Ask me to explain your math homework

Sophielight said:
If Sophielight has 5 chocolate bars and katsucats asks for 2. How many were left?

Sophielight said:
Okay, here's something fun.

Derive a) total sum of squares, b) sum of squares due to model, and c) sum of squares due to error from Yi=a+bXi+Ei. And finally D) explain these things in relation to linear regression.

katsucats said:
5.

2, yours truly has a kind heart.

So assuming you have a dataset Y= WX + b, but the data doesn't fit exactly on the line. There is some small error (residual) such that r = WX + b - y.

The problem is that the equation given isn't convex, so there's no easy way to minimize with respect to r, so we square both sides. First let WX+b = y_pred, because we are predicting a linear solution. Then we have
r^2 = (y_pred - y)^2

Then, we scale it to get the mean, and that's how you get the standard linear regression cost function (1/m) sum (y_pred - y)^2

Aster- said:


thanks my dude, if you don't wanna or don't know how to it's fine i wasn't planning on doing it anyway lol

Sophielight said:

Yi = a+bXi+Ei = Y_pi+Ei
Yi-e(Y) = Y_pi+Ei-e(Y)

katsucats said:
Okay, I don't have a protractor, but if you draw the triangle (scale 1) real quick, you'd see that it looks like a right angle. We can prove that.

Since A = (-4, -6) and C = (6, -2), if we want the slope of the line, then we solve for

slope = (y1-y2)/(x1-x2), right?

So the slope of AC is (-6-(-2))/(-4-6) = -4/(-10) = 4/10

On the other hand, since B is (2, 8) the slope of BC is (8-(-2))/(2-6) = -10/4

Notice that since 10/4 is the negative reciprocal of -4/10, they must be perpendicular, so the angle ACB (m<c) must be 90 degrees.

Now notice Pythagorea's principle a^2 + b^2 = c^2 applies for any right angle. Well, if we take y1-y2, that's always the length of a vertical line, and x1-x2 is always the length of a horizontal line, so there is always a right angle around every sloped line. Case in point:

For AC, y1-y2 = 4, and x1-x2 = 10, so you can imagine a horizontal line of length 10 going along under AC, and a vertical line of length 4 on the side, with AC being the hypotenuse. Therefore, 10^2 + 4^2 = AC^2. So the length of AC is sqrt(116)

Similarly for BC, y1-y2 = -10, and x1-x2 = 4, so (-10)^2 + 4^2 = BC^2 (since negative numbers squared become positive). So the length of BC is sqrt(116)

Since both AC and BC are the same, then the other 2 angles must also be the same. The sum of angles in a triangle is 180 degrees. So

m<a + m<b + m<c = 180

We know m<3 is 90, so

m<a + m<b + 90 = 180
m<a + m<b = 90

We know m<a and m<b must be the same (due to congruency), so
m<a = m<b

Plug that in above, and get

m<a = 90 - m<b
m<a = 90 - m<a
2 m<a = 90
m<a = 90/2 = 45

m<b = m<a = 45

Now all that remains is line AB, which is the hypotenuse of ABC. So
sqrt(116)^2 + sqrt(116)^2 = AB^2
2 * sqrt(116)^2 = AB^2

Take sqrt of both sides and get

sqrt(2) * 116 = AB

And there you have it for scale 1.

For any other scale, you just multiply each coordinate and length by the scale number (but not the angles, they stay the same).

Aster- said:
woah dude thanks!

I didn't expect to actually learn anything through text but i think i got it down too

katsucats said:
I have a bit of a machine learning background, but not much of a frequentist stats background, so after looking up the definition, I'm assuming you're looking for



Y_i - E[Y] = r_i + (Y_pi - E[Y])
Y_i - E[Y] = (Y_i - Y_pi) + (Y_pi - E[Y])
SST = SSE + SSM

Alex260002 said:
I hope my translation doesn't change the meaning of this senior high school problem.
Find the recurrence formula for the following series:
I(n)=Indefinite integral (e^x)(sin(x))^n dx

Sophielight said:


Not really, since the main issue why I presented this problem is because I can't seem to see how

Σ(Yi-e[Y])^2 = Σ(Y_pi+Ei-e[Y])^2 = Σ(Y_pi-e[Y])^2 + ΣEi^2

Without assuming cov(Y_p,E)=e(Y_p,E)-e(Y_p)e(E)=0

Of course, its ultimate basis is cov(X,E)=0. But the problem this is associated to is a biased ols estimator.

Now, if there is no way around making this assumption, does it mean endogeneity also makes the goodness-of-fit measure wrong?

Comic_Sans said:
What is the shortest string containing all permutations of a set of n elements?

ScarletCelestial said:
@katsucats

Know how logs work? As in: log(n)? (Computer Science A-Level student who can't algebra(?) to save her life. [If wrong about algebra, substitute it for mathematics])

katsucats said:
Since you're CS, I'm assuming you mean log-base2. It means
log_2(n) = x
2^x = n

Are equivalent. Typically in non-CS math, the base is e (natural log),
log_e(n) = ln(n) = x
e^x = n

are equivalent. In CS, logs are useful to talk about the complexity of binary algorithms, where you eliminate half of the input n on every iteration, so the number of iterations is log(n).

katsucats said:
traed, what are you going to do, whip out a computer in the middle of the game and enter all the cards you've seen? I don't see the point of this. The whole point of counting is to get a simple approximation that can be done without attracting too much attention.

traed said:

That is no problem with online casinos. A small number use webcam feeds instead of a program. Down side is they play very slowly compaired to a physical casino.

katsucats said:
There are bots that count this for you. At least there is for Hold 'Em.

traed said:

I was considering writing my own program if I don't find one that uses a true count method. I only found ones that don't so far.

katsucats said:
In all honesty, online gambling platforms know cards are countable, so it's likely that they shuffle the cards after every few rounds. Unless you know how often they shuffle, this is entirely pointless.

traed said:

I suppose that's true. I'm sure at least a couple are vulnerable.

katsucats said:
Not vulnerable enough. I suggest Texas Hold 'Em if you want to game the system, since you're not playing against the house but other players, it's less rigged. Back in the day we used Poker Trackers, which automatically tracked all the hands seen and calculated all the odds.

Calculating the odds isn't particularly difficult with a calculator. You have to know all the remaining cards in the deck to see how likely it is you'd get any number, and how likely it is the house will get any number. But it's wasted effort IMO. The blackjack dealer is never going to let you game the system. In a live casino, where the number of hands dealt between shuffles are publicized and known, it is possible to count the cards to get a slight advantage.