4/2/09

3.2 Measuring Business Cycles
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10 comments:

  1. AnonymousApril 9, 2009 at 2:56 AM

    Hi, I'm very confused about this point. I have the 100 realizations of the economy of 120 quarters each. Now, should i compute the statistics for each realization of the economy? should i do the same with the correlation coefficients? what about the lags? how many lags should we considered? should i have to do this for each realization? because in that case it will be a very large number of statistics.
    I will appreciate any help with these, thanks
    Lian

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  2. jdgApril 13, 2009 at 10:21 AM

    Hi Lian!

    This is a rough sketch of the algorithm:

    Assume you are only interested in three time series: (y,c,i)

    1. Obtain a sample for {y,c,i} from t=0 to t=119
    2. Log your sample {log y, log c, log i}
    3. HP1600 your sample and keep the deviations: {dy, dc, di}
    4. Compute the standard deviations of the three time series:
    \sigma(dy), \sigma(dc), \sigma(di)
    5. Compute the correlation coefficients:
    \rho(dy,dy), \rho(dy,dc), \rho(dy,di)
    6. Compute the correlations of three leads and three lags of each series
    and output:
    \rho(dy,dx-3), \rho(dy,dx-2), \rho(dy,dx-1), \rho(dy,dx+1), \rho(dy,dx+2),
    \rho(dy,dx+3),
    7. Store the 12 statistics that you have just computed.
    8. Obtain another independent sample.
    9. Repeat.
    10. When you have 100 samples, compute and report the averages and
    the standard deviations of the 12 statistics.


    Good luck! :-)

    ReplyDelete
  3. AnonymousApril 15, 2009 at 2:22 AM

    the part upto getting the standard deviations and thus the volatilities is pretty straight forward. However, the part of computing the correlation of y with the lags of the other variables repeated a over a 100 samples is pretty complicated. In fact, I am slightly lost about how I should be proceeding on that front.

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  4. AnonymousApril 15, 2009 at 9:56 AM

    First, the calculation of correlations (with lags) are not any different from the calculation of other statistics. So, having said that, let's see if this helps. The problem is the following: in each iteration i=1,2,...,100, you get a number of time series. Among them dy and dx. To calculate their contemporaneous correlation, you simple write corr(dy,dx) in your code. To calculate the correlation of dy with dx lagged n periods, you write corr(dy(1+n:N),dx(1:N-n)), where N=length(dy)=length(dx)=120. Similarly for lags of dy. So, for each i, you calculate those correlations for n=0,1,2,3. As you do with the standard deviations, you store each one of these correlations. In the end, after the 100th iteration, you should have 100 of each one of those correlations as well as standard deviations for each variable. With that at hand, you simply take their averages and standard deviations. That's all.
    MT

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  5. jdgApril 15, 2009 at 10:06 AM

    This is my explanation. I was typing it while Manuel was writing his. That shows how much we both value your welfare. Essentially we say the same things, so take your pick:

    Say your sample has 10 observations of two variables y, x
    and that you want to compute the correlations with y of three leads
    and three lags of x. For the first lag do the following:

    Define y^ = y4, y5, y6, y7
    Define x1 = x3, x4, x5, x6

    Compute corr(y^, x1)

    Repeat for the remaining lags and leads of x
    And you are done!

    Note that y^ never changes and that x changes every time.

    ReplyDelete
  6. jdgApril 15, 2009 at 10:10 AM

    Oops! x1 is the first LEAD of x. Not the first LAG. I am sorry about that.

    ReplyDelete
  7. AnonymousApril 15, 2009 at 11:42 AM

    Hi,
    I talked to Javier yesterday and he told me that the results of HW3.2c should be the same as the Hansen's table. Now, My results are pretty similar, except for the investment standard deviation tat i get 0.61 and hansen has 0.51, is this correct? or did i have something wrong? also the other numbers are pretty similar but some are not exactly the same. (Consider that for the Average and Standard deviation of the Standard deviation you have to multiply what you get by 100 to get hansen's numbers).
    Thanks
    Lian

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  8. AnonymousApril 15, 2009 at 12:06 PM

    Recall that those figures are sample averages so they might be different. Actually, if you run your code (i.e., do the simulations) more than once, you're going to get different numbers. For the same reason, your results and Hansen's are going to be somewhat different.
    However, they should be reasonably close. In other words, you should not be able to reject the hypothesis that those two sample averages (yours and Hansen's) are the same.
    MT

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  9. jdgApril 15, 2009 at 1:02 PM

    If you do exactly as Hansen did (100 independent samples of 115 observations), you should get exactly the same results. Up to perhaps rounding errors. If you are getting exact solutions for y and h, you must get an exact solution for y/h. Did you divide before you log and detrend?

    ReplyDelete
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