MICRO-(c) Consider the random walk process Y_t=Y_t

Sure, let's break down the problem step by step.

Solution

a

Given the random walk process

Y_{t}=Y_{t-1}+\epsilon_{t}

, where

\epsilon_{t}

is a zero mean white noise process with variance

\sigma_{\epsilon}^{2}

, we need to find the variance of the sample mean

\bar{Y}_{n}=\sum_{t=1}^{n} Y_{t} / n

b

The variance of the sample mean

\bar{Y}_{n}

can be expressed as

V\left[\bar{Y}_{n}\right] = \frac{1}{n^2} V\left[\sum_{t=1}^{n} Y_{t}\right]

c

Since

Y_{t}

is a random walk,

Y_{t} = Y_{0} + \sum_{i=1}^{t} \epsilon_{i}

. Therefore,

V\left[Y_{t}\right] = V\left[Y_{0} + \sum_{i=1}^{t} \epsilon_{i}\right] = \sum_{i=1}^{t} V\left[\epsilon_{i}\right] = t \sigma_{\epsilon}^{2}

d

Now,

V\left[\sum_{t=1}^{n} Y_{t}\right] = \sum_{t=1}^{n} V\left[Y_{t}\right] = \sum_{t=1}^{n} t \sigma_{\epsilon}^{2} = \sigma_{\epsilon}^{2} \sum_{t=1}^{n} t = \sigma_{\epsilon}^{2} \frac{n(n+1)}{2}

e

Therefore,

V\left[\bar{Y}_{n}\right] = \frac{1}{n^2} \cdot \sigma_{\epsilon}^{2} \cdot \frac{n(n+1)}{2} = \frac{\sigma_{\epsilon}^{2} (n+1)}{2n}

f

As

n \rightarrow \infty

,

V\left[\bar{Y}_{n}\right] \rightarrow \frac{\sigma_{\epsilon}^{2}}{2}

. This indicates that the variance of the sample mean does not go to zero as

n

increases, implying that the sample mean

\bar{Y}_{n}

is not a reliable estimator of

E\left[Y_{t}\right]

for a random walk process

Answer

The variance of the sample mean

\bar{Y}_{n}

is

\frac{\sigma_{\epsilon}^{2} (n+1)}{2n}

, and as

n \rightarrow \infty

, it approaches

\frac{\sigma_{\epsilon}^{2}}{2}

. This means that the sample mean is not a reliable estimator of the expected value of

Y_{t}

in a random walk process.

Key Concept

Variance of the sample mean in a random walk process

Explanation

The variance of the sample mean

\bar{Y}_{n}

does not go to zero as the sample size

n

increases, indicating that the sample mean is not a reliable estimator of the expected value of

Y_{t}

in a random walk process.

Solution

a

Given the process

Z_{t}=\epsilon_{t}-\theta \epsilon_{t-12}

, where

\epsilon_{t}

is a zero mean white noise process with variance

\sigma_{\epsilon}^{2}

, we need to derive the autocorrelation function of

Z_{t}

b

The autocorrelation function

\rho_k

of

Z_t

is defined as

\rho_k = \frac{\gamma_k}{\gamma_0}

, where

\gamma_k

is the autocovariance function at lag

k

and

\gamma_0

is the variance of

Z_t

c

First, calculate the variance

\gamma_0

of

Z_t

:

\gamma_0 = E[Z_t^2] = E[(\epsilon_t - \theta \epsilon_{t-12})^2] = E[\epsilon_t^2] + \theta^2 E[\epsilon_{t-12}^2] = \sigma_{\epsilon}^2 + \theta^2 \sigma_{\epsilon}^2 = (1 + \theta^2) \sigma_{\epsilon}^2

d

Next, calculate the autocovariance

\gamma_k

for

k \neq 0

:

\gamma_k = E[Z_t Z_{t-k}] = E[(\epsilon_t - \theta \epsilon_{t-12})(\epsilon_{t-k} - \theta \epsilon_{t-12-k})]

For

k = 12

:

\gamma_{12} = E[(\epsilon_t - \theta \epsilon_{t-12})(\epsilon_{t-12} - \theta \epsilon_{t-24})] = -\theta \sigma_{\epsilon}^2

For

k = 0

:

\gamma_0 = (1 + \theta^2) \sigma_{\epsilon}^2

For

k \neq 0, 12

:

\gamma_k = 0

e

Finally, the autocorrelation function

\rho_k

is:

\rho_k = \frac{\gamma_k}{\gamma_0}

For

k = 12

:

\rho_{12} = \frac{-\theta \sigma_{\epsilon}^2}{(1 + \theta^2) \sigma_{\epsilon}^2} = \frac{-\theta}{1 + \theta^2}

For

k \neq 0, 12

:

\rho_k = 0

Answer

The autocorrelation function of

Z_t

is

\rho_{12} = \frac{-\theta}{1 + \theta^2}

for

k = 12

and

\rho_k = 0

for

k \neq 0, 12

.

Key Concept

Autocorrelation function of a time series process

Explanation

The autocorrelation function measures the correlation between values of the process at different times. For the given process

Z_t

, the autocorrelation is non-zero only at lag 12, indicating a specific periodic relationship.

Solution

a

Define the process: The given process is

X_{t}=\phi X_{t-4}+\epsilon_{t}

, where |\phi|<1 and

\epsilon_{t}

is a zero mean white noise process with variance

\sigma_{\epsilon}^{2}

b

Calculate the mean: Since

\epsilon_{t}

is a zero mean white noise process, the mean of

X_{t}

is also zero

c

Calculate the variance: The variance of

X_{t}

can be derived using the fact that

\epsilon_{t}

is white noise with variance

\sigma_{\epsilon}^{2}

. The variance of

X_{t}

is given by

\sigma_{X}^{2} = \frac{\sigma_{\epsilon}^{2}}{1-\phi^2}

d

Define the autocovariance function: The autocovariance function

\gamma_k

for lag

k

is defined as

\gamma_k = \mathbb{E}[(X_t - \mu)(X_{t-k} - \mu)]

. Since

\mu = 0

, this simplifies to

\gamma_k = \mathbb{E}[X_t X_{t-k}]

e

Calculate the autocovariance for lag 0: For

k=0

,

\gamma_0 = \mathbb{E}[X_t^2] = \sigma_{X}^{2} = \frac{\sigma_{\epsilon}^{2}}{1-\phi^2}

f

Calculate the autocovariance for lag 4: For

k=4

,

\gamma_4 = \mathbb{E}[X_t X_{t-4}] = \phi \mathbb{E}[X_{t-4}^2] = \phi \sigma_{X}^{2} = \phi \frac{\sigma_{\epsilon}^{2}}{1-\phi^2}

g

Calculate the autocovariance for other lags: For

k \neq 0, 4

,

\gamma_k = 0

because

X_t

and

X_{t-k}

are uncorrelated for these lags

h

Define the autocorrelation function: The autocorrelation function

\rho_k

is given by

\rho_k = \frac{\gamma_k}{\gamma_0}

i

Calculate the autocorrelation for lag 0: For

k=0

,

\rho_0 = \frac{\gamma_0}{\gamma_0} = 1

j

Calculate the autocorrelation for lag 4: For

k=4

,

\rho_4 = \frac{\gamma_4}{\gamma_0} = \phi

k

Calculate the autocorrelation for other lags: For

k \neq 0, 4

,

\rho_k = 0

Answer

The autocorrelation function of the process

X_{t}=\phi X_{t-4}+\epsilon_{t}

is given by: -

\rho_0 = 1

-

\rho_4 = \phi

-

\rho_k = 0

for

k \neq 0, 4

Key Concept

Autocorrelation Function

Explanation

The autocorrelation function measures the correlation between values of the process at different times. For the given process, the autocorrelation is non-zero only at lags 0 and 4, reflecting the structure of the process.

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