Consider the random variable $$y = c^T x$$Then the mean of $y$ is $$E(y) = c^T E(x) = c^T \mu$$ and variance of $y$ is $$var(y) = E(y - c^T \mu)^2 = E(y^2) - (c^T \mu)^2 = E(y^2) + c^T \mu\mu^T c \... View answer Accepted answer 5 votes It doesn't have to be thru Yule-Walker. You can use the OLS approach, namely,$$X_t = \pmb X_{t-1}^T \pmb \phi + n_t , \qquad t = p+1,p+2,\ldots$$where \pmb X_{t-1} = [X_{t-1} \ldots X_{t-p} ]and ... View answer Accepted answer 3 votes PDF We can first derive the PDF using convolution of two PDFs: Case 1: If 0 \leq y \leq 1, then f_{X_1}(y - x_2) = 1 if  0\leq x_2 \leq y, and f_{X_1}(y-x_2) = 0 if x_2 > y. This means ... View answer Accepted answer 2 votes Adding a constant X\beta to a normally distributed random variable is again normal. The mean of the y = X\beta + \epsilon is:$$E(y) = E(X\beta + \epsilon) = E(X\beta) + E(\epsilon) = X\beta + 0 ...

I will share a picture with you to clear your ambiguities. Assume you've got the training data in 2D space that are labeled either red or green. On the left figure, you've got a test data point (in ...

Given \begin{equation} X \sim {{\chi}_{2n}}^2 \end{equation} and \begin{equation} Y \sim P(\lambda) \end{equation} Let's prove (the complement of your expression) \begin{equation} P(X < 2\lambda)...

$y = X \beta + \epsilon$ is your model. After estimating $\beta$, which is denoted as $\hat{\beta}$, using any estimation method, you could then use it to estimate your $y$, i.e. $\hat{y} = X\hat{\... View answer 1 votes Part$(a)$seems legit. Computing$E(\hat{\beta})$Taking it from where you stopped \begin{equation} \hat{\beta}=\frac{\sum y_i+2y_ix_i}{\sum1+4x_i+4x_i^2} \end{equation} Then \begin{equation}... View answer Accepted answer 0 votes Using the definition of inner product $$y^Ty =\overbrace{\sum_{t=1}^{T} y^2_t}^{\textsf{given}}$$$$\hat {\beta^T}X^Ty = \hat{\beta}_1\overbrace{\sum_{t=1}^{T} x_{t1} y_t}^{\textsf{given}} + \hat{\... View answer 0 votes Disclaimer: This is not my proof, but rather an understanding of one of Lukács theorems. First transform$X_k$using orthogonal matrix$V$so that it contains entries equal to$\frac{1}{ \sqrt{n}}\$ ...