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The notation $E(f)$ is most commonly used in probability theory and this means simply just $\int_{X}fd\mu$ where $(X,\mu)$ is your measure space (it is called expectation). Therefore the only difference between those two relations is the fact that the second uses real scalar product. In other words $g$ is real if and only if $\overline{g(t)}=g(t)$ for all ...


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3) Assuming that $y$ and $x$ are continuous, you probably expected to estimate the regression model $$ y = \beta_0 + \beta_1x + \beta_2z + \epsilon, $$ and report the $\beta_1$ with its standard error, while "controlling" for $z$, i.e., (probably), by estimating the model you are addressing the effect of $x$ on $y$ in the presence of $z$. And when ...


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Your idea is correct, but you have some typo and some mistake. From the first equation we have: $$ \phi_1=\rho(1)(1-\phi_2) $$ and, substituting, the second equation becomes: $$ \rho(2)=\frac{\rho^2(1)(1-\phi_2)^2+\phi_2[1-\rho(1)(1-\phi_2)]}{1-\phi_2} $$ that, for $1-\phi_2 \ne 0$, gives: $$ ...


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I think you can constrain $P[A,B,C]$ to an interval such as $$\max(0,\, P[A,B]+P[A,C]-P[A], \,P[A,B]+P[B,C]-P[B], \,P[A,C]+P[B,C]-P[C]) \\ \le P[A,B,C] \le \\ \min(P[A,B],\,P[A,C],\,P[B,C],\,1+P[A,B]+P[A,C]+P[B,C]-P[A]-P[B]-P[C]).$$ As an example, if $P[A]=P[B]=P[C]=\frac12$ and $P[A,B]=P[A,C]=P[B,C]=\frac14$ then $0 \le P[A,B,C] \le \frac14$.


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I use the fact that any linear combination of normal random variables is a normal random variable, regardless of independence. For a proof, see the answers to this question. As shown in the second answer to this linked question, by characteristic functions, the variance of $Y = X_1+X_2$ is shown to be $$\sigma^2_Y = \sigma^2_{X_1} + \sigma^2_{X_2} + ...



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