Instead, once you know that either 1, 2, or 3 is the heavy ball, just weigh 1 v 2.

edit: sorry we're solving different problems should read more carefully.

What you wrote looks reasonable.

Trying to help a friend out with this, but I'm not sure how to standardize this data (ie comparing two normal distributions if they have different arithmetic means and/or standard deviations)

we have this data on 2 types of bacteria:
-Initial Number of Bacteria at Time Zero
-Number of Human Cells at Time Zero (Cells that will be infected)
-Multiplicity of Infection (Initial Number of Bacteria divided by Number of Human Cells)
-Number of Bacteria after 3 hours (Time 3-hr Supernatant)
-Number of Bacteria attached and invaded into Human Cells after 3 hours (Time 3-hr Cells)
-Number of Bacteria invaded after 6 hours (GEN)
-Number of Bacteria attached (Time 3-hr Cells minus GEN)

We are looking at:
1) Adhesiveness- how well the bacteria sticks to the surface of the human cell
2) Invasiveness- how well the bacteria gets inside the human cell

I'm somewhat confused because if this was a straight stats problem it'd be cake but I don't see how I can compare/configure any means, standard deviations, or normal-distribution graphs with this info.

thanks

Stats 410 question

Suppose X and Y are independent exponentially distributed random variable with mean 5 and 2 respectively. Find the pdf of w=X+Y

Calculate P[w>10]

we have just covered convolution, so this problem can be done using the cdf method or convolution. My integrals are comign out weird.

Thanks

edit: Sorry I missed the sticky. Here is my attempt

i used convolution to get
Integral from 0 to Infinity e^(-x/2) e^(x-w)/5) w.r.t. x

this gives 10/3 e^(-w/5) as a pdf. What do you guys think?

Can someone tell me how tricky this is?

----
Prove that the function on the open unit disk...

f(b, a) = ∥(b - a) / (1 - āb) ∥

...is a metric.
----

I don't want to spend a lot of time on it, but I'd like to understand the basic idea. (Viz.---how to show f(b,c) ≤ f(b,a) + f(a,c). Inequalities own my soul.)

if you are trying to prove that a sequence doesn't converge, then you can pick an epsilon and show the inequality can't be satisfied. since the definiton says the inequality holds for all epsilon, finding an epsilon that doesn't work is enough to show that the sequence doesn't converge

Hello.
I am doing a project on the construction of the real numbers from the rationals, using Cauchy sequences. More info here: http://en.wikipedia.org/wiki/Constru...e_real_numbers

I'm currently trying to show that the operation of division (or equivalently, taking the reciprocal) is well defined for all non-zero equivalence classes.

I think I need to show that given (a_n) -> (b_n), (a_n)-/-> 0, we can conclude that 1/(a_n) -> 1/(b_n).

I'm sure this is fairly easy, but I just can't see how to do it without (a_n) or (b_n) having a limit.

Any help much appreciated, thanks.

Are the a_n and b_n sequences? What is the equivalence relation '->' (assuming that was an equivalence relation?).

I thought initially it might be 'converge to the same limit', but this doesn't make sense after you said you couldn't do it without them having a limit

ah wait nm should have checked the link:

Two Cauchy sequences are called equivalent if and only if for every rational ε > 0, there exists an integer N such that |xn -yn|<ε for all n > N.

Back to thinking about it

edit: why can't we drop all the terms for which a_n or b_n are zero and do a term-by-term division?

They're Cauchy sequences of rationals. Two such sequences (a_i), (b_i) are equivalent if for any rational e > 0 there is some n such that for all m > n, |a_m - b_m| < e.

The question is how do you prove that (a_i/b_i) is also a Cauchy sequence, etc.

Let me re-frame that question. Suppose ϕ_a(z) is the Mobius transformation ϕ_a(z) = (z-a) / (1-āz). Then why does...

|ϕ_z(x)| ≤ |ϕ_y(x)| + |ϕ_z(y)|

...for all x, y, z ∈ↁ, the unit disk?

In general, can anything interesting be said about how conformal self-maps of the unit disk affect the Euclidean norm of points in the disk? Is there some relationship to the characteristic constant?

WTF, why can't I find anything on the entire interwebs about this?!

What do you get if you write x = a+bi, y = c+di, z = e+fi and then expand the r.h.s of that inequality.
(I have no idea, it's just what I would try if I was lacking any insight).
Perhaps you can prove the triangle inequality by brute force like that for this map?

I don't think that's a feasible strategy; the expression is extremely unwieldy. Since I've become obsessed with this problem (extra credit in a class where I don't need the points), I've spent hours manipulating the equation without any progress.

I've dug up only one suggestive fact during my search. Under the Poincaire metric on the disk, one has...

d_ρ(z, w) = d_ρ(0, |ϕ_z(w)|)

...as shown on pages 47-48 of Complex Analysis: The Geometric Perspective. But my problem comes from a Real Analysis class, so I doubt that's even relevant here. :/

In other words, we know from complex analysis:
---
d_ρ(x, z) ≤ d_ρ(x, y) + d_ρ(y, z)
=>
d_ρ(0, |ϕ_z(x)|) ≤ d_ρ(0, |ϕ_y(x)|) + d_ρ(0, |ϕ_z(y)|)
=>
½log[(1+|ϕ_z(x)|) / (1-|ϕ_z(x)|)] ≤ ½log[(1+|ϕ_y(x)|) / (1-|ϕ_y(x)|)] + ½log[(1+|ϕ_z(y)|) / (1-|ϕ_z(y)|)]
...or written another way...
tanh^-1|ϕ_z(x)| ≤ tanh^-1|ϕ_y(x)| + tanh^-1|ϕ_z(y)|
---

And we want to prove:
---
|ϕ_z(x)| ≤ |ϕ_y(x)| + |ϕ_z(y)|
---

You also used the notation ϕ_a(z) = (z - a) / (1 - āz), so that f(b, a) = |ϕ_a(b)|. First show algebraically that

ϕ_a(ϕ_b(z)) = wϕ_c(z),

where

and

Note that |w| = 1. Use this to show that

Finally, given z₁, z₂, and z₃, show that there exists a map of the form wϕ_a(z) sending z₂ to 0 and z₃ to some real number r > 0. Use this to show that it suffices to prove

f(z, r) ≤ f(z, 0) + f(0, r).

Then prove this last bit, which should be easy.

jason1990 -

Thank you, thank you, thank you! Words fail, so I must turn to emoticons...!!!

Quick question. By "should be easy", did you mean "easy for jason1990"? Or easy for a beginner like me? Because I've spent the last five hours on this line, with no apparent progress.

as;dlfkja;sdlfkja;sldkfj;aslkjdfasd;lkfjas;dlkjfa; sldkjfas;ldkjf

Heck, I can't even prove that for a, b ∈ (0, 1):

|(a-b)/(1-ab)| ≤ a + b

Which surely begs the question of why I bother living. WTF is the point?

You want to make some reductions. First, you may as well assume that a>b. Then the inequality is just (a-b)/(1-ab) \leq a+b. Multiplying through by 1-ab (which is positive), you get the inequality a-b \leq (a+b)(1-ab) = a+b-a^2b-ab^2, which is equivalent to proving 0 \leq 2b-a^2b+b^2a. Can you see why that final inequality is true?

I didn't even try it. I just figured it was a lot easier than the original. Now that I've tried it... it's a lot harder than I remembered it. Maybe there is a slicker way, but here's something. Be sure to check the details because it could be wrong. Let z = x + iy. Assume x > 0. (Hopefully, you can handle x < 0 on your own.) Then

Since f(1) = 0, this implies f(r) > 0 for all r in (0,1). Hence,

Now check that if 0 < A < B < C, then (B - A)/(C - A) < B/C. (You can check this with calculus like above.) Applying this to the above gives

Edit: Okay, calculus is obviously not necessary for (B - A)/(C - A) < B/C. Also,

which is in the unit disk, so the numerator must be smaller than the denominator, making calculus unnecessary even for that.

(Red '+' should be '-', yes?) Is it just because 2b-a^2b-b^2a = b(1-a^2)+b(1-ba)? Meh...I simply MUST learn basic inequality skills. Are these good first books: Inequalities by Hardy, Littlewood, Polya; and The Cauchy-Schwarz Master Class by Steele? Would you recommend others?

jason1990 -

That explains a lot...I'm too frustrated to focus very well right now, but the basic theme seems clear. I will try to work through it on my own tomorrow. (I did recognize the x < 0 case is trivial, but algebraic manipulation didn't seem to help in the least when x > 0.)

If I ever learn this material, I will seek out people to help in your name.

Or just...

|1-zr|^2 - |z-r|^2 = (1-|z|^2)(1-r^2) > 0

...yes? I think the (B - A)/(C - A) < B/C step is most opaque to me, since I completely failed to notice it after five hours.

This looks right, and it shows that ϕ_r(z) is in the unit disk. A similar inequality holds when r is complex. But this should really be proven prior to this last step we are discussing. If nothing else, z in this last step is really wϕ_z2(z₁), and we would like to know that |z| < 1.

I had a friend who worked on some very complicated research, but his papers were always much shorter than mine. I asked him why he thought that was. His answer was that much of his work involved long strings of equalities. According to him, most of these could be omitted and left to the reader, because there are only a limited number of ways to get from one end of a string of equalities to the other. My work, on the other hand, involves long strings of inequalities. And, again according to him, these need to be explicitly described to the reader, because the reader would not be able to easily reproduce them.

I am not sure how much truth is in my friend's comments. I do, however, feel there is a lot of art in doing inequalities. Whatever the case may be, there is no denying that inequalities are at the heart of analysis.

3.4.15
Let Y be the random variable described in question 3.4.3. Define W = 3Y+2. Find f_W(w). For which values of w is f_W(w) does not equal 0.

3.4.3
let f_Y(y)=(3/2)y^2, -1<=y<=1

I guess my problem here is he part define W=3Y+2 am I substituting the equation in 3.4.3 for Y in 3.4.15?

IM confused on multiplying cycles

2 examples

(1,4,5,6)(2,1,5)

A={1,2,3,4,5,6}

multiplying we get

(1 2 3 4 5 6)
(6 4 3 5 2 1)

2ND EXAMPLE

(2,1,5)(1,4,5,6)

multiplying we get

(1 2 3 4 5 6)
(4 1 3 2 6 5)

same set A, If you could explain this to me it would be great, my book is a little vague.

(1,4,5,6)(2,1,5)

Always go right to left in terms of cycles. In each individual cycle go from left to right.

Start on the right hand side. 2 goes to 1, where does 1 go in the next cycle? It goes to 4.

1 goes to 5, then 5 goes to 6, and 6 doesn't go anywhere. So 1 goes to 6

5 goes to 2, and 2 doesn't go anywhere else. So 5 goes to 2.

3 goes to itself since it does not appear in any of the cycles.
4 goes to 5.
6 goes to 1.

(1 2 3 4 5 6)
(6 4 3 5 2 1)

You can always check by re-writing the cycles like
(1,4,5,6) =
1,2,3,4,5,6
4,2,3,5,6,1

(2,1,5)=
1,2,3,4,5,6
5,1,3,4,2,6

Then multiplying, but you will find with some practice (1,4,5,6)(2,1,5) is much faster.