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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=\...


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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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