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If (t1,y1),(t2,y2),(t3,y3),...,(tn,yn)(t_1, y_1), (t_2, y_2), (t_3, y_3), ..., (t_n, y_n) denote the time series and yty_t are the trend values of the variable y, then ∑(y−yt)\sum(y - y_t), the sum of the deviations of y from their corresponding trend value is equal to:

Solution

✅ Correct Option: 3

Given a time series with data points (t1,y1),(t2,y2),(t3,y3),...,(tn,yn)(t_1, y_1), (t_2, y_2), (t_3, y_3), ..., (t_n, y_n) where yty_t represents the trend values of variable yy.


When a trend line is fitted using the least squares method, it has a fundamental property: the sum of deviations of actual values from their corresponding trend values equals zero.


The expression ∑(y−yt)\sum(y - y_t) represents the sum of all deviations, where:

yy = actual values

yty_t = trend values (values on the fitted trend line)

(y−yt)(y - y_t) = deviation at each point


The trend line is constructed such that positive deviations (where actual values exceed trend values) are exactly balanced by negative deviations (where actual values fall below trend values).

This occurs because:

∑(y−yt)=∑y−∑yt\sum(y - y_t) = \sum y - \sum y_t

=∑y−∑yt= \sum y - \sum y_t

=0= 0


Therefore, ∑(y−yt)=0\sum(y - y_t) = 0

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