 \sim N(\mu_1, \sigma_1^2), \quad X_2 \sim N(\mu_2, \sigma_2^2)$ are independent, and $c$ is a constant. Solution Let $f(w, x_1,x_2)$ denote the joint pdf ...

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You can use BinormalDistribution instead of two univariate normal distributions: TransformedDistribution[Sqrt[x1^2 + x2^2], {x1, x2} ~Distributed~ BinormalDistribution[{0, 0}, {σ, σ}, 0]] (* RayleighDistribution[σ] *) TransformedDistribution[Sqrt[x1^2 + x2^2], {x1, x2} ~Distributed~ BinormalDistribution[{m, m}, {σ, σ}, 0]] (* RiceDistribution[Sqrt[2] m, ...

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You can get the desired result in two steps: d1 = TransformedDistribution[x1^2 + x2^2, {Distributed[x1, NormalDistribution[0, σ]], Distributed[x2, NormalDistribution[0, σ]]}] (* ExponentialDistribution[1/(2 σ^2)] *) d2 = TransformedDistribution[Sqrt[z], Distributed[z, d1]] (* RayleighDistribution[σ] *) PDF[d2,r] \$\begin{cases} \frac{r ...

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Fixed in 10.1 (windows) code S = {{2, 0.5}, {0.5, 1}}; Covariance[BinormalDistribution[{0, 0}, {Sqrt[2], 1}, 0.5/Sqrt[2]]]; Covariance[MultinormalDistribution[{0, 0}, S]]; Mean[BinormalDistribution[{0, 0}, {Sqrt[2], 1}, 0.5/Sqrt[2]]]; Mean[MultinormalDistribution[{0, 0}, S]]; NExpectation[x*y, {x, y} \[Distributed] BinormalDistribution[{0, 0}, ...

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The distribution of the means of sufficiently large sets of random drawings from a given distribution (well, many of them) approaches a normal distribution with a mean equal to the mean of the given distribution and a standard deviation equal to the standard deviation of the given distribution divided by the square root of the sample size -1. Using this ...

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