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# st: Why does this scalar calculation return the wrong value when using time series operators?

 From Aaron Kirkman To statalist@hsphsun2.harvard.edu Subject st: Why does this scalar calculation return the wrong value when using time series operators? Date Tue, 6 Nov 2012 15:25:06 -0600

```Dear Statalist,

I'm performing a simple linear regression on time series data and
calculating the t-statistic for coefficients afterwards. However, I
noticed that when using time series operators, the t-statistic always
calculates to be one, even though the values from the regression are
correct. For example, this code:

----------
clear all

local y0 0
set obs 200
quietly {
gen y = .
gen t = _n
tsset t

replace y = cond(t == 1, `y0', L.y + rnormal())
regress D.y L.y
}

di _b[L.y]
di _se[L.y]
di _b[L.y] / _se[L.y]

scalar t = _b[L.y] / _se[L.y] // t = Beta / SE

di t
----------

outputs the following:

-.02092465     // _b[L.y]
.01391362      // _se[L.y]
-1.5038971     // _b[L.y] / _se[L.y]
1                   // scalar "t"

The first three numbers are the correct values from the regression,
but the calculation for the t-statistic is incorrect. If I remove the
time series operators from the code and instead refer to observations
numbers (I would prefer to use time series operators, but just as an
example), the resulting t-statistic is correct:

----------
clear all
local y0 0
set obs 200
quietly {
gen y = .
gen ly = .
gen dy = .

replace y = cond(_n == 1, `y0', y[_n - 1] + rnormal())
replace ly = y[_n - 1]
replace dy = y - ly

regress dy ly
}

di _b[ly]
di _se[ly]
di _b[ly] / _se[ly]

scalar t = _b[ly] / _se[ly] // t = Beta / SE
di t
----------

This code outputs the correct t-statistic of  -1.5038971

-.02092465     // _b[L.y]
.01391362      // _se[L.y]
-1.5038971     // _b[L.y] / _se[L.y]
-1.5038971     // scalar "t"

I read through "[U] 13.5 Accessing coefficients and standard errors"
and --help scalar--, but I don't see anything in either of those
manuals that would cause the problem. Any ideas?

Aaron Kirkman
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```