.ipynb .pdf

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Part 1

In this part you will simulate a sample of 4 observations from a stationary AR(1) process with $\alpha =0.8$, ${\sigma }^2=1$

# TODO
# import numpy

Generate a $4D$ vector of innovations using gen = np.random.default_rng(100)

# TODO
#eps = ...

Using ${\epsilon }_0$, generate $z_0$ from the stationary distribution of $\mathbf{z}$

Hint: see here

Using a for loop, generate the remaining 3 realizations of $\mathbf{z}$

# TODO
#for ....

As discussed in a lecture (see again here) there is a linear relationship between $\mathbf{z}$ and $\mathit{\epsilon }$, In particular, there is a $(T+1)\times (T+1)$ matrix $\mathit{A}$ such that

$$\begin{array}{r}\underset{\mathbf{z}}{\underbrace{\left[\begin{array}{c}{z}_0\\ z_1\\ z_2\\ z_3\\ z_4\\ ⋮\\ z_T\end{array}\right]}}=\underset{\mathit{A}}{\underbrace{\left[\begin{array}{cccccccc}\frac{1}{\sqrt{1-{\alpha }^2}}& 0& 0& 0& \cdots & 0& 0& 0\\ \frac{\alpha }{\sqrt{1-{\alpha }^2}}& 1& 0& 0& \cdots & 0& 0& 0\\ \frac{{\alpha }^2}{\sqrt{1-{\alpha }^2}}& \alpha & 1& 0& \cdots & 0& 0& 0\\ ⋮& ⋮& ⋮& ⋮& \cdots & ⋮& ⋮& ⋮\\ \frac{{\alpha }^T}{\sqrt{1-{\alpha }^2}}& {\alpha }^{T-1}& {\alpha }^{T-2}& {\alpha }^{T-3}& \cdots & {\alpha }^2& \alpha & 1\end{array}\right]}}\underset{\mathit{\epsilon }}{\underbrace{\left[\begin{array}{c}{\epsilon }_0\\ {\epsilon }_1\\ {\epsilon }_2\\ {\epsilon }_3\\ ⋮\\ {\epsilon }_{T-1}\\ {\epsilon }_T\end{array}\right]}}\end{array}$$

Build the $4x4$ matrix $\mathit{A}$, and confirm that the equality $\mathbf{z}=\mathit{A}\mathit{\epsilon }$ holds

# TODO
#A =

#np.testing.assert_array_almost_equal(z, A @ eps)

An alternative (and perhaps simpler) way to represent the linear relationship between $\mathbf{z}$ and $\mathit{\epsilon }$, is with a matrix $\mathit{B}$, such that

$$\begin{array}{r}\underset{\mathit{B}}{\underbrace{\left[\begin{array}{cccccccc}\sqrt{1-{\alpha }^2}& 0& 0& 0& \cdots & 0& 0& 0\\ -\alpha & 1& 0& 0& \cdots & 0& 0& 0\\ 0& -\alpha & 1& 0& \cdots & 0& 0& 0\\ ⋮& ⋮& ⋮& ⋮& \cdots & ⋮& ⋮& ⋮\\ 0& 0& 0& 0& \cdots & 0& -\alpha & 1\end{array}\right]}}\underset{\mathit{z}}{\underbrace{\left[\begin{array}{c}{z}_0\\ z_1\\ z_2\\ ⋮\\ z_T\end{array}\right]}}=\underset{\epsilon }{\underbrace{\left[\begin{array}{c}{\epsilon }_0\\ {\epsilon }_1\\ {\epsilon }_2\\ ⋮\\ {\epsilon }_T\end{array}\right]}}\end{array}$$

Using pen and paper, confirm for yourself that this relationship indeed holds. Then, build the matrix $\mathit{B}$ and check programmatically if it is invertible (the matrix rank is equal to the number of columns/rows, i.e. it is full rank)

Hint: check the documentation of Numpy for a function that returns the rank of a matrix. Use an assertion to confirm that $\mathit{B}$ is full rank.

# TODO
#B =

# assert full rank...

Verify that the equality $\mathbf{z}={\mathit{B}}^{-1}\mathit{\epsilon }$ holds

# TODO
#np.testing.assert_array_almost_equal(...)

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Part 2