Hi, I am reading the CFAI book about this and I am confused. The first assumption mentions that “The requirement does not exclude X from being raised to a power other than 1”. In that case, does that mean that the equation is still linear? For example if y = b0 + b1*x^2, how can that be linear? Am I missing something here? Thanks.
Linear regression can be used to estimate the values of b0 and b1 from the measured data. This model is non-linear in the “x” variable, but it is linear in the parameters b0 and b1; if we take regressors xi = (x^0, x^1, x^2) = (1, xi0,bi1), the model takes on the standard form y= b1xi1
When we say that a regression model is linear, we mean that that the model is linear in the parameters. It may or may not be linear in the independent variable. Examples: y = beta0 + beta1.x^3 Linear (as all paramters are raised to the power of 1) y=beta0 + beta1^3. x non-linear (the second paramter is cubed) y= beta0 + beta1.beta2.x non-linear (the second and third parameters are multiplied)
Thanks for the replies. I am just confused because I picture the equation in terms of graphs. Since b0 and b1 are just constants, raising the constant to any power would result in another constant. However, if the independent variable is raised to a power > 1, the graph would look like a curve and not a line anymore. So, I am confused why it is still considered linear.
^^^ I havent cracked the books yet, but from my grad econ days, I would have to disagree with Hockeytime. It is actually the opposite of what you posted. d31dy, you are correct in that it wont look like a line, but a curve mirrored through the Y axis. The modeling of the regression line can still be linear though.
Ah yes, it shows that I havent been in grad school for a few years… and that there will be more theory tested in level2. The graphs are not linear with respect to X, but we are testing for B so that is the variable when discussing linearity, it is the statistic that we are measuring with minimal variance(E). The X values are known numbers, not estimates. Though I must disagree with the third example, you cant have a Beta without a corresponding X variable, unless it is B(0), intercept.
My third model for regressors is beta0 + beta1.beta2.x there is an x after beta 2. As you mentioned, there can be only one constant parameter.
You would have to have an X2 to match with the Beta2. Each Beta has to correspond to an X value in this form of the equation. You cant have 2 seperate Beta vectors and only one X matrix of values. Properly written in this format would be B0+B1B2X2+B2X2
In case any clarification is needed: y = b0 + b1*x^2 Y has a NON-LINEAR relationship with X, however it has a LINEAR relationship with X^2. i.e. transform xi-> zi by squaring each xi and regress y on z. if u draw a scatter of x against y, youll get a curve, but if you plot x^2 against y, youll get a line.
Thanks for all the clarifications. I was originally confused because I can’t picture how to use a line to “best fit” a curve. But if we do the transformation first, then it all makes sense.
BizBanker you are absolutely right that I cannot have a beta vector without a regressor matrix. I should have written: y = b0+ b1b2X1+ b2X2 However, in your example you are writing y= B0+B1B2X2+B2X2 you are using the same regressor matrix for the second and third terms of the regressor. This will result in blatant multicollinearity. The two matrices should be not be expressed as a linear combination of each other.
Ha! You got me on that! Yeah looks my retarded ass cant translate numerical analysis to QWERTY. Though i dont miss the pages of chicken scratch to write a proof.