Recent content by TheBestMilk

Never mind. I figured out I could estimate the log-linear model in levels which would allow me a comparison that fits with the generic linear regression. I couldn't figure out how to delete the post. Thanks anyways!

I'm not sure if this is the right place for this question, but it was on the comparison between different model's AIC/SBC values. I ran a linear regression and got an AIC/SBC of .743/.768. When I ran the same regression in log-linear form I ended up with an AIC/SBC of -7.559/-7.534. My...

Yep, you're right. Forgot to carry that negative sign through.

Hey, Sorry it's been a few days. I had a friend in from out of town and a test in another class I was studying for. This is what I managed to come up with: \int^{u}_{0}\int^{y_{1}/2}_{0} dy_{2}dy_{1} + \int^{2u}_{u}\int^{y_{1}/2}_{y_{1}-u} dy_{2}dy_{1} = \int^{u}_{0}...

Thank you very much. That makes perfect sense to me. Next time I'll draw the graph better to scale, which I think was throwing off where the y2 = y1 - u line was above the xaxis. Thanks!

Ah, it seems I wrote the original bounds incorrectly. Sorry about that. The bounds are actually: 0≤y2≤1, 0≤y1≤2, and 2y2≤y1. Thanks!

Homework Statement I have the bounds, 0≤y_{1}≤2, 0≤y_{2}≤1, and 2y_{2}≤y_{1}. I now have a line u=y_{1}-y_{2} and I'm trying to find the area such that y_{2}≥y_{1}-u. The integral comes down to two parts, the first of which I'm stuck on (when 0≤y1≤1). I'm pretty sure I have one way setup...

Homework Statement I'm working a problem, and I've come to taking the derivative with respect to ln(x): \frac{\partial ln(x^{c})}{\partial ln(p_{x})} Homework Equations ln(x^{c})=ln(p^{2}_{y}I)+ln(p_{x}+p_{y})-ln(p_{x})-ln(p_{y}) The Attempt at a Solution I've worked it out...

That makes a lot of sense now. Thank you for walking me through that!

Sorry about the confusing terms, but I did mean to say negative infinity as opposed to 0. Given the eigenvalues, the eigenvectors come out to y = 1/2*x and y = -1/2*x. I'm assuming I did the math right. That does make sense, I believe. I had thought that could be the case originally, but...

Well, the steady point, (2,2) will be a saddle solution. Not only because it can be seen due to the slopes in the quadrants, but also because the eigenvalues of the matrix formed by x',y' are +1/2 and -1/2 - thus giving us a saddle point. Given your dy/dx example, as x -> positive infinity...

Homework Statement Given x' = y - 2 y' = \frac{1}{4}x-\frac{1}{2} Draw the phase plane. The Attempt at a Solution First I found what I believe are the nullclines by taking x'=0 and y'=0. x' = 0 = y -2 so y = 2. y' = 0 = \frac{1}{4}x-\frac{1}{2} so x = 2. This...