The denominator makes sense to me. I still don't understand the numerator though. Isn't it:

[m(p-p)q(p)]' (mq(p)) - [m(p-p)q(p)] (mq(p))' = 0?

Denominator(p) * numerator'(p) - numerator(p) * denominator'(p)

= mq(p) * [mq(p)) + q'(p)*m*0] - ...[0 * denominator'(p)]

= mq(p) * mq(p)

So, if I understand correctly:

[m(x - p)]' at x = p is m?

I don't understand why.

[mx - mp]' = m

Then that gets multiplied by q(p). Check mine again as I added more detail.

Now I've got it. For some reason, I kept trying to evaluate x = p before differentiating, as you could see here:

Rather:

[mx - mp]' = mx' - mp' = m - 0

Easy.

Thanks!

Microtheory type question because I know almost none. I have a utility function in two variables t, tax rate, and a, income.

v(t, a) = L(t) + h(n - L(t)) + (1-t)(a - n)

What I know: h(.) is well behaved and strictly concave. L(t) is decreasing in t by the concavity of h(.)

I need to prove that this implies a single preference order. Basically, a person with lower income will always prefer a tax rate at least as high as a person with higher income and a person with higher income will always prefer a tax rate at least as low as a person with lower income.

It has been hinted to me that the answer lies somewhere in the cross-partial derivative, d^2 v(t, a)/dtda to relate changes in income to changes in preferred tax rate. I got -1 for this, which I think is right but I'm also totally willing to accept that I did something wrong. anyone who knows anything about what this means and is allegedly telling me would be appreciated

Here's what I've done so far.

1. Got PDF of Fault1 in terms of R
2. Got PDF of Fault2 in terms of R
3. Got PDF of AreaSource in terms of R

Now what do I do?

I know there are ranges in which 1,2,or all sources are included.

For example, for 0<R<30, it's only AreaSource.

But for 30<R<40, its Area and F2.

etc.

So my idea, for example with the 30<R<40 range, was to add the two PDFs, weighting them by the Rates per year that I am given.

i.e. PDF for 30<R<40 is 0.5*PDF(Area)+0.1*PDF(F2) [should i subtract out 0.5PDFArea*0.1PDF(F2)???]. Then how do I normalize it? I'm having trouble because I know CDF for the entire distribution needs to go from 0 at R = 0 to 1 at R = sqrt(40^2+160^2), but I can't figure out if my process is correct or how to normalize stuff to make that work

If each of your pdfs are normalized so they integrate to 1, then they are conditional on the given fault being the one that occurred, so then just weight them by the probabilities that a random quake was caused by that fault.

f(R) = f(R|fault1)*P(fault 1) + f(R|fault2)*P(fault 2) + f(R|area)*P(area)

= f(R|fault1)*3/9 + f(R|fault2)*1/9 + f(R|area)*5/9

Same for the F(R).

Ok. Yeah each of my individual PDF are normalized so they integrate to 1. Thanks!

I have a triangle with vertices (0,0), (a,0), (0,b).

I need f(X|Y< b/2). I'm having trouble formulating it so I get the distribution.

So far I already have f(XY), f(X), f(Y), f(X|Y=y)

f(X|Y<b/2)*f(Y<b/2) = f(X)

f(X|Y<b/2) = f(x) / f(Y< b/2).

If we're assuming that f(x,y) = 1/2 ab on the triangle and 0 everywhere else, then f(Y<b/2) = 3ab/8. Otherwise in general

f(Y<b/2) = integral of f(y) from y= -infinity to y=b/2. Integrate from y=0 if f(y) only non-zero on triangle.

We are just learning about improper integrals and, thus far, the convention is that an integral is improper if the interval of integration extends infinitely in either direction or if their is an infinite discontinuity. We've learned to evaluate both kind of improper integrals using limits. Pretty straight forward.

It seems though that our technique for evaluating these limits can only handle one or the other type of improper integral. Is there a technique for, say, integrating 1/x from -1 to infinity?

In principle you can divide the integration interval into parts, integrate them separately, and add the results. So you'd say



Then you have just one limit in each. However, all the individual limits must converge, which they don't in this case. So this integral isn't well-defined.

Yeah Ace -- with multiple "bad" things to worry about (meaning limits of +/- infinity or values in the domain where the integrand blows up), you need to split the integral into pieces, each of which only has one problem spot (that will be handled by taking a limit).

So if we were trying to integrate 1/x^2 from -inf to inf, we'd have 3 problems: +inf, -inf, and 0.

So I'd write:


Each of these integrals is improper because of exactly one point, and each of these integrals must converge for the large improper integral to converge.

edit: and in this case, the 'b' and 'c' integrals diverge, so the whole thing diverges.

Thanks, that makes sense.

I knew how to deal with both cases seperately and also cases where we need to split something like -infinity to infinity into two seperate intervals, but I'm still at the point in my development where when I see something like--GASP--the necessity to take the limit as MULTIPLE placeholders go to different places I instantly think "I've stumbled on something clearly outside the realm of mathematical possibiliy, this will require magic".

I need to stop doing that and start thinking more creatively.

Edited the above. Assuming f(x,y) = 2/(ab), not ab/2. I changed f(Y<b/2) to P(Y<b/2) since this is a probability rather than a pdf. Also this assumes you mean f(x,y) is non-zero on the region bounded by the triangle. Did you mean that it is non-zero only on the perimeter of the triangle? In that case you would take the probability that the point was on each side times the probability that y < b/2 on that side, and sum.

P(Y<b/2) = 1/(a+b+sqrt(a^2+b^2)*[a*1 + b*1/2 + sqrt(a^2+b^2)*1/2].

Note that the integral of 1/x from 1 to infinity diverges. That means that if you had to paint the area under the graph of 1/x from 1 to infinity with a coat of paint of some constant thickness, you'd have to paint an infinite area, so no matter how thin the coat, you'd need an infinite volume of paint. But now suppose we take that graph and rotate it around the x axis so that we will generate a volume that looks like an infinitely long horn. The radius of the horn is r = 1/x which changes with x, so the volume of that horn is the integral of pi*r^2 = pi/x^2 from 1 to infinity which converges to pi. You could fill up that whole volume of the horn with just pi units of paint! This seems paradoxical because surely if you can fill up the whole volume, then you will have painted under the whole graph because that whole area is contained inside the volume of the horn. So what gives?

The fact is that it IS possible to paint an infinite area with a finite volume of paint. We just can't do it with a coat of constant thickness no matter how thin we make it. It's only possible if we make the thickness of the paint decrease as a function of x. Moreover, it must decrease faster than the area increases. When we fill the horn with paint, the thickness will automatically decrease in thickness at the proper rate so that we can paint the infinite area with a finite volume of paint.

Similarly, the area of the inner surface of the horn is infinite, so it would require an infinite volume of paint if the thickness of the coat were constant. But when we fill the horn with paint, clearly the inner surface gets painted. The thickness of the paint is decreasing and not constant. This horn has infinite surface area but finite volume! See Gabriel's Horn.

Ha. Mind just blown.

I got 2/(ab) for f(x,y).

It's just uniformly distributed inside the triangle for this problem

The equation I used for f(X|Y) was

f(X|Y) = f(x,y)/f(y)

So I thought to do this I would need to do

f(X|Y<b/2) = f(x,y<b/2)/f(y<b/2)

for f(x,y<b/2) what I did was compute the difference in the areas, area of entire triangle = ab/2, area of triangular region above was ab/8, so the difference in those was 3*ab/8, giving me f(x,y<b/2) = 8/(3ab). Did you mean f(x,y<b/2) = 8/(3ab) ?

for f(y<b/2) is where I get stuck, because I don't know how to make it not come out to be a probability. P(Y<b/2) is 3/4 I think.

I was kind of thinking I needed to do some sort of piecewise solution because f(X|Y<b/2) is a rectangle for 0<x<a/2 and then a triangle from a/2 < x < a

Yes, P(Y<b/2) = 3/4, and that is a probability not a density. We want

f(x|Y<b/2)*P(Y<b/2) = f(x)

f(x|Y<b/2) = f(x) / P(Y<b/2) = 4/3*f(x)

You should have gotten

f(x) = -2/a^2 x + 2/a.

That is the integral of f(x,y)dy for y=0 to y = -b/a x + b.

So f(x|Y<b/2) = -8/(3a^2) x + 8/(3a)

I still didn't have this right.

That's not f(x), it is f(x AND Y<b/2) = f (x,Y<b/2).

f(x,Y<b/2) = f(x,Y<b/2) / P(Y<b/2)

same as you had for the numerator, but the denominator is total probability P(Y<b/2) = 3/4.

For 0 <= x <= a/2:

f(x,Y<b/2) = integral f(x,y) for y=0 to y=b/2

= 2/(ab) * b/2 = 1/a.


For a/2 <= x <= a:

f(x,Y<b/2) = integral f(x,y) for y=0 to y = -b/a x + b

= 2/(ab) * (-b/a x + b) = -2/a^2 x + 2/a.

These make sense because for x = 0 this is 1/a = f(0)/2 for the f(x) we got before which it should be since Y < b/2 half the time. For x >= a/2, this is the same f(x) as we got before which it should be since Y is always <b/2 here.

So dividing by 3/4 gives

0 <= x <= a/2:

f(x|Y<b/2) = 4/(3a).


a/2 <= x <= a

f(x|Y<b/2) = -8/(3a^2) x + 8/(3a).

basic abs alg:

WTS: If every element of a group G is its own inverse, G is abelian.

PF: Take two arbitrary elements, x and y, in G. (xy)^2 = xyxy = e

So (xyxy)y = y = xyx

Multiply through by x

xy = xxyx = yx, as desired

amirite?

yes

xyxy = e

xyxyyx = yx

xy = yx

thanks guys. bruce, never asked, what's your background re: math?