/sci/ - Science & Math

So I actually cheated and gave a look at it while posting it. The PDF of the <span class="math">Z=X/Y[/spoiler] is given by what's called a ratio distribution and is (c.f. http://en.wikipedia.org/wiki/Ratio_distribution#Uniform_ratio_distribution):
<span class="math">1/2[/spoiler] for <span class="math">0<z<1[/spoiler]
<span class="math">\frac{1}{2z^2}[/spoiler] for <span class="math">z\geq 1[/spoiler]
<span class="math">0[/spoiler] otherwise.

From there, we write the probabilities for even and odd values of <span class="math">\lfloor z\rfloor[/spoiler] (where I use the <span class="math">\lfloor \rfloor[/spoiler] for the closest integer and not the floor) as sums over <span class="math">(0,1/2)\cup (3/2,5/2)\cup (7/2,9/2)\cup \dots[/spoiler] and its complement in <span class="math">\mathbb{R}^+[/spoiler]. Hopefully the simplifications aren't too complicated.

For <span class="math">k>1[/spoiler], let we consider
<div class="math">\int_{\frac{2k +1}{2}}^{\frac{2k +3}{2}}\frac{ \mathrm{d}z}{2z^2}</div>
The sum of these for all <span class="math">k>1[/spoiler] will be the contribution of <span class="math">(3/2,5/2)\cup(7/2,9/2)\cup \dots\cup(2k+1,2k+3) \cup\dots[/spoiler] in the probability we're trying to find. The contribution of <span class="math">(0,1/2)[/spoiler] is easier since the PDF is constant equal to <span class="math">1/2[/spoiler] there (so the integral is <span class="math">1/4[/spoiler], which is the probability to obtain <span class="math">0[/spoiler] as the closest integer to <span class="math">x/y[/spoiler]).

> posted 5 minutes
> already cheating

Would you at least ATTEMPT the fucking problem before giving up?

>>4332868
I'm attempting. I read the problem, looked at wikipedia to know if the distribution for what I call <span class="math">z[/spoiler] had a name and a simple PDF, and posted it right after. I took maybe a headstart of 3 minutes compared to /sci/, and used a wikipedia page. That's hardly gamebreaking.

So anyway, I've "cheated" again now since I used wolfram to compute the integral because I'm too lazy to do the simplifications by myself:
<div class="math">\int_{\frac{2k +1}{2}}^{\frac{2k +3}{2}}\frac{ \mathrm{d}z}{2z^2}= \frac{2}{4k^2+8k+3}</div>

So what we want now is
<div class="math">\frac{1}{4} + \sum_{k=1}^{\infty} \frac{2}{4k^2+8k+3}</div>

>>4332875
> excuses

>>4332875
The problem seems to be to find the PDF in the first place because the rest is trivial. Remember you are not allowed to use the internet or books in real putnam.

>>4332883
Okay. Sorry about spoiling the PDF then. We still don't have the proof about that the PDF is correct though, so even if I gave the answer, we can still work on the proof.

Also it looks like I've screwed something up about the rest of the work, because the result I get in the end is actually greater than one...

>>4332881
<div class="math">\sum_{k=1}^{\infty} \frac{2}{4k^2+8k+3}=\sum_{k=1}^{\infty} \left( \frac{1}{2k+1} -\frac{1}{2k+3} \right)</div>
Which is almost the leibniz series for pi.

So let's try to show that the PDF is what I wrote in the first answer.

We are writing <span class="math">z=x/y[/spoiler]. Now we want the PDF <span class="math">p_Z(z)[/spoiler] at a given point <span class="math">z[/spoiler]. To do that, we write <span class="math">x[/spoiler] in terms of <span class="math">y[/spoiler] knowing <span class="math">z[/spoiler] (I guess we could write <span class="math">y[/spoiler] in terms of <span class="math">x[/spoiler] too but there would be a quotient, let's work with a product instead). <span class="math">x=yz[/spoiler]. So now, we use the PDFs of <span class="math">X[/spoiler] and <span class="math">Y[/spoiler]:
<span class="math">p_Z(z)=\int yP_X(yz)P_Y(y) \mathrm{d}y[/spoiler]

For <span class="math">z<0[/spoiler], either <span class="math">y>0[/spoiler] and then <span class="math">yz<0[/spoiler], <span class="math">P_X(yz)=0[/spoiler], or <span class="math">y<0[/spoiler] and then <span class="math">P(y)=0[/spoiler], so anyway the product of the two is <span class="math">0[/spoiler], and so is <span class="math">p_Z(z)[/spoiler]. Also, f.u., non-measurable cases.

From there yeah, the fact that I spoiled two cases will probably help a bit so do the rest, my bad... (but it shouldn't have been hard anyway).

>>4332926

fuck off.

>>4332926
fuck off

Might as well just call these stickies TN5's notebook. He logs his every thought in these threads.

> OH GOODIE A NEW PUTNAM THREAD
> Has a violent seizure on the keyboard.
> 2408tyghaf0GH)*ghf0a4ghf0........
> Wait, I think I messed that up a little.
> TEEEEEEE HEEEEE!

>>4332926
FUCK OFF

>>4332926
>non-measurable
I meant "zero measure".

First of all, let's put some bounds on the integral. We want <span class="math">p_Y(y)\neq 0[/spoiler] so we need <span class="math">y\in(0,1)[/spoiler].

So for <span class="math">0\leq z\leq 1[/spoiler], <span class="math">p_Z(z)=\int_0^1 yP_X(yz) \mathrm{d}y[/spoiler]
And because <span class="math">z\in(0,1)[/spoiler], the fact that <span class="math">y\in(0,1)[/spoiler] also gives <span class="math">yz\in(0,1)[/spoiler], so <span class="math">p_Z(z)=\int_0^1 y \mathrm{d}y = 1/2[/spoiler].

Also:
>>4332936
I guess I could try to go more to the point.

>>4332929
>>4332933
>>4332939
Cool, I have a fan club. Do you read the Putnam thread because you want to participate (in which case, by all means, do it, and constructively tell me if I'm annoying, I listen to intelligible criticism), or just to spit on me (in which case, well do it anyway, it's not like a I care).

>>4332881
Here is your mistake in the integral your even numbers are the 2k. In the sum you sum over the k not the 2k.

>>4332944
Thanks! I feel a bit dumb...

>>4332945
Anyway, when <span class="math">z>1[/spoiler], the upper bound of the integral can be reduced to <span class="math">1/z[/spoiler] (otherwise <span class="math">p_Z(z)=0[/spoiler]):
<div class="math">p_Z(z)=\int_0^{1/z} y \mathrm{d}y=\frac{1}{2z^2}</div>

Actually finding the PDF was not too hard.

I'm done with /sci/.

Goodbye.

For <span class="math">x,y\in(0,1)[/spoiler] and <span class="math">k\in\mathbb{Z}[/spoiler], <span class="math">x/y\in[2k-1/2,2k+1/2)[/spoiler] iff <div class="math">0<x<y/2,\quad k=0,</div> <div class="math">(2k-1/2)y \leq x < (2k+1/2)y.\quad k\geq1.\quad\quad\quad (I_k)</div> (The inequality cannot hold for negative <span class="math">k[/spoiler].) We must find <span class="math">P(I_0\cup I_1\cup\cdots)[/spoiler]. These events are mutually exclusive, so <div class="math"> P(I_0\cup I_1\cup\cdots)= P(I_0)+\sum_{k=1}^\infty P(I_k).</div> <span class="math">P(I_0) = \int_0^1 y/2\,dy=1/4[/spoiler]. <span class="math">P(I_k) = \int_0^1\frac{x}{2k-1/2} - \frac{x}{2k+1/2}= (\frac{1}{4k-1} - \frac{1}{4k+1}[/spoiler].

So
<span class="math">P(I_0\cup I_1\cup\cdots)= 1/4 + (1/3 - 1/5) + (1/7 - 1/9) +\cdots[/spoiler]
<span class="math">= 5/4 - (1 -1/3 + 1/5 - 1/7 + \cdots)[/spoiler]. I couldn't at first remember if the series in parentheses is the series for <span class="math">tan^{-1}(1)[/spoiler] or <span class="math">sin^{-1}(1)[/spoiler]. But if it were the latter, the probability would be negative ( <span class="math">5/4 - \pi/2 <0[/spoiler]), so that makes me confident it's the former.

So the answer is: <span class="math">5/4 - \pi/4[/spoiler].

For <span class="math">x,y\in(0,1)[/spoiler] and <span class="math">k\in\mathbb{Z}[/spoiler],
<span class="math">x/y\in[2k-1/2,2k+1/2)[/spoiler] iff <div class="math">0<x<y/2,\quad k=0,</div>
<div class="math">(2k-1/2)y \leq x < (2k+1/2)y,\quad k\geq1.\quad\quad\quad (I_k)</div> (The inequality cannot hold for negative <span class="math">k[/spoiler].)
We must find <span class="math">P(I_0\cup I_1\cup\cdots)[/spoiler]. These events are mutually
exclusive, so <div class="math"> P(I_0\cup I_1\cup\cdots)= P(I_0)+\sum_{k=1}^\infty
P(I_k).</div> <span class="math">P(I_0) = \int_0^1 y/2\,dy=1/4[/spoiler]. <div class="math">P(I_k) =
\int_0^1\frac{x}{2k-1/2} - \frac{x}{2k+1/2}= \frac{1}{4k-1} -
\frac{1}{4k+1}.</div> So
<span class="math">P(I_0\cup I_1\cup\cdots)= 1/4 + (1/3 - 1/5) + (1/7 - 1/9) +\cdots[/spoiler]
<span class="math">= 5/4 - (1 -1/3 + 1/5 - 1/7 + \cdots)[/spoiler].

I couldn't at first remember if the series in parentheses is the series for <span class="math">tan^{-1}(1)[/spoiler] or <span class="math">sin^{-1}(1)[/spoiler]. But if it were the latter, the probability would be negative ( <span class="math">5/4 - \pi/2 <0[/spoiler]), so that makes me confident it's the former.

So the answer is: <span class="math">5/4 - \pi/4[/spoiler].

>>4332850
You don't really need to consider the ratio distribution for this problem. All you need to know is that <span class="math">(X,Y)[/spoiler] is uniformly distributed in the unit square <span class="math">(0,1)\times(0,1)[/spoiler]. You eliminate the pesky ratio by multiplying both sides of the inequality <span class="math"> 2k -1/2 \leq x/y < 2k+1/2[/spoiler] by <span class="math">y[/spoiler].

>>4333106
>both sides of the inequality
I meant inequalitIES.

Answer is 1/4 + pi/4, sol'n in a sec

theres no such thing as random so this problem is not solution

QED

>>4334556
Disregard that, i suck cocks

Using the ratio distribution, you can break the positive real line to two parts; the real numbers that are 'closer' to even integers and real numbers that are 'closer' to odd integers.

Suppose n is an even integer; we know that any real number that is within +/-1/2 from n is 'closer' to n than either odd numbers that bound n.

From my image, you can see these areas in blue superimposed on the PDF of the ratio distribution.

Anyway, follow the rest of my solution and you'll come to (using Leibniz' Formula for pi) (1/4) + (1 - pi/4)

So P = 5/4 - pi/4, as in >>4333047

>>4334583
Well done!

>>4332838

yo dog,

what about that post i put up yesterday, you know answering your question by saying: N/S, its an improper(ly) (worded) question?

ill at random trolls u untillz y'all recognize (jpegs? i got em) btw this is the real anon's new mug. f masks we are going full blown costum

like them

like your question

derp? what is the probability of 2 probabilites herp? realz numbers plz. ty.

true answer: n/s (true solution is not a real number, therefore irrelevant becasue is it not math it is gamble.


stop pretending like your smart by moving around mnumbers numbers and recognize please. your pride is not amusing.

haha he says you do grandpa's maths books

>>4333047
>>4334624
nice, guys

>>4335092
Sucking at math is a forced meme at best. Go to bed.

It can greatly vary. What are the measurements between? If its 1/10 then the probability would be much greater then say 1/1000000.

>>4336655
The problem is well defined and has a definite answer. In fact, it has already been answered in the thread.

50% probability.

>>4337259
How cool is it that you're not only wrong, but wrong a day after the correct answer has been posted 3 times by 3 different guys, each with their proof, while you're only posting a trivial guess with no justification? Man I wish trolling came that easy to me. You must get a lot of pussy with that humor.

>>4337283
It's pretty cool actually.

Hello my good friends from /sci/.

I want to learn Java and LabView, so which books do you guys recommend?

Info about myself: I know some C programming.

>>4337999

oh my, I'm sorry. My mistake.