Gujarati - Question on Regression

[Stuti]

83.2 If the functional form is given by Y = B1 + B2*X, the population regression function (PRF) ….
a. Has one set of values for each SRF; e.g., if five SRFs exist, then five PRFs exist
b. Passes, in several points, through the conditional means of (Y) values
c. Passes, in several points, through the conditional means of (X) values
d. Must be observable in order to conduct a regression

I am not able to understand the solution?

 

[Stuti]

Another question:

87.2 Let Y(t) = the S&P 500 Index and let X(t) = the three-month Treasury bill rate. Assume our linear regression model finds the following relationship: Y(t) = -15 + 26*[1/X(t)]. For example, if X(t) = 2.0%, then Y(t) = -15 + 26/2% = 1,285. If the Treasury bill rate starts at 3.0%, what is the average decline in the S&P Index (predicted by the model) if the rate were to increase 1%; i.e., from 3% to 4%? (note: do you know why this is a valid "linear" regression?)
a. -289
b. -489
c. -650
d. -28,889

Solution:

87.2 A. -289
The rate of change is given by dY(t)/dX(t) = -B2/X(t)^2 = -B2*[1/X(t)^2].
... Note this is the same as observing that the slope coefficient--as first partial derivative--is the rate of change in a more typical linear regression where, if Y(t) = B2*X(t) + B1, d(Yt)/d(Xt) = B2.
Where B2 = 26 and if X(t) = 3%, then the rate of change (first derivative) = -26/3%^2 = -28,888.89.
That is, for each 1.0 unit change in the T-bill rate the S&P Index declines -28,889.
As 1.0 unit = 100%, a 1% change implies a 28,889/100 = 289 index decline.
… this is why we often see 10,000 in the denominator (e.g., for DV01): 1 basis point is 1/100th of 1%, which is 1/100th of 1.0 (100%), so 1 basis point is 1/10,000th of 1.0 unit.
... regression is linear because it is linear in the parameters; it is okay that it is non-linear in the explanatory variable X.

Query - Even if we understand that the decline in S&P is determined by the model how can we say the statement that That is, for each 1.0 unit change in the T-bill rate the S&P Index declines -28,889. when we calculating @ 3% here?

I am getting confused with the terms or the language used

 

[David Harper CFA FRM]

Hi @Stuti

• The regression Y = B1 + B2*X can be expressed as Y = E(Y | X) = B1 + B2*X. In the PRF, (Y) is a straight line with several values. Each value of Y is a function of X, and specifically, each value of Y is a "conditional mean," it is the expected value conditional on a given value of X. So we can say the line is defined by the set of conditional means of Y (expected values of Y conditional on a value of a given X)
• As a first step, you can ignore the 3%, right? We start with a linear (regression) function: y = a + b*x^(-1). Then we take the first derivative of (y) with respect to (x): dy/dx = -b*x^(-2) per the http://en.wikipedia.org/wiki/Power_rule. Where does this get us? This first derivative, -b*x^(-2), is the slope or rate of change; just as (m) is the slope in y = b + m*x. It gives the change in (y) per unit change in (x). The x-axis is %, so a unit is 1.0 = 100%. This slope, -b*x^(-2), varies with each value of x. As x = 3%, it equals -28,889; for example, at x = 4%, dy/dx = -b*x^(-2) = -26*4%^(-2) = 16,250.
  
  it helps me to notice that this is similar to retrieving slope (m) from y = bx + a, where the first derivative is simple. But it's still the power rule! Note dy/dx = (1)*b*x^0 = 1*b*1 = b. Only in this case, we are starting with y = a + b*x^(-1).

 

[AVasa7074]

Hi,

This is a very silly question.

In 87.2, can we just not take the difference of Y(t) when X(t)=3 and 4. In which situations do we take the first derivative vs differencing?