```

```{solution-start} ar1p_ex1

:class: dropdown

```

Here is one solution:

```{code-cell} python3

from numba import njit

from scipy.special import factorial2

@njit

def sample_moments_ar1(k, m=100_000, mu_0=0.0, sigma_0=1.0, seed=1234):

np.random.seed(seed)

sample_sum = 0.0

x = mu_0 + sigma_0 * np.random.randn()

for t in range(m):

sample_sum += (x - mu_star)**k

x = a * x + b + c * np.random.randn()

return sample_sum / m

def true_moments_ar1(k):

if k % 2 == 0:

return std_star**k * factorial2(k - 1)

else:

return 0

k_vals = np.arange(6) + 1

sample_moments = np.empty_like(k_vals)

true_moments = np.empty_like(k_vals)

for k_idx, k in enumerate(k_vals):

sample_moments[k_idx] = sample_moments_ar1(k)

true_moments[k_idx] = true_moments_ar1(k)

fig, ax = plt.subplots()

ax.plot(k_vals, true_moments, label="true moments")

ax.plot(k_vals, sample_moments, label="sample moments")

ax.legend()

plt.show()

```

```{solution-end}

```

```{exercise}

:label: ar1p_ex2

```

```{exercise}

:label: ar1p_ex3

In the lecture we discussed the following fact: for the $AR(1)$ process

$$

X_{t+1} = a X_t + b + c W_{t+1}

$$

with $\{ W_t \}$ iid and standard normal,

$$

\psi_t = N(\mu, s^2) \implies \psi_{t+1}

= N(a \mu + b, a^2 s^2 + c^2)

$$

Confirm this, at least approximately, by simulation. Let

- $a = 0.9$

- $b = 0.0$

- $c = 0.1$

- $\mu = -3$

- $s = 0.2$

First, plot $\psi_t$ and $\psi_{t+1}$ using the true

distributions described above.

Second, plot $\psi_{t+1}$ on the same figure (in a different

color) as follows:

1. Generate $n$ draws of $X_t$ from the $N(\mu, s^2)$

distribution

1. Update them all using the rule

$X_{t+1} = a X_t + b + c W_{t+1}$

1. Use the resulting sample of $X_{t+1}$ values to produce a

density estimate via kernel density estimation.

Try this for $n=2000$ and confirm that the

simulation based estimate of $\psi_{t+1}$ does converge to the

theoretical distribution.

```

## Solutions

```{solution-start} ar1p_ex1

:class: dropdown

```

```{code-cell} python3

from numba import njit

from scipy.special import factorial2

@njit

def sample_moments_ar1(k, m=100_000, mu_0=0.0, sigma_0=1.0, seed=1234):

np.random.seed(seed)

sample_sum = 0.0

x = mu_0 + sigma_0 * np.random.randn()

for t in range(m):

sample_sum += (x - mu_star)**k

x = a * x + b + c * np.random.randn()

return sample_sum / m

def true_moments_ar1(k):

if k % 2 == 0:

return std_star**k * factorial2(k - 1)

else:

return 0

k_vals = np.arange(6) + 1

sample_moments = np.empty_like(k_vals)

true_moments = np.empty_like(k_vals)

for k_idx, k in enumerate(k_vals):

sample_moments[k_idx] = sample_moments_ar1(k)

true_moments[k_idx] = true_moments_ar1(k)

fig, ax = plt.subplots()

ax.plot(k_vals, true_moments, label="true moments")

ax.plot(k_vals, sample_moments, label="sample moments")

ax.legend()

plt.show()

```

```{solution-end}

```

```{solution-start} ar1p_ex2

:class: dropdown

```

```

```{exercise}

:label: ar1p_ex3

In the lecture we discussed the following fact: for the $AR(1)$ process

$$

X_{t+1} = a X_t + b + c W_{t+1}

$$

with $\{ W_t \}$ iid and standard normal,

$$

\psi_t = N(\mu, s^2) \implies \psi_{t+1}

= N(a \mu + b, a^2 s^2 + c^2)

$$

Confirm this, at least approximately, by simulation. Let

- $a = 0.9$

- $b = 0.0$

- $c = 0.1$

- $\mu = -3$

- $s = 0.2$

First, plot $\psi_t$ and $\psi_{t+1}$ using the true

distributions described above.

Second, plot $\psi_{t+1}$ on the same figure (in a different

color) as follows:

1. Generate $n$ draws of $X_t$ from the $N(\mu, s^2)$

distribution

1. Update them all using the rule

$X_{t+1} = a X_t + b + c W_{t+1}$

1. Use the resulting sample of $X_{t+1}$ values to produce a

density estimate via kernel density estimation.

Try this for $n=2000$ and confirm that the

simulation based estimate of $\psi_{t+1}$ does converge to the

theoretical distribution.

```

```{solution-start} ar1p_ex3

:class: dropdown

```