I'm not totally sure how you arrived at that, so I will just start from the beginning.

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Let u = 1/(s0*t-1).

You were reviewing g(s0,t), and trying to write each in the form a+bu for constants a and b.

The first component was 1+2u, so far so good.

What about the second component, say (s0-t)/(s0*t - 1) ?

We do polynomial division (wiki link here).

We want to force a factor of s0 on the t term of the numerator, that way we can divide out the t term. To that end, we write the numerator -t+s0 = (-1/s0)(s0*t - 1) + 1/s0 + s0.

So the whole thing becomes (-1/s0) + (s0 + 1/s0)u. That's of the form you wanted.

Oh duh, I see what I was doing now. I think you made a typo though-- shouldn't

and

Yes sorry it is a mistake.

Prove that a rectangle R = [a1, b1] x ... x [an, bn] c Rn is a closed set.

If it matters, book definition of closed is: Suppose S c Rn. If S has the property that every convergent sequence of points in S converges to a point in S, then we say S is closed. That is, S is closed if the following is true: Whenever a convergent sequence x_k --> a has the property that x_k is an element of S for all k (which are natural numbers), then a is an element of S as well.

We also have the proposition that the subset S c Rn is closed iff its complement Rn - S is open.


Also, I have another question that is somewhat general. How would you go about drawing f<x, y> = xy. I guess you draw level sets but I have trouble going from those to the actual picture in R3 for certain functions.

1. Take a converging sequence x_n in the rectangle with limit x. Each component x_n^(i) is going to converge to the component of its limit x^(i). But each component x_n^(i) was chosen from a closed set [a_i,b_i], and therefore its limit x^(i) is contained in it.


2. Mmmm I'll let Wyman handle this haha, I know how I would graph it but...

I get (pi r^2 + w h + 2h r + 2 w r) / pi R^2 for small circle radius r, big circle radius R and rectangle w x h. For any given orientation, the numerator is the size of the "hit area" and the denominator is the size of the possible area. So ~5.6% in this case

I am experiencing strong feelings of regret wrt taking this class. It was probably a bad idea to take Calculus III my first semester and Calculus of Several Variables my 8th (last) semester.

I have two easy probability questions, but I have problems explaining why they are not to be the solved in the same way. Hopefully you guys can help:

Question A:
Three people, A, B and C draw from a finite (>5) population of red/white balls without replacement, i.e. according to the hypergeometric distribution. Say we have x red and y white balls. A draws 1, B draws 2, C draws 3.
What is the probability of C drawing precisely 1 red ball?

Question B:
You have two urns, Urn 1 and Urn 2, with red and white balls. Say that we have respectively x red and y white in Urn 1 and v red and z white in Urn 2.
You take one ball from Urn 1 and place it in Urn 2. Now you draw from Urn 2 - what is the probability of drawing a red ball from Urn 2?

When we know x, y, z and v the questions are easy. But in Question A we can ignore the draws of A and B (right?) and only consider C's draw as if neither A or B had drawn before. In Question B we have to take into account the first draw (to see this, take as an example v=0, so no red balls a priori in Urn 2).

How do I best explain this asymmetry of the questions? In both questions we can use the law of total probability to find the conditional probabilities and sum them up, but in Question A, it does not seem to be needed.

In the first question, the unconditional probability that C draws red doesn't depend on when C draws because the probability of a red ball is the same for any 3 balls no matter when they are drawn, just as your chance of being dealt aces doesn't depend on what position you're in. But in the second question, the unconditional probability that C draws red will not in general be the same after the ball from urn 1 is transferred than it would be if C drew before that ball was transferred.

To draw z = xy, you can draw level curves. It's certainly what I would do. You could also check out Wolfram Alpha if you need some help, but this picture really didn't help me visualize it.

www.wolframalpha.com/input/?i=xy-z+%3D+0

The way I think about it is that I think about y = 1/x (or 1 = xy), y = 10/x, and xy=0. With those 3 pictures in mind, you can imagine how the level curve changes as z changes and try to fill in the 3D picture of sheets in your mind.

Hey,

got a quick question for exams tomorrow

if we have to maximize/minimize a function

f(x,y) on one constraint like x^2+y^2<=1

then we first check the gradient

then we make our lagrange thing with f_y=C*g_y etc, and we can say the maxima has to be on the constraint so x^2+y^2=1

now my question is:

if we have an additional constraint like y<=0.5 can we say that y=0.5 when we take the lagrange multipliers, so that the extrema has to be on both boundaries?

Here's how to plot z=xy in R. You can rotate it to get different perspectives by playing with theta and phi. This is just with the base package. There are packages you can download which would allow you to automate that.

Many thanks for your answer. It was what I gathered. To sum up; the reason that the conditional and unconditional probabilities are the same in the first question is that all three players draw from the same distribution? I mean, if player A drew from another (say he is twice as likely to draw a red ball) distribution, we would need to use the conditional probability, right?

already found out

The conditional and unconditional probabilities are not the same in either question. In the first question, the conditional probabilities would be the probabilities that C draws red given that we know what the other players drew. That is not the same as the unconditional probability that C draws red which is what we want to know. Unconditional means it is the overall probability, not the probability given what the other players drew. The unconditional probability is the same whether C draws 1st, 2nd, or 3rd. You could evaluate it using the conditional probabilities given every possible draw for the other players, but you'd get the same answer as if you just ignored the other players. You're getting 3 balls selected from the same distribution no matter when they draw. The distribution is the same because we don't know what they drew. If we knew what they drew, then we'd have a conditional probability which would be different. The second problem is different because the mere fact that a ball is transferred changes the distribution that you are drawing from even when you don't know what the transferred ball is. That is, it changes the unconditional probability that you draw red.

The first problem is like when you have a 9/47 chance of completing a flush draw on the turn. You don't have to consider every possible number of flush cards your opponents may have been dealt. If you knew what some of them were, the probability would be different, and that's the conditional probability which would be different, but you don't know what they are, so you can ignore them. If you attempted to account for every possible number of flush cards that could be in your opponent's hands, you would get the same answer 9/47.


No, if player A drew from another distribution, it wouldn't change the probability that C draws a red. That's just the same as C going before A.

Thanks again for taking the time explaining it. When I said "conditional" probability I meant "P(C draws 1 red | A,B draws first)" and by unconditional I meant "P(C draws 1 red)". We are asked for the first, but can do with the last, since the two are equal in the first question.

Wait, so are you saying that if A draws from a different distribution - let's say that there is a 99% chance that he draws a red ball - this will not affect "P(C draws 1 red | A,B draws first)"?


EDIT: I am being a bit intentionally thick here, because for some reason I can not fully get the rigorous argument for why the two questions are not the same. The intuitive part of it (which you also explained very well) is OK.

I could also ask a different way. Let's rephrase the exercise and say that we have two persons, A (draws 1) and C (draws 3). What is it that makes

P(C draw 1 red when going first) = P(C draw 1 red when going after A)?

We can write the latter using the law of total probability,

P(C draw 1 red when going after A) = P(C draw red|A 0 red)*P(A 0 red) + P(C draw red|A 1 red)*P(A 1 red).

My question becomes: what is the property that allows us to say

P(C draw red|A 0 red)*P(A 0 red) + P(C draw red|A 1 red)*P(A 1 red) = P(C draw 1 red when going first)?

OK, if that's what you meant then you are correct that the conditional and unconditional probabilities are the same in the first question because the condition that A,B draws first doesn't change the probability since we don't know what they drew. But in the second question they are different because P(C draws 1 red | ball is transferred) is not the same as P(C draws red) if he just drew a ball without a ball being transferred, even though we still don't know what was transferred.

Just take the simple case where there is 1 red ball and 1 white ball, and we each draw 1. It doesn't matter if you go first or second, your unconditional probability of drawing red is 1/2. What actually happens when I go first is that 1/2 the time I draw the red and you have 0 chance, and the other 1/2 the time I draw white and you are guaranteed to draw red, but since you don't know what I drew, the fact that I drew first makes no difference, and your probability is 1/2 whether I go first or second. It would only make a difference if I showed you what I drew. Now suppose I draw from another distribution that has 2 reds and 1 white, and I put that ball in the urn with the 1 red and 1 white that you draw from. Now 2/3 of the time you will have a 2/3 chance of drawing red, and 1/3 of the time you will have a 1/3 chance of drawing red, so your probability has now become 2/3*2/3 + 1/3*1/3 = 5/9, so it got better. That's like the second problem where the probability changes because adding a ball from another distribution changes the distribution you draw from. On the other hand, if the other distribution had 1 red and 1 white, then it wouldn't change your probability because 1/2 the time you would have 2/3, and 1/3 the time you would have 1/3, so your probability is 1/2*2/3 + 1/2*1/3 = 1/2 just like when you go first.


If A draws from another distribution it doesn't matter what he draws. Or are you saying that he is putting that ball into the urn before C draws? In that case it would be different as above.

We can show algebraically that these are the same. Say there are x red and y white and each draw 1. Then if C goes first

P(C draws red) = x/ (x+y).

If C goes second, then

P(C draws red) = x/ (x+y) * (x-1)/(x+y-1) + y/(x+y) * x/(x+y-1)

= x(x-1+y)/[(x+y)(x+y-1)]

= x/(x+y). QED

You can extend this for drawing any number of balls.

Sorry, I might be confusing you with my choice of words. When I am saying that A draws from a different distribution, I mean that A has a different probability of drawing a red ball. Let me try again:

We have two people, A and C. On a table we have an urn with 25 white balls and 5 red balls. A draws 1 (without replacement!) and after that C draws one. We want to find the probability that C draws a red ball.

Setting 1:
P(A draws red) = #red/#total.
P(C draws red) = #red/#total.

P(C draw 1 red and draws first) = 5/30.
P(C draw 1 red | A draws first) = P(C draw 1 red | A draws 0 red)*P(A draws 0 red) + P(C draw 1 red | A draws 1 red)*P(A draws 1 red)
=(5/29)*(25/30) + (4/29)*(5/30) = 5/30.

Setting 2:
Say A knows that the right side of the urn contains more red balls than the left. A decides to draw from the right side (changing his chances of getting a red ball).
P(A draws red) = 2*#red/#total.
P(C draws red) = #red/#total.

P(C draw 1 red and draws first) = 5/30.
P(C draw 1 red | A draws first) = P(C draw 1 red | A draws 0 red)*P(A draws 0 red) + P(C draw 1 red | A draws 1 red)*P(A draws 1 red)
=(5/29)*(20/30) + (4/29)*(10/30) = 14/87.


We see that in Setting 1 we have:
P(C draw 1 red | A draws first) = P(C draw 1 red and draws first)

And in setting 2 we have
P(C draw 1 red | A draws first) != P(C draw 1 red and draws first)


and my question becomes why this is? It seems that (and I am convincing myself of that this is the right thought process now) the thing that makes us able to not consider A's draw in Setting 1 is, that she draws "from the same distribution" (i.e. with the same probability) as C. Is this correct?


In your poker example of getting dealt aces it would translate to: You are equally likely of getting dealt aces in the SB and on the button. But if for some reason MP is twice as likely to get dealt an ace than the rest of you, this obviously is not the case any more.

Yes, thank you. And the property (in the general case) that makes your calculation go through, is that A and C picks a red ball with the same probability distribution (cf my post above)?

Right, the difference is that when they draw from the same urn, the probability of a ball being red does not depend on the order in which it is drawn. The first ball drawn, the 3rd ball drawn, the last ball drawn, all have the same probability of being red x/(x+y). So balls drawn don't change the probability of future balls being red. So it doesn't matter if A and B go first. But if A draws from a part of the distribution that he knows has more reds than are present in the whole distribution, then that no longer holds since he is reducing the proportion of reds on average, and he is reducing the probability of each future ball being red. Similarly when a ball is drawn from a different distribution and transferred over, obviously that different distribution can influence the probability of future balls being red because it can change the distribution.

Suppose we had 100 ping pong balls numbered 1 to 100. You draw 2 balls with replacement. The average sum of the numbers drawn will be the sum of the averages for each ball which will be 50.5 + 50.5 = 101. What would be the average sum of 2 balls drawn without replacement?

EDIT: UPDATE... I just realized A ranges from [0,inf] and O ranges from [0,2*pi], so I'm re-checking with this new info
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2nd Update. Ok, so I did

f(A) = int( [(A*exp(-A^2/(2*s^2)))/(2*pi*s^2)] , O , 0, 2*pi) and that gives me back joint pdf*2pi (makes sense)

f(A) = (A*exp(-A^2/(2*s^2)))/s^2

now for f(O) I do

f(O) = int( [(A*exp(-A^2/(2*s^2)))/(2*pi*s^2)] , A , 0, inf)

and I get 1/(2*pi) !!!

Nice.

so those 2 marginals make sense, and when they multiply times each other I get the Joint PDF, which answers #4 (they are S.I.).

As for answer to 3, looks like O (theta), follows the uniform distribution from O = [0,2pi]. Also makes sense. Also I guess A follows ______________. This is the only thing I don't know at this point.

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Update 3: Looks like A follows the RAYLEIGH distribution. Awesome. Glad I was able to figure this one out on my own. Maybe there is hope for me after all in this class.

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My differential equations professor taught us a method to solve special 2nd order equations that meet certain criteria. I can't match the answer in the book on this one. Here goes:

y'' + y' = e^(-t)

let v' = y''
let v = y'

v' + v = e^(-t)

integrating factor is e^t

(v' + v)e^t = e^(-t)e^t

(ve^t)' = 1

ve^t = t + c

v = te^(-t) + ce^(-t)

Now integrate to find y

y = integral of [te^(-t) + ce^(-t)]dt with respect to t

y = integral of [te^(-t)]dt + c_1*e^(-t) + c_2

I use parts on this integral and get:

PARTS:
u = t
du = dt

dv = e^(-t)dt
v = -e^(-t)

y = -te^(-t) - integral of [-e^(-t)]dt + c_1*e^(-t) + c_2

y = -te^(-t) - e^(-t) + c_1*e^(-t) + c_2

The book just says:

y = -te^(-t) + c_1*e^(-t) + c_2

Am I doing parts wrong? Where's the -e^(-t) term going? I did this problem 3 times now.

Apologies if the format is difficult to follow. I can fix it if need be.

Thank you!

BruceZ and masque de Z,

Here is the solution I finally arrived at today after going to office hours about the tornado rectangle circle problem, I worked it on scratch paper and scanned it to show you both if you were still curious about how we were supposed to do the problem (I believe this is how we were intended to solve it), since my picture probably explains better than I could with words.

The only thing is that I don't this method capture the scenario where the tornado hits via a corner touching (i.e. 25 < R <= 25.1 in polar coordinates)