On 12/18/2021 12:05 PM, Julio Di Egidio wrote:
> On Friday, 17 December 2021 at 18:39:42 UTC+1,
> Mike Terry wrote:
>> On 17/12/2021 08:16, Julio Di Egidio wrote:

> That was quite uncalled for, as usual with you:

You -- or someone using your account -- called for it here:

>>> I am not the expert here and if you guys keep
>>> not answering I'll have to go back reading definitions:

On Saturday, 18 December 2021 at 21:00:25 UTC+1, Jim Burns wrote:
> On 12/18/2021 12:05 PM, Julio Di Egidio wrote:
> > On Friday, 17 December 2021 at 18:39:42 UTC+1,
> > Mike Terry wrote:
> >> On 17/12/2021 08:16, Julio Di Egidio wrote:
>
> > That was quite uncalled for, as usual with you:
> You -- or someone using your account -- called for it here:
> >>> I am not the expert here and if you guys keep
> >>> not answering I'll have to go back reading definitions:

*That* was uncalled for, but of course you just rewrite history,
you other resident retarded systematic lying piece of shit...

*Lying Cunt Alert*

Julio

On Wednesday, December 15, 2021 at 1:42:01 PM UTC-6, ju...@diegidio.name wrote:
> On Wednesday, 15 December 2021 at 20:31:18 UTC+1, FromTheRafters wrote:
> > Julio Di Egidio used his keyboard to write :
> > > On Wednesday, 15 December 2021 at 04:53:00 UTC+1, RichD wrote:
> > >> A one meter glass rod falls and breaks in two, at
> > >> a random point. What's the expected ratio of the longer
> > >> to shorter piece?
> > >>
> > >> This is a schoolboy homework problem, nothing deep.
> > >> But it's also a trick question, which makes it mildly amusing -
> > >
> > > The expected value of that ration is 1/2.
> > >
> > > I disagree with Mike's approach, to begin with because if things blow up then
> > > it's not a probability distribution to begin with. Here I'd simply start
> > > from the expected value of the break point position, which I see no reason to
> > > think should be other than the midpoint, then the expected ratio you are
> > > asking about is simply 1/2.
> > I would say that the piece on average breaks in half yielding a one to
> > one ratio with no 'longer' or 'shorter' piece to actually compare.
> Yep, sorry of course I did mean a ratio of 1/1 in that case...
> > An expected value is not always a possible value.
> That's another story.
>
> Julio

Hi, Julio, sorry I dissent, in that I have had some actual physics encounter with this problem. That the glass rod will break at the tip end where it most like impacts the ground concrete or floor. So I would say that a ratio of 1 to 10 in distance so if a 100 cm glass rod hit the floor, that you have two pieces of length 10 cm and 90 cm.

I think the problem here is a divorce between math idealisms and that of actual physical reality.

This is my opinion and other than actual experiments has no supporting evidence.

In fact I had about 4 glass rods all of which were tip end broken off.

It is extremely difficult to get a glass rod to land parrallel to the floor of concrete. And most often the glass rod, one of its ends impacts the floor first and receives the most physical shock wave, breaking at the end.

One of my glass prisms is broken at the end, and I assume it was dropped on the floor.

On Saturday, 18 December 2021 at 22:57:53 UTC+1, Archimedes Plutonium wrote:
> On Wednesday, December 15, 2021 at 1:42:01 PM UTC-6, ju...@diegidio.name wrote:
> > On Wednesday, 15 December 2021 at 20:31:18 UTC+1, FromTheRafters wrote:
> > > Julio Di Egidio used his keyboard to write :
> > > > On Wednesday, 15 December 2021 at 04:53:00 UTC+1, RichD wrote:
> > > >> A one meter glass rod falls and breaks in two, at
> > > >> a random point. What's the expected ratio of the longer
> > > >> to shorter piece?
> > > >>
> > > >> This is a schoolboy homework problem, nothing deep.
> > > >> But it's also a trick question, which makes it mildly amusing -
> > > >
> > > > The expected value of that ration is 1/2.
> > > >
> > > > I disagree with Mike's approach, to begin with because if things blow up then
> > > > it's not a probability distribution to begin with. Here I'd simply start
> > > > from the expected value of the break point position, which I see no reason to
> > > > think should be other than the midpoint, then the expected ratio you are
> > > > asking about is simply 1/2.
> > > I would say that the piece on average breaks in half yielding a one to
> > > one ratio with no 'longer' or 'shorter' piece to actually compare.
> > Yep, sorry of course I did mean a ratio of 1/1 in that case...
> > > An expected value is not always a possible value.
> > That's another story.
> >
> Hi, Julio, sorry I dissent,

I proposed 1 and I still don't see how it can it be otherwise, but I won't be sure about it until I see it actually done or do it myself.

> in that I have had some actual physics encounter with this problem. That the glass rod will break at the tip end where it most like impacts the ground concrete or floor. So I would say that a ratio of 1 to 10 in distance so if a 100 cm glass rod hit the floor, that you have two pieces of length 10 cm and 90 cm.

Fair enough, but we just weren't given that information here, so we are assuming it can break anywhere equally likely. If we change the distribution, we would indeed get something along the lines that you say.

> I think the problem here is a divorce between math idealisms and that of actual physical reality.

It simply is not a physics problem; moreover, physical or not, there do exists situations where *we* just don't know better, still we must learn to work with the info we have...

Julio

On Saturday, December 18, 2021 at 4:06:41 PM UTC-6, ju...@diegidio.name wrote:
> On Saturday, 18 December 2021 at 22:57:53 UTC+1, Archimedes Plutonium wrote:
> > On Wednesday, December 15, 2021 at 1:42:01 PM UTC-6, ju...@diegidio.name wrote:
> > > On Wednesday, 15 December 2021 at 20:31:18 UTC+1, FromTheRafters wrote:
> > > > Julio Di Egidio used his keyboard to write :
> > > > > On Wednesday, 15 December 2021 at 04:53:00 UTC+1, RichD wrote:
> > > > >> A one meter glass rod falls and breaks in two, at
> > > > >> a random point. What's the expected ratio of the longer
> > > > >> to shorter piece?
> > > > >>
> > > > >> This is a schoolboy homework problem, nothing deep.
> > > > >> But it's also a trick question, which makes it mildly amusing -
> > > > >
> > > > > The expected value of that ration is 1/2.
> > > > >
> > > > > I disagree with Mike's approach, to begin with because if things blow up then
> > > > > it's not a probability distribution to begin with. Here I'd simply start
> > > > > from the expected value of the break point position, which I see no reason to
> > > > > think should be other than the midpoint, then the expected ratio you are
> > > > > asking about is simply 1/2.
> > > > I would say that the piece on average breaks in half yielding a one to
> > > > one ratio with no 'longer' or 'shorter' piece to actually compare.
> > > Yep, sorry of course I did mean a ratio of 1/1 in that case...
> > > > An expected value is not always a possible value.
> > > That's another story.
> > >
> > Hi, Julio, sorry I dissent,
> I proposed 1 and I still don't see how it can it be otherwise, but I won't be sure about it until I see it actually done or do it myself.
> > in that I have had some actual physics encounter with this problem. That the glass rod will break at the tip end where it most like impacts the ground concrete or floor. So I would say that a ratio of 1 to 10 in distance so if a 100 cm glass rod hit the floor, that you have two pieces of length 10 cm and 90 cm.
> Fair enough, but we just weren't given that information here, so we are assuming it can break anywhere equally likely. If we change the distribution, we would indeed get something along the lines that you say.
> > I think the problem here is a divorce between math idealisms and that of actual physical reality.
> It simply is not a physics problem; moreover, physical or not, there do exists situations where *we* just don't know better, still we must learn to work with the info we have...
>
> Julio

Yes, so true, Julio, that so many of these questions in sci.math are so vague they are almost not worth replying. Perhaps 80% of questions put to sci.math are too vague.

But the interesting part of this question at least for me is the idea that when something falls in the shape of a rod, that one of the endpoints becomes point of impact.

So if we have thousands of rod shaped objects falling from a building window onto the concrete pavement. I would hazard to guess that if we could slow motion photography the moment of impact, that over 95% hit the pavement from one of the ends. And all dropped rods starting out as parallel to pavement.

Now whether in a vacuum, started out parallel and by the time it impacts whether it is the same 95% I do not know.

And a very exciting question is whether we electrify the glass rod before we drop it, whether the electrification makes the impact be 100% at one of the ends. For I am thinking that the electricity in the ambient air causes the rod shaped object to orient itself to hitting the pavement at one of its ends.

Thinking about airplanes when they "fall from the sky" that their moment of impact is usually nose first on impact. Yet built aerodynamically, one would think the falling airplane can glide to a landing, but no, they smack into the ground nose first. Of course I am speaking of airplanes all of which lost complete control of the craft.

AP

On 12/18/2021 4:47 PM, Julio Di Egidio wrote:
> On Saturday, 18 December 2021 at 21:00:25 UTC+1,
> Jim Burns wrote:
>> On 12/18/2021 12:05 PM, Julio Di Egidio wrote:
>>> On Friday, 17 December 2021 at 18:39:42 UTC+1,
>>> Mike Terry wrote:
>>>> On 17/12/2021 08:16, Julio Di Egidio wrote:

>>> That was quite uncalled for, as usual with you:
>>
>> You -- or someone using your account --
>> called for it here:
>>
>>>>> I am not the expert here and if you guys keep
>>>>> not answering I'll have to go back reading definitions:
>
> *That* was uncalled for, but of course you just rewrite
> history, you other resident retarded systematic
> lying piece of shit...
>
> *Lying Cunt Alert*
>
> Julio

Rewriting history? Or is it... AN ALTERNATE TIMELINE

https://youtu.be/I9xN0Ol5vZQ?t=104

My theory is that Julio is wearing a goatee now.

The ease with which a high diver of huge height can straighten out so as to land in water with head first, not a belly flop or a back flop is suggestive of a electricity and magnetism entrance into water or even a glass rod falling from heights. That somehow the EM forces within the rod and the Earth's gravity favors a perpendicular impact.

AP

On 12/15/2021 7:54 PM, RichD wrote:
> On December 15, FromTheRafters wrote:
>>>> A one meter glass rod falls and breaks in two, at
>>>> a random point. What's the expected ratio of the longer
>>>> to shorter piece?
>>
>> I would say that the piece on average breaks in half yielding a one to
>> one ratio with no 'longer' or 'shorter' piece to actually compare.
>
> No.
> There's a long and short piece, every time. They represent two distinct
> random variables, with distinct distributions.

it is not two random variables, just one say L1, the other is L2 = TL - L1 where TL is total length

>
> Distinct, but certainly not independent - independence isn't a necessary
> condition, in this case.
>
>
> --
> Rich

the break could happen with equal probability at any point along the rod

that is equivlent to picking a random number out of an interval, say 0 to 1, and asking what the distribution of lengths are.

On Saturday, December 18, 2021 at 11:11:19 PM UTC-6, Serg io wrote:
> On 12/15/2021 7:54 PM, RichD wrote:
> > On December 15, FromTheRafters wrote:
> >>>> A one meter glass rod falls and breaks in two, at
> >>>> a random point. What's the expected ratio of the longer

Well, it is like Julio said-- vague, vague question.

Why even bother with talking about a glass rod falling and breaking if you then add the idea of "random point"

Why not just say, in all that vagueness, why not just say you have a glass rod laying across the toilet bowl and where is the rod most covered in piss or shit.

Or have the rod be the clothes line and at what random point is the first bird likely to land on.

Or have a rod and what is the first random point your toddler takes a stick and beats the rod.

> >>>> to shorter piece?
> >>
> >> I would say that the piece on average breaks in half yielding a one to
> >> one ratio with no 'longer' or 'shorter' piece to actually compare.
> >
> > No.
> > There's a long and short piece, every time. They represent two distinct
> > random variables, with distinct distributions.
> it is not two random variables, just one say L1, the other is L2 = TL - L1 where TL is total length
> >
> > Distinct, but certainly not independent - independence isn't a necessary
> > condition, in this case.
> >
> >
> > --
> > Rich
> the break could happen with equal probability at any point along the rod
>
> that is equivlent to picking a random number out of an interval, say 0 to 1, and asking what the distribution of lengths are.

So, Julio is right and correct, and that we have to ask where in the world do we get such idiots framing a idiotic question? Is there some graduate school for Idiotic Vague questions. For which we have a Serg io _robustly_ interested in idiotic vague questions.

On Tuesday, December 14, 2021 at 9:53:00 PM UTC-6, RichD wrote:
> A one meter glass rod falls and breaks in two, at
> a random point. What's the expected ratio of the longer
> to shorter piece?
>
> This is a schoolboy homework problem, nothing deep.
> But it's also a trick question, which makes it mildly amusing -
>
> --
> Rich

AP is fascinated with the question of forming the most obtuse and mindless question in math possible. And the winner of the most obtuse question has to lead many people astray by just the sheer number of people it draws in.

So in the above Rich obtuse question we can all gather that Rich was borne in a generation where they taught him at school that "no question is a dumb question" just to get kids to raise questions.

But in truth and reality, perhaps over 50% of questions are dumb questions that the person never thought through their question.

In Rich's above, he mixs and mingles the physics law of gravity, with a dumb and worthless and fakery math concept of "random number".

Rich was of the generation also where they fobbed off this crazy insane concept of "random number". There never existed a concept of random number. There is truly a concept of "given any number" but there never was a dumb pitiful concept of a random number, which only exists in the minds of dumb people in math.

So Rich has a nice problem of a "mindless question of mixing Gravity law with the insane random number".

Can AP think of a more dumber question that has plenty of people falling heads over heals backwards in answering the dumb garbled mixed up question?

I remember in the heyday of Usenet sci.math, sci.physics in the 1990s when someone-- I believe in Poland listed a sheet of funny jokes on mathematics. One of them went like this---

As he quartered the circle into 7 equal parts.

So taking that as a question to sci.math.

Can you quarter the Circle into 7 equal parts?

On Sunday, December 19, 2021 at 3:53:42 PM UTC-6, Archimedes Plutonium wrote:
> On Tuesday, December 14, 2021 at 9:53:00 PM UTC-6, RichD wrote:
> > A one meter glass rod falls and breaks in two, at
> > a random point. What's the expected ratio of the longer
> > to shorter piece?
> >
> > This is a schoolboy homework problem, nothing deep.
> > But it's also a trick question, which makes it mildly amusing -
> >
> > --
> > Rich
>
> AP is fascinated with the question of forming the most obtuse and mindless question in math possible. And the winner of the most obtuse question has to lead many people astray by just the sheer number of people it draws in.
>
> So in the above Rich obtuse question we can all gather that Rich was borne in a generation where they taught him at school that "no question is a dumb question" just to get kids to raise questions.
>
> But in truth and reality, perhaps over 50% of questions are dumb questions that the person never thought through their question.
>
> In Rich's above, he mixs and mingles the physics law of gravity, with a dumb and worthless and fakery math concept of "random number".
>
> Rich was of the generation also where they fobbed off this crazy insane concept of "random number". There never existed a concept of random number. There is truly a concept of "given any number" but there never was a dumb pitiful concept of a random number, which only exists in the minds of dumb people in math.
>
> So Rich has a nice problem of a "mindless question of mixing Gravity law with the insane random number".
>
> Can AP think of a more dumber question that has plenty of people falling heads over heals backwards in answering the dumb garbled mixed up question?
>
> I remember in the heyday of Usenet sci.math, sci.physics in the 1990s when someone-- I believe in Poland listed a sheet of funny jokes on mathematics. One of them went like this---
>
> As he quartered the circle into 7 equal parts.
>
> So taking that as a question to sci.math.
>
> Can you quarter the Circle into 7 equal parts?

Oh yes, well, recently in sci.math, Dan Christensen came up with a dumber than usual question of what happens when a function runs out of numbers and you apply that function to those numbers.

Of course another Dan Christensen foible was when he declared a vertex point has a derivative.

On December 18, Archimedes Plutonium wrote:
> The ease with which a high diver of huge height can straighten out so as
> to land in water with head first, not a belly flop or a back flop is suggestive of
> a electricity and magnetism entrance into water or even a glass rod falling from heights.

Not so, platypus breath.

It's a demonstration of the Falling Cat theorem, mastered by
divers and cats. Whose brain size, compared to yours, are liketh
unto a gnu compared to a lemming.

--
Rich

On Sunday, December 19, 2021 at 6:03:47 PM UTC-6, RichD wrote:
> On December 18, Archimedes Plutonium wrote:
> > The ease with which a high diver of huge height can straighten out so as
> > to land in water with head first, not a belly flop or a back flop is suggestive of
> > a electricity and magnetism entrance into water or even a glass rod falling from heights.
> Not so, platypus breath.
>
> It's a demonstration of the Falling Cat theorem, mastered by
> divers and cats. Whose brain size, compared to yours, are liketh
> unto a gnu compared to a lemming.
>
>
> --
> Rich

Aw, don't be sore, rich, I am hoping you can supply us with far more dumb mangled questions than your one above.

For what you taught the world, is that a dumb foolish idiotic question, more than likely contains a concept that is itself a piece of shit-- random number

Diego Garcia <dg@chaos.info> wrote:
> On Wed, 15 Dec 2021 13:52:30 -0600, Serg io wrote:
>> 1
> Yes. 1 is the correct answer.
>
> Breaking the rod into two pieces is equivalent to picking a point at random
> along its length.
>
> The density function is uniform, and the expectation value of a uniform
> distribution, with length = 1, is 1/2.
>
> Thus, the ratio 1/2 / 1/2 = 1.
>
> Someone should do a computer simulation to verify this.

Computer simulation (below) indicates that the previously-mentioned
2(log(2)-1/2), or about 0.38629436, is plausible. Actually, a simple
calculation proves that the expected value of the ratio of the shorter
length to the longer (when a rod of given length breaks at a uniformly
random point) is less than 1. For this proof, let X denote a random
variate uniformly distributed on (0,1); let F(X) = ratio of min part
length to max part length = min(X,1-X)/max(X,1-X); and let
???{a,b}(F(x)) represent the integral of F over interval [a, b]. (Note,
we use ???{a,b}(F(x)) = ???{a,c}(F(x)) + ???{c,b}(F(x)) in following.)

Now E(F(X)) = ???{0,1}(F(X)) = ???{0,1/3}(F(X)) + ???{1/3,2/3}(F(X)) +
???{2/3,1}(F(X)) < 1/3 * (1/2 + 1 + 1/2) = 2/3 < 1, completing proof
that the expected value of the ratio is less than 1.

The inequality arises because F(X) < 1/2 in the outer thirds of (0,1),
and F(X) < 1 in the inner third. One can extend this technique
further, eg we can break the integral up across 9 intervals and note
that 1/9*(1/8+2/7+3/6+4/5+1+4/5+3/6+2/7+1/8) < 0.4912698413 < 1/2.

Anyhow, the simple calculation shows E(F(X)) < 2/3, and a slightly
longer calculation shows it's less than 1/2. To get the earlier-
mentioned correct answer, we need to evaluate ???{0,1}(F(X)), which
equals ???{0,1/2}(F(X)) + ???{1/2,1}(F(X)). The first of those is
???{0,1/2}(X/(1-X)) which by change of variables U = 1-X is equal to
???{1/2,1}((1-U)/U) = ???{1/2,1}((1-X)/X) = ???{1/2,1}(F(X)) =
(ln(1)-ln(1/2)) - 1/2 = ln(2)-1/2, whence E(F(X)) = 2(log(2)-1/2).

Here's some computer code (in Julia language) to try to estimate the
expected value of F via a Monte-Carlo-like method:

F(X) = min(X,1-X)/max(X,1-X)
expectF(tries) = sum(F(rand()) for i in 1:tries)/tries

Here are some example executions of function expectF (in Julia REPL):

julia> expectF(1000000)
0.3864327640365364
julia> expectF(10000000)
0.38632295692699936
julia> expectF(10000000)
0.3864170969570848
julia> 2log(2)-1
0.3862943611198906

Here's basically the same code in Python, plus one execution:

from random import random
def F(X): return min(X,1-X)/max(X,1-X)
def expectF(T): return sum(F(random()) for i in range(T))/T

In [5]: expectF(1000000)
Out[5]: 0.38630706073483484

On Sunday, December 19, 2021 at 11:15:23 PM UTC-6, James Waldby wrote:
> Diego Garcia <d...@chaos.info> wrote:
> > On Wed, 15 Dec 2021 13:52:30 -0600, Serg io wrote:
> >> 1
> > Yes. 1 is the correct answer.
> >
> > Breaking the rod into two pieces is equivalent to picking a point at random
> > along its length.
> >
> > The density function is uniform, and the expectation value of a uniform
> > distribution, with length = 1, is 1/2.
> >
> > Thus, the ratio 1/2 / 1/2 = 1.
> >
> > Someone should do a computer simulation to verify this.
> Computer simulation (below) indicates that the previously-mentioned
> 2(log(2)-1/2), or about 0.38629436, is plausible. Actually, a simple
> calculation proves that the expected value of the ratio of the shorter
> length to the longer (when a rod of given length breaks at a uniformly
> random point) is less than 1. For this proof, let X denote a random
> variate uniformly distributed on (0,1); let F(X) = ratio of min part
> length to max part length = min(X,1-X)/max(X,1-X); and let
> ???{a,b}(F(x)) represent the integral of F over interval [a, b]. (Note,
> we use ???{a,b}(F(x)) = ???{a,c}(F(x)) + ???{c,b}(F(x)) in following.)
>
> Now E(F(X)) = ???{0,1}(F(X)) = ???{0,1/3}(F(X)) + ???{1/3,2/3}(F(X)) +
> ???{2/3,1}(F(X)) < 1/3 * (1/2 + 1 + 1/2) = 2/3 < 1, completing proof
> that the expected value of the ratio is less than 1.
>
> The inequality arises because F(X) < 1/2 in the outer thirds of (0,1),
> and F(X) < 1 in the inner third. One can extend this technique
> further, eg we can break the integral up across 9 intervals and note
> that 1/9*(1/8+2/7+3/6+4/5+1+4/5+3/6+2/7+1/8) < 0.4912698413 < 1/2.
>
> Anyhow, the simple calculation shows E(F(X)) < 2/3, and a slightly
> longer calculation shows it's less than 1/2. To get the earlier-
> mentioned correct answer, we need to evaluate ???{0,1}(F(X)), which
> equals ???{0,1/2}(F(X)) + ???{1/2,1}(F(X)). The first of those is
> ???{0,1/2}(X/(1-X)) which by change of variables U = 1-X is equal to
> ???{1/2,1}((1-U)/U) = ???{1/2,1}((1-X)/X) = ???{1/2,1}(F(X)) =
> (ln(1)-ln(1/2)) - 1/2 = ln(2)-1/2, whence E(F(X)) = 2(log(2)-1/2).
>
> Here's some computer code (in Julia language) to try to estimate the
> expected value of F via a Monte-Carlo-like method:
>
> F(X) = min(X,1-X)/max(X,1-X)
> expectF(tries) = sum(F(rand()) for i in 1:tries)/tries
>
> Here are some example executions of function expectF (in Julia REPL):
>
> julia> expectF(1000000)
> 0.3864327640365364
> julia> expectF(10000000)
> 0.38632295692699936
> julia> expectF(10000000)
> 0.3864170969570848
> julia> 2log(2)-1
> 0.3862943611198906
>
> Here's basically the same code in Python, plus one execution:
>
> from random import random
> def F(X): return min(X,1-X)/max(X,1-X)
> def expectF(T): return sum(F(random()) for i in range(T))/T
>
> In [5]: expectF(1000000)
> Out[5]: 0.38630706073483484

James Waldby, willfully plunging into the depths of ludicrous silliness.

So then James is saying that in Newton's law of gravity, the number 0.386... is somehow fundamental to F = G*m_1*m_2 / r^2

Now if RichD had started his silly question -- "
A one meter glass rod falls and breaks in two, at
a random point. What's the expected ratio of the longer
to shorter piece?

This is a schoolboy homework problem, nothing deep.
But it's also a trick question, which makes it mildly amusing -"

rather instead to be this:

"A one meter rod where a math person can make a mark anywhere on that rod, at
a random point. What's the expected ratio of the longer
to shorter piece?

This is a math professors out at flying a kite homework problem, nothing deep.
But it's also a trick question, which makes it mildly amusing -

On Mon, 20 Dec 2021 05:15:13 +0000, James Waldby wrote:

>
> Computer simulation (below) indicates that the previously-mentioned
> 2(log(2)-1/2), or about 0.38629436, is plausible.
>

That is the expected value of the short/long ratio.

The problem is solved by considering the expected value of a function
of a random variable. In this case, the random variable is x, and the function
is f(x) = x/(1-x) or f(x) = (1-x)/x for the short/long ratio. (For long/short just invert.)

Solve via integration:

integrate(x/(1-x), x, 0, 1/2) + integrate((1-x)/x, x, 1/2, 1) =

(2*log(2)+1)/2+(2*log(2)-1)/2-1 = 0.3862943611198907

For the long/short ratio, the integrals become divergent.

This is verified by the C code simulation using the GNU Scientific Library
to produce random numbers from a uniform distribution (code below).

Results:

N: average short/long long/short

10000: 0.501162 0.380338 20.6597
100000: 0.498832 0.38641 24.7128
1000000: 0.500327 0.386576 31.973
10000000: 0.499986 0.386274 32.0904

The average of N draws is 1/2.
The average of short/long is 0.386
The average of long/short diverges to infinity

C code

#define _GNU_SOURCE
#define HAVE_INLINE
#define GSL_RANGE_CHECK_OFF
#define GSL_DISABLE_DEPRECATED
#include <stdio.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <unistd.h>
#include <gsl/gsl_rng.h>
#include <gsl/gsl_randist.h>

int main(int argc, char **argv)
{ int fd1;
unsigned long int seed, i, n;
double rnd_val, a, b;
double ratsl=0.0, ratls=0.0;
double avgsl=0.0, avgls=0.0;
double prob = 0.0;
gsl_rng *r;

n=strtoul(argv[1], NULL, 10);
a=atof(argv[2]);
b=atof(argv[3]);

// initialize generator
seed=0; fd1=open("/dev/urandom", O_RDONLY); read(fd1, &seed, 8); close(fd1);
r=gsl_rng_alloc(gsl_rng_mt19937);
gsl_rng_set(r,seed);

for(i=1UL;i<=n;i++)
{ // rnd_vali=gls_rng_get(r);
rnd_val=gsl_ran_flat(r, a, b);
if(rnd_val<=0.5) { ratsl=rnd_val/(1.0-rnd_val); ratls=(1.0-rnd_val)/rnd_val; }
else { ratls=rnd_val/(1.0-rnd_val); ratsl=(1.0-rnd_val)/rnd_val; }
avgsl=avgsl+ratsl; avgls=avgls+ratls;
prob = prob+rnd_val;
}

avgsl=avgsl/(double)n; avgls=avgls/(double)n;
prob=prob/(double)n;
fprintf(stdout, "%ld: %g %g %g\n", n, prob, avgsl, avgls);

gsl_rng_free(r);
return 0;
}

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On Tuesday, 21 December 2021 at 11:37:06 UTC+1, Diego Garcia wrote:
> On Mon, 20 Dec 2021 05:15:13 +0000, James Waldby wrote:
> >
> > Computer simulation (below) indicates that the previously-mentioned
> > 2(log(2)-1/2), or about 0.38629436, is plausible.
> >
> That is the expected value of the short/long ratio.
>
> The problem is solved by considering the expected value of a function

Thanks for posting it.

> of a random variable. In this case, the random variable is x, and the function
> is f(x) = x/(1-x) or f(x) = (1-x)/x for the short/long ratio. (For long/short just invert.)
>
> Solve via integration:
>
> integrate(x/(1-x), x, 0, 1/2) + integrate((1-x)/x, x, 1/2, 1) =
>
> (2*log(2)+1)/2+(2*log(2)-1)/2-1 = 0.3862943611198907

Being a rookie in probability, I am quite puzzled by all this.

Let e = E(x/(1-x)) = ~0.38, the "corresponding x" is x = e/(1-e) = ~0.63.

How do we interpret that result, if that "corresponding x" means anything at all?
Is it from the left or from the right end of the rod? Or, both are valid?
(But I guess the same could be asked of the orientation of that ratio already.
Is there a way to fix it or is it just both, as in an absolute value somewhere?)

I am also puzzled by the fact that the "corresponding x" is not the mid point, but taking x alone as the random variable, meaning where we expect the rod to break, that is the mid-point, so a ratio of 1/1. I cannot reconcile the two x values: the very meaning of what we are computing above.

If anybody finds the time and inclination to explain some basics here, that would be very much appreciated.

> For the long/short ratio, the integrals become divergent.
>
> This is verified by the C code simulation using the GNU Scientific Library
<snip>

Julio