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It all follows from the properties of multivariate normals. Since $X_i$ are independent and normally distributed, they're jointly normal, which means any of their linear combination is also jointly normal with them. So, $p_{\mathbf{X},Z_N}(\mathbf{x},z)$ is a multivariate normal, which in turn means $p_{X_i,Z_N}(x,z)$ is multivariate normal with $$\mu=\... 2 It is just the elementary inequality$$P(A)+P(B)-1\le P(A\cap B)\le \sqrt{P(A)P(B)}$$for events A=\{X\le x\} and B=\{Y\le y\}. There is no need to go into distributions. Since P(A\cap B)\le P(A) and P(A\cap B)\le P(B), we have$$(P(A\cap B))^2\le P(A)P(B)$$And P(A^c\cup B^c)\le P(A^c)+P(B^c)=1-P(A)+1-P(B) implies$$P(A\cap B)=1-P(A^c\cup B^c)\...

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The accepted $X$ can be written as $$X=Y_1\mathbb I_{U_1\le f_X(Y_1)/c f_Y(Y_1)}+Y_2\mathbb I_{U_1> f_X(Y_1)/c f_Y(Y_1)}\mathbb I_{U_2\le f_X(Y_2)/c f_Y(Y_2)}+\cdots$$ It is therefore the transform of the whole sequence $(Y_1,U_1,Y_2,U_2,Y_3,\ldots)$ and not of a single pair $(Y_1,U_1)$. To derive the distribution of such an $X$, one cannot proceed by a ...

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