# Thread: Time Series: Theory and Methods

1. ## Time Series: Theory and Methods

Can someone help me with the follow textbook question from Time Series: Theory and Methods by Brockwell.

$\displaystyle Suppose\, that\, m_t = c_0 + c_1 + c_2t^2, \,t=0,\, \pm1,\, \pm2, ...$

a) $\displaystyle Show \, that\, m_t = \sum^2_{i=-2} a_im_{t+1} = \sum^3_{i=-3} b_im_{t+1}, \,t=0,\, \pm1,\, \pm2, ...$

$\displaystyle \, where \, a_2=a_{-2}=\frac{3}{35}, \, a_1=a_{-1}=\frac{7}{35}, a_0=\frac{17}{35}, \, b_3=b_{-3}=\frac{2}{21},\, b_2=b_{-2}=\frac{3}{21},\, b_1=b_{-1}=\frac{6}{21}, b_0=\frac{7}{21}$
I can do this with ease.

But i am stuck with this one :

b. $\displaystyle Suppose\, that\, X_t =m_t + Z_t \,where\, (Z_t, \, t=0,\, \pm1,\, \pm2, ...),$ $\displaystyle \, is\, a\, sequence\, of\, independent\, normal\, random\, variables,\, each\, with\, mean\, 0\, and\, variance\, \sigma^2.$
$\displaystyle \, Let\, U_t = \sum^2_{i=-2}{a_iX_{t+i}}\, and\, V_t=\sum^3_{i=-3}{a_iX_{t+i}}.$

$\displaystyle i Find\, the\, means\, and\, variances\, of\, U_t\, and\, V_t\,$

$\displaystyle ii. Find\, the\, Correlations\, between\, U_t\,and\,U_{t+1}\,and\,between\,V_t\,and\,V_{t+1}$

Thank=yoU!!

Thankyou!!

2. ## Re: Time Series: Theory and Methods

Can anyone help me with finding the mean and variances for bi?

3. ## Re: Time Series: Theory and Methods

Originally Posted by lpd
Can someone help me with the follow textbook question from Time Series: Theory and Methods by Brockwell.

$\displaystyle Suppose\, that\, m_t = c_0 + c_1 + c_2t^2, \,t=0,\, \pm1,\, \pm2, ...$

a) $\displaystyle Show \, that\, m_t = \sum^2_{i=-2} a_im_{t+1} = \sum^3_{i=-3} b_im_{t+1}, \,t=0,\, \pm1,\, \pm2, ...$

$\displaystyle \, where \, a_2=a_{-2}=\frac{3}{35}, \, a_1=a_{-1}=\frac{7}{35}, a_0=\frac{17}{35}, \, b_3=b_{-3}=\frac{2}{21},\, b_2=b_{-2}=\frac{3}{21},\, b_1=b_{-1}=\frac{6}{21}, b_0=\frac{7}{21}$
I can do this with ease.

But i am stuck with this one :

b. $\displaystyle Suppose\, that\, X_t =m_t + Z_t \,where\, (Z_t, \, t=0,\, \pm1,\, \pm2, ...),$ $\displaystyle \, is\, a\, sequence\, of\, independent\, normal\, random\, variables,\, each\, with\, mean\, 0\, and\, variance\, \sigma^2.$
$\displaystyle \, Let\, U_t = \sum^2_{i=-2}{a_iX_{t+i}}\, and\, V_t=\sum^3_{i=-3}{a_iX_{t+i}}.$

$\displaystyle i Find\, the\, means\, and\, variances\, of\, U_t\, and\, V_t\,$

$\displaystyle ii. Find\, the\, Correlations\, between\, U_t\,and\,U_{t+1}\,and\,between\,V_t\,and\,V_{t+1}$

Thank=yoU!!

Thankyou!!

$\displaystyle U_t=\sum_{i=-2}^2 X_{t+1}=\sum_{i=-2}^2\left( m_{t+i}+Z_{t+i}\right)=\sum_{i=-2}^2 m_{t+i}+ \sum_{i=-2}^2Z_{t+i}$

So the mean is:

$\displaystyle \overline{U_t}=\sum_{i=-2}^2 m_{t+i}$

and:

$\displaystyle {\text{Var}}(U_t) = 5 \sigma^2$

CB

4. ## Re: Time Series: Theory and Methods

Originally Posted by CaptainBlack
$\displaystyle U_t=\sum_{i=-2}^2 X_{t+1}=\sum_{i=-2}^2\left( m_{t+i}+Z_{t+i}\right)=\sum_{i=-2}^2 m_{t+i}+ \sum_{i=-2}^2Z_{t+i}$

So the mean is:

$\displaystyle \overline{U_t}=\sum_{i=-2}^2 m_{t+i}$

and:

$\displaystyle {\text{Var}}(U_t) = 5 \sigma^2$

CB
Thanks for that...

So let me try... could you check my working. I am a tad lost still...

$\displaystyle U_t=\sum_{i=-2}^2 a_i X_{t+1}=\sum_{i=-2}^2\left( a_i m_{t+i}+ a_iZ_{t+i}\right)=\sum_{i=-2}^2 a_i m_{t+i}+ \sum_{i=-2}^2 a_i Z_{t+i}$

So the mean is:

$\displaystyle \overline{U_t}=\sum_{i=-2}^2 a_i m_{t+i} = m_t$ from part a.

and:
$\displaystyle Var[U_t] = Var[m_t] + Var[\sum^2_{i=-2}a_iZ_{t+i}] = 0 + (\sum^2_{i=-2}a_i)^2Var[Z_{t+i}]$

$\displaystyle Var[U_t] = 0 + 1 \times \sigma^2 = \sigma^2$

I'm not sure how I can get 5. What am I doing wrong? ($\displaystyle {\text{Var}}(U_t) = 5 \sigma^2$)

Or can I do something like...
$\displaystyle Var(U_t) = E(U_t^2) - (E(U_t))^2?$
$\displaystyle (E(U_t))^2 = m_t^2$
$\displaystyle E(U_t^2) = (m_t + 2\sum^2_{i=-2}a_iZ_{t+i})^2 = m_t^2 + m_t\sum^2_{i=-2}a_iZ_{t+i}) + (\sum^2_{i=-2}a_iZ_{t+i})^2$

$\displaystyle Var(U_t) = m_t^2 + 2m_t\sum^2_{i=-2}a_iZ_{t+i} + (\sum^2_{i=-2}a_iZ_{t+i})^2 - m_t^2$
$\displaystyle Var(U_t) = 2m_t\sum^2_{i=-2}a_iZ_{t+i} + (\sum^2_{i=-2}a_iZ_{t+i})^2$
$\displaystyle Var(U_t) = 2m_t\sigma^2 + \sigma^4$
$\displaystyle Var(U_t) = \sigma^2(2m_t + \sigma^2)=2m_t\sigma^2 + \sigma^4$

5. ## Re: Time Series: Theory and Methods

Originally Posted by lpd
Thanks for that...

So let me try... could you check my working. I am a tad lost still...

$\displaystyle U_t=\sum_{i=-2}^2 a_i X_{t+1}=\sum_{i=-2}^2\left( a_i m_{t+i}+ a_iZ_{t+i}\right)=\sum_{i=-2}^2 a_i m_{t+i}+ \sum_{i=-2}^2 a_i Z_{t+i}$

So the mean is:

$\displaystyle \overline{U_t}=\sum_{i=-2}^2 a_i m_{t+i} = m_t$ from part a.

and:
$\displaystyle Var[U_t] = Var[m_t] + Var[\sum^2_{i=-2}a_iZ_{t+i}] = 0 + (\sum^2_{i=-2}a_i)^2Var[Z_{t+i}]$

$\displaystyle Var[U_t] = 0 + 1 \times \sigma^2 = \sigma^2$

I'm not sure how I can get 5. What am I doing wrong? ($\displaystyle {\text{Var}}(U_t) = 5 \sigma^2$)
the 5 was because I had used unit weights, what you have is nearly correct at the next to last line above

$\displaystyle Var[U_t] = Var[m_t] + Var\left[\sum^2_{i=-2}a_iZ_{t+i}\right] = 0 + \sum^2_{i=-2}a_i^2Var[Z_{t+i})]$

.............$\displaystyle =\sum^2_{i=-2}a_i^2 \sigma^2=\sigma^2 \sum^2_{i=-2}a_i^2$

We are using the independednce of the $\displaystyle Z$ s to replace the variance of the sum by the sum of the variances

Or can I do something like...
$\displaystyle Var(U_t) = E(U_t^2) - (E(U_t))^2?$
$\displaystyle (E(U_t))^2 = m_t^2$
$\displaystyle E(U_t^2) = (m_t + 2\sum^2_{i=-2}a_iZ_{t+i})^2 = m_t^2 + m_t\sum^2_{i=-2}a_iZ_{t+i}) + (\sum^2_{i=-2}a_iZ_{t+i})^2$

$\displaystyle Var(U_t) = m_t^2 + 2m_t\sum^2_{i=-2}a_iZ_{t+i} + (\sum^2_{i=-2}a_iZ_{t+i})^2 - m_t^2$
$\displaystyle Var(U_t) = 2m_t\sum^2_{i=-2}a_iZ_{t+i} + (\sum^2_{i=-2}a_iZ_{t+i})^2$
$\displaystyle Var(U_t) = 2m_t\sigma^2 + \sigma^4$
$\displaystyle Var(U_t) = \sigma^2(2m_t + \sigma^2)=2m_t\sigma^2 + \sigma^4$
You lose expectation operators part way through that should still be there, and have more than one operation that looks dubious to me.

CB

6. ## Re: Time Series: Theory and Methods

Originally Posted by CaptainBlack
the 5 was because I had used unit weights, what you have is nearly correct at the next to last line above

$\displaystyle Var[U_t] = Var[m_t] + Var\left[\sum^2_{i=-2}a_iZ_{t+i}\right] = 0 + \sum^2_{i=-2}a_i^2Var[Z_{t+i})]$

.............$\displaystyle =\sum^2_{i=-2}a_i^2 \sigma^2=\sigma^2 \sum^2_{i=-2}a_i^2$

We are using the independence of the $\displaystyle Z$ s to replace the variance of the sum by the sum of the variances

CB
Oh I See. So, $\displaystyle \sigma^2 \sum^2_{i=-2}a_i^2$ would look something like $\displaystyle \sigma^2 (2(\frac{3}{35})^2+ 2(\frac{12}{35})^2+ (\frac{17}{35})^2)$

How would I tackle the next part.
bii. $\displaystyle Find\,the\,correlations\, between\, U_{t}\, and\, U_{t+1}\,$
Do I do something like $\displaystyle Cov(U_t, U_{t+1})$?

$\displaystyle =E[(U_t-E(U_t))(U_{t+1}-E(U_{t+1}))]$
I should get something like...

$\displaystyle =E((\sum^2_{i=-2}a_i Z_{t+i})$$\displaystyle (\sum^2_{i=-2}a_{i+1} Z_{t+1+i})) Then, is the correlation just equal to zero? 7. ## Re: Time Series: Theory and Methods Originally Posted by lpd Oh I See. So, \displaystyle \sigma^2 \sum^2_{i=-2}a_i^2 would look something like \displaystyle \sigma^2 (2(\frac{3}{35})^2+ 2(\frac{12}{35})^2+ (\frac{17}{35})^2) How would I tackle the next part. bii. \displaystyle Find\,the\,correlations\, between\, U_{t}\, and\, U_{t+1}\, Do I do something like \displaystyle Cov(U_t, U_{t+1})? \displaystyle =E[(U_t-E(U_t))(U_{t+1}-E(U_{t+1}))] I should get something like... \displaystyle =E((\sum^2_{i=-2}a_i Z_{t+i})$$\displaystyle (\sum^2_{i=-2}a_{i+1} Z_{t+1+i}))$
Then, is the correlation just equal to zero?
I will assume that you have a definition of the correlation something like:

$\displaystyle {\rm{Cor}}(U,V)=\frac{{\rm{E}}[(U-\overline{U})(V-\overline{V})]}{\sigma_U \sigma_V}$

essentially the normalised covariance.

CB

8. ## Re: Time Series: Theory and Methods

Originally Posted by CaptainBlack
I will assume that you have a definition of the correlation something like:

$\displaystyle {\rm{Cor}}(U,V)=\frac{{\rm{E}}[(U-\overline{U})(V-\overline{V})]}{\sigma_U \sigma_V}$

essentially the normalised covariance.

CB
I'm stuck on bii now.

Basically I got up to here... after a few cancellations in expanding the covariance equation out. $\displaystyle Cov(U_t, U_{t+1}) = ... = E[(\sum_{i=-2}^2 a_i Z_{t+i})(\sum_{i=-2}^2 a_{i+1} Z_{t+i+1})]$

How do I expand this out :$\displaystyle E((\sum_{i=-2}^2 a_i Z_{t+i})(\sum_{i=-2}^2 a_{i+1} Z_{t+i+1}))$

Can i say something like... $\displaystyle \sum_{i=-2}^2 a_i \sum_{i=-2}^2 a_{j+1}E[Z_{t+i}]E[Z_{t+j}]=0\,if\, i\neq j?$ $\displaystyle where\, j=i+1$
And then get something like...

$\displaystyle E[\sum_{i=-2}^1 a_i a_{i+1} Z^2_{t+i+j}] = (\sum_{i=-2}^1 a_i a_{i+1}) \sigma^2$?

9. ## Re: Time Series: Theory and Methods

Originally Posted by lpd
I'm stuck on bii now.

Basically I got up to here... after a few cancellations in expanding the covariance equation out. $\displaystyle Cov(U_t, U_{t+1}) = ... = E[(\sum_{i=-2}^2 a_i Z_{t+i})(\sum_{i=-2}^2 a_{i+1} Z_{t+i+1})]$

How do I expand this out :$\displaystyle E((\sum_{i=-2}^2 a_i Z_{t+i})(\sum_{i=-2}^2 a_{i+1} Z_{t+i+1}))$
Because the $\displaystyle Z$s are zero mean and independent $\displaystyle E(Z_m Z_n)=0$ if $\displaystyle m\ne n$ and $\displaystyle =\sigma^2$ if $\displaystyle m=n$

$\displaystyle E\left[\left(\sum_{i=-2}^2 a_i Z_{t+i}\right)\left(\sum_{i=-2}^2 a_{i+1} Z_{t+i+1}\right)\right]=E\left[(a_{-2}Z_{t-2}+..+a_2Z_2)(a_{-1}Z_{t-1}+..+a_2Z_3) \right]$

.......... $\displaystyle =\sigma^2\sum_{i=-1}^2a_i$

CB