New answers tagged regression
0
In this sort of regression problem, $X_i$ may be random in the sense that if you take another sample, all the $X_i$ values change, but one behaves as if one seeks the conditional expected value of $Y_i$ given $X_i$, so that in effect $X_i$ is treated as if it were constant rather than random. And the $\beta$s are also being treated as constant.
0
If $Y_i=\beta_0+\beta X_i+\epsilon_i$, where $\beta_0$ and $\beta$ are constants and $\epsilon_i$ is an "error" random variable with mean $0$, then $E(Y_i)=\beta_0+\beta E(X_i)$.
1
This is a problem of model comparison. A standard approach would be to fit the two competing models with same data and compare their performance on a test set.
You could do least squares fit, or if you have better idea about the noise distribution, you can use maximum likelihood estimation.
2
A better method is to plot the points $(\log x,\log y)$ for all data points $(x,y)$. If the relationship is roughly quadratic, $y\approx Cx^2$, then $\log y\approx 2\log x+\log C$, which is a linear equation in $\log x$ and $\log y$. Straight lines are easier to detect in a point plot, and if you get a straight line with slope $2$, then you can be pretty ...
0
It looks like your software is (implicitly?) assuming that your data are normal distributed and is then trying to estimate parameters for a pair of normal distributed random variables that would looks like your dataset.
Since a normal distribution doesn't have a true maximum and minimum, the software doesn't know that 0 and 1 are supposed to be special ...
0
For Q1-b), take that formula $b_1 = (X_1'X_1)^-1(X_1'Y) $ and now plug in the true value for Y using model 2 and look what happens in the limit. You should end up with something like:
$E[b1]=\beta_1 + \beta_2 * b_{12} $ where $ b_{12} $ is the coefficient a regression of $X_2$ on $X_1 $ (Note, I'm taking that formula from memory, so it might not be exactly ...
0
Here is another answer to your question. You've correctly noted that the condition E[u|x]=0 --- called "mean independence" of u and x --- is weaker than full-on independence of u and x. Bruce Hansen's notes on Econometrics give a particularly simple counterexample. (The link to the book is here: http://www.ssc.wisc.edu/~bhansen/econometrics/. See pages ...
0
[update] Upps, just saw your comment at @Nameless, that you have 10000 variables. So I think, that covariaton-approach is useless too. Should I delete the answer? [/update]
One method needs only the inversion of the covariation matrix, don't know whether this is already smart enough?
Construct the datamatrix $D$ with the top row from the ...
0
Here's what I would do. (I am assuming for the moment your $f$ is linear, so we can use OLS. If that's not the case, you have to use nonlinear LS or something appropriate.)
I would throw your 5 sets of data together. Now if you estimate your $a$ coefficients on that big data set, your $a_3,a_4$ will be "global" in the sense you want. But so will be $a_1$ ...
0
Well, in matrix form, the vector of OLS coefficients is (see wikipedia)
$$\beta=(X'X)^{-1}X'y,$$
where $'$ is the matrix transpose. Should be easy to calculate quickly in, e.g., matlab. Then, all you have to do is compute
$$RSS=\sum_{i=1}^n (y-x_i'\beta)^2=(y-X\beta)'(y-X\beta).$$
That does all the steps you mentioned above, but really, it's just two ...
Top 50 recent answers are included
