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is, $$\hat y = \Phi(\Phi^\top\Phi)^{-1}\Phi^\top y$$ But why isn't $\Phi(\Phi^\top\Phi)^{-1}\Phi^\top$ simply the identity matrix? … Observe, $\Phi(\Phi^\top\Phi)^{-1}\Phi^\top = \Phi(\Phi^{-1}{\Phi^\top}^{-1})\Phi^\top = I I = I$ Thanks! …

From Wikipedia, Gaussian Copula, it states that a Gaussian Copula can be defined as: $$ C_P(u_1, \ldots, u_d) = \boldsymbol{\Phi}_P(\Phi^{-1}(u_1), \ldots, \Phi^{-1}(u_d)), $$ where $\boldsymbol{\Phi … In that respect, it appears that for simulation, we are doing something like: $$ C_P(u_1, \ldots, u_d) = \boldsymbol{\Phi}_P(\Phi(u_1), \ldots, \Phi(u_d)), $$ instead. …

(y_i) $$ take derivative with respect to $\phi$ and set to 0 $$ \sum_i^n\frac{y_i\phi^{y_i-1}(1-\phi)^{1-y_i}-(\phi^{y_i})(1-y_i)(1-\phi)^{-y_i}}{\phi^{y_i}(1-\phi)^{1-y_i}} = 0 $$ there are 2 cases $y_i … had $$ y=\begin{bmatrix}1\\1\\0\end{bmatrix} $$ then we have $$ \frac{1}{\phi}+\frac{1}{\phi}-\frac{1}{1-\phi}=0 $$ $$ \phi=\frac{2}{3} $$ which is correct and what the indicator function would give but …

Assume that $\phi(\cdot)$ is an increasing function on $(0,\infty)$. Show that for each $t>0$, $P(|X| \ge t) \le E_\phi(X) / \phi(t)$. … The answer states that $P(|X| > t) \le 2E(\phi(X))/ \phi(t)$, but a comment implies that $P(|X| > t) \le E(\phi(X))/ \phi(t)$, which is what we are actually trying to show. …

Given that $\phi : \mathcal{X} → \mathcal{X}′$, prove that $k(u, v) = \tilde{k}(\phi(u), \phi(v))$. … I've seen similar proofs where if $\phi : \mathcal{X} → \mathcal{X}$, the transformation is simply in a transformation into a different kernel in the same domain (e.g. here, page 18). …

Bishop: Here $\mathbf\Phi$ represents the design matrix and $\beta$ is precision which is positive. My question is: why is $\mathbf\Phi^\top\mathbf\Phi$ a positive definite matrix? … So, can you please help me prove that $\mathbf\Phi^\top\mathbf\Phi$ is positive definite? Thanks a lot. …

I am frustrated of looking for a simple explanation of this term $\phi$-divergence, but I cannot find any. …

I have to prove the following: Let $\phi$ and $\Phi$ denote the standard normal density and distribution functions respectively. … Prove that $$\lim_{x\to\infty}{1-\Phi(x)\over \phi(x)/x}=1$$ I am not able to start. $\Phi$ is $\int \phi(x)dx$. How can I calculate the limit without L'Hopital's rule? …

Problem Derive that the asymptotic distribution of $\hat{\phi}:=[\hat{\phi}_1, \cdots, \hat{\phi}_p]$ follows, $$ \sqrt{n}(\hat{\phi} - \phi) \sim N(0, \sigma_\epsilon^2 \Gamma^{-1}) $$ where $\Gamma … But I cannot proceed to show the result I want, $$ \sqrt{n}(\hat{\phi} - \phi) \sim N(0, \sigma_\epsilon^2 \Gamma^{-1}) $$ …

Attempt 1: whuber suggestion \begin{align} E[\phi(\hat \alpha X)] & = E[E[\phi(\hat \alpha X)|X]] \\ & \le E[\phi(E[\hat \alpha X|X])] \\ & = E[\phi(X E[\hat \alpha|X])] \\ & = E[\phi(\alpha X)]. … When we do this we get $$ E[\phi(\hat \alpha X)] = E[E[\phi(\hat \alpha X)|X]] = E[E[\phi(\alpha X)|X]] = E[\phi(\alpha X)]. $$ So in this approach, it seems $E[\phi(\hat \alpha X)] = E[\phi(\alpha X)] …

I was looking at AR processes in Chris Brooks (2008) and at one point, when deriving the variance of an AR(1) process he writes this part $(1- \phi L)^{-1}* u_{t}$ as $(1+ \phi L+\phi^{2}L + \cdots)* u …

we have persp(x,y,z ,theta =, phi =) , now I want to know that if I want to imagine a 3d plot from a certain angle then where should I expect the theta and phi angle. …

How to calculate $\int_{-\infty}^{\infty} \Phi(y+c)^{k-1}d(\Phi(y))$? When I check the list of integrals of Gaussian functions, I only find $k-1=2$. …

So $$\mathbf\Phi^T\mathbf\Phi=\bigl(\mathbf\phi(\mathbf x_1),\mathbf\phi(\mathbf x_2),\ldots,\mathbf\phi(\mathbf x_N)\bigr) \left( {\begin{array}{*{20}{c}} \mathbf\phi^T(\mathbf x_1)\\ \mathbf\phi^T(\mathbf … x_2)\\ \vdots\\ \mathbf\phi^T(\mathbf x_N) \end{array}} \right) =\sum\limits_{n=1}^N \mathbf\phi(\mathbf x_n)\mathbf\phi^T(\mathbf x_n). …

For the standard normal distribution $\phi(x)$, we can see that $\phi'(x)=-x\phi(x)$. Put differently, $\frac{\mathrm{d}\ln(\phi(x))}{\mathrm{d} x}= -x $. …