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Math Help - Time Series: Theory and Methods

  1. #1
    lpd
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    Time Series: Theory and Methods

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

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

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

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

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

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

    Thank=yoU!!

    Thankyou!!
    Last edited by lpd; August 17th 2011 at 06:23 PM.
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  2. #2
    lpd
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    Re: Time Series: Theory and Methods

    Can anyone help me with finding the mean and variances for bi?
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  3. #3
    Grand Panjandrum
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    Re: Time Series: Theory and Methods

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

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

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

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

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

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

    Thank=yoU!!

    Thankyou!!

    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:

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

    and:

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

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

    Quote Originally Posted by CaptainBlack View Post
    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:

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

    and:

    {\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...

    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:

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

    and:
    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}]

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

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

    Or can I do something like...
    Var(U_t) = E(U_t^2) - (E(U_t))^2?
    (E(U_t))^2 = m_t^2
    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

    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
    Var(U_t) = 2m_t\sum^2_{i=-2}a_iZ_{t+i} +  (\sum^2_{i=-2}a_iZ_{t+i})^2
    Var(U_t) = 2m_t\sigma^2 + \sigma^4
    Var(U_t) = \sigma^2(2m_t + \sigma^2)=2m_t\sigma^2 + \sigma^4
    Last edited by lpd; August 20th 2011 at 02:29 AM.
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  5. #5
    Grand Panjandrum
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    Re: Time Series: Theory and Methods

    Quote Originally Posted by lpd View Post
    Thanks for that...

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

    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:

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

    and:
    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}]

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

    I'm not sure how I can get 5. What am I doing wrong? ( {\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

    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})]

    ............. =\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 Z s to replace the variance of the sum by the sum of the variances

    Or can I do something like...
    Var(U_t) = E(U_t^2) - (E(U_t))^2?
    (E(U_t))^2 = m_t^2
    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

    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
    Var(U_t) = 2m_t\sum^2_{i=-2}a_iZ_{t+i} +  (\sum^2_{i=-2}a_iZ_{t+i})^2
    Var(U_t) = 2m_t\sigma^2 + \sigma^4
    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
    Last edited by CaptainBlack; August 20th 2011 at 03:10 AM.
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  6. #6
    lpd
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    Re: Time Series: Theory and Methods

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

    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})]

    ............. =\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 Z s to replace the variance of the sum by the sum of the variances

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

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

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

    =E((\sum^2_{i=-2}a_i Z_{t+i}) (\sum^2_{i=-2}a_{i+1} Z_{t+1+i}))
    Then, is the correlation just equal to zero?
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  7. #7
    Grand Panjandrum
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    Re: Time Series: Theory and Methods

    Quote Originally Posted by lpd View Post
    Oh I See. So, \sigma^2 \sum^2_{i=-2}a_i^2 would look something like \sigma^2 (2(\frac{3}{35})^2+ 2(\frac{12}{35})^2+ (\frac{17}{35})^2)

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

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

    =E((\sum^2_{i=-2}a_i Z_{t+i}) (\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:

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

    essentially the normalised covariance.

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

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

    {\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. 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 : 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... \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? where\, j=i+1
    And then get something like...

    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?
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  9. #9
    Grand Panjandrum
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    Re: Time Series: Theory and Methods

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

    Basically I got up to here... after a few cancellations in expanding the covariance equation out. 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 : E((\sum_{i=-2}^2 a_i Z_{t+i})(\sum_{i=-2}^2 a_{i+1} Z_{t+i+1}))
    Because the Zs are zero mean and independent E(Z_m Z_n)=0 if m\ne n and =\sigma^2 if m=n

    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]

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

    CB
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