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

On 15/12/2021 03:52, 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 -
>

Nobody's answered, so here's my thinking:

When you divide by a length that can be arbitrarily small (as we must do
here) there's no upper bound on the result so there is the possibility
that the expectation may not exist (or put another way, it might be
infinite). Or it might still be finite, depending on the exact integral
for the expectation. (I just said all that as a pre-warning.)

In your case, let the breakage point be distance x from one end - so the
ratio of longer to shorter is going to be greater than (C/x) for a
suitable constant C, and we will have an integral of (C/x) for an
interval [0, something > 0]. This integral is infinite, as the
indefinite integral is C log(x) which goes to -oo as x goes to zero.

So the expectation is going to be infinite for your problem. I may have
been tricked by the "trick question" aspect though?

Mike.

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.

Julio

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. An
expected value is not always a possible value.

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

On 12/14/2021 9:52 PM, 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

1

next...

RichD <r_delaney2001@yahoo.com> 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?

Ah "random", without specifying the distribution. ;-)

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

Put the as yet unbroken rod on a table, with what is going to be the
shorter end at the left.

The breakage point is going to be uniformly distributed along that left
hand half. The expected break point will be in the middle of that left
half, giving a ratio of 3 to 1.

So, my answer is 3.

> --
> Rich

--
Alan Mackenzie (Nuremberg, Germany).

On 12/15/2021 3:16 PM, Alan Mackenzie wrote:
> RichD <r_delaney2001@yahoo.com> 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?
>
> Ah "random", without specifying the distribution. ;-)

he could have left off "at a random point.

but notice that one expects no really short piece is broken off an end, say 1/20th of total length...

because glass rod break point need some mass on both sides ? breaks into similar mass lengths ?

this implies it is not an even distribution, unless the rod is extremly thin.

what about short FAT rods ?

>
>> This is a schoolboy homework problem, nothing deep.
>> But it's also a trick question, which makes it mildly amusing -
>
> Put the as yet unbroken rod on a table, with what is going to be the
> shorter end at the left.
>
> The breakage point is going to be uniformly distributed along that left
> hand half. The expected break point will be in the middle of that left
> half, giving a ratio of 3 to 1.
>
> So, my answer is 3.

given the assumptions "breakage point is going to be uniformly distributed along that left
hand half"

>
>> --
>> Rich
>

On 15/12/2021 21:16, Alan Mackenzie wrote:
> RichD <r_delaney2001@yahoo.com> 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?
>
> Ah "random", without specifying the distribution. ;-)

Yes, that's a bit of a flaw in the telling of the story. Sometimes
people are berated for pointing this out, but that's completely wrong,
because they are quite right! Different distributions give different
answers.

It's like there's an unspoken rule for these types of problem where the
obvious uniform distribution is to be assumed. (Sometimes people think
this applies even when there is no such thing as a uniform distribution
that it could apply to!! But in this problem the assumption is natural
enough.)

>
>> This is a schoolboy homework problem, nothing deep.
>> But it's also a trick question, which makes it mildly amusing -
>
> Put the as yet unbroken rod on a table, with what is going to be the
> shorter end at the left.

Fair enough.

>
> The breakage point is going to be uniformly distributed along that left
> hand half. The expected break point will be in the middle of that left
> half, giving a ratio of 3 to 1.
>
> So, my answer is 3.

That's not going to be right - what you're arguing would work if the
measurement were linear as a function of x, meaning that the average
measurement will be the value taken in the middle of the interval. But
our measurement is the ratio (longest piece/shortest piece) which is
extremely biased towards the lowest values of x, where the ratio grows
like 1/x.

Mike.

On December 15, Mike Terry 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 -
>
> When you divide by a length that can be arbitrarily small (as we must do
> here) there's no upper bound on the result so there is the possibility
> that the expectation may not exist (or put another way, it might be
> infinite).
> In your case, let the breakage point be distance x from one end - so the
> ratio of longer to shorter is going to be greater than (C/x) for a
> suitable constant C
> This integral is infinite, as the indefinite integral is C log(x)
hmmm... my first reaction is, no problem, the integral is weighted by the
distribution. But that's uniform, which doesn't ameliorate the problem.

Surprising, isn't it? The ratio of short piece to long is well defined, but the
converse blows up. I tricked myself! Good catch.
Re-formulate: What's the expected ratio of the shorter to longer piece?

It appears you're well on the way to the solution. A mere schoolboy task,
as promised.

--
Rich

On December 15, Alan Mackenzie 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 -
>
> Put the as yet unbroken rod on a table, with what is going to be the
> shorter end at the left.
> The breakage point is going to be uniformly distributed along that left
> hand half. The expected break point will be in the middle of that left
> half, giving a ratio of 3 to 1.

You fell into the trap.
Which is better than others, who missed the point by a mile.
Read the question again, more carefully.

However, 3:1 is the correct answer... to a different question...

--
Rich

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.

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

--
Rich

On December 15, RichD wrote:
> Read the question again, more carefully.

The revised version: what's the expected ratio of the shorter to longer piece?

--
Rich

On 16/12/2021 01:38, RichD wrote:
> On December 15, Mike Terry 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 -
>>
>> When you divide by a length that can be arbitrarily small (as we must do
>> here) there's no upper bound on the result so there is the possibility
>> that the expectation may not exist (or put another way, it might be
>> infinite).
>> In your case, let the breakage point be distance x from one end - so the
>> ratio of longer to shorter is going to be greater than (C/x) for a
>> suitable constant C
>> This integral is infinite, as the indefinite integral is C log(x)
>
> hmmm... my first reaction is, no problem, the integral is weighted by the
> distribution. But that's uniform, which doesn't ameliorate the problem.
>
> Surprising, isn't it? The ratio of short piece to long is well defined, but the
> converse blows up. I tricked myself! Good catch.
>
> Re-formulate: What's the expected ratio of the shorter to longer piece?

I'll take your rewording as asking for
E( short length / long length)
which won't have any divergence issues as the random variable is bounded.

But I'll have to look at it tomorrow, and I'm out most of the day so
probably someone else will have got the answer by then...

Quick note: the expectation can't be derived by dividing
E(short length) / E(long length).
(even if they were independent, which they're not..)

Mike.

On Thursday, 16 December 2021 at 03:48:05 UTC+1, Mike Terry wrote:
> On 16/12/2021 01:38, RichD wrote:
> > On December 15, Mike Terry 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 -
> >>
> >> When you divide by a length that can be arbitrarily small (as we must do
> >> here) there's no upper bound on the result so there is the possibility
> >> that the expectation may not exist (or put another way, it might be
> >> infinite).
> >> In your case, let the breakage point be distance x from one end - so the
> >> ratio of longer to shorter is going to be greater than (C/x) for a
> >> suitable constant C
> >> This integral is infinite, as the indefinite integral is C log(x)
> >
> > hmmm... my first reaction is, no problem, the integral is weighted by the
> > distribution. But that's uniform, which doesn't ameliorate the problem.
> >
> > Surprising, isn't it? The ratio of short piece to long is well defined, but the
> > converse blows up. I tricked myself! Good catch.
> >
> > Re-formulate: What's the expected ratio of the shorter to longer piece?

One problem is insoluble and the reciprocal is trivial? Doesn't sound right.

> I'll take your rewording as asking for
> E( short length / long length)
> which won't have any divergence issues as the random variable is bounded.
>
> But I'll have to look at it tomorrow, and I'm out most of the day so
> probably someone else will have got the answer by then...
>
> Quick note: the expectation can't be derived by dividing
> E(short length) / E(long length).
> (even if they were independent, which they're not..)

The expectation of a quotient is not the quotient of the expectations.

But can't we write our problem as E(x/(L-x))? Why is that two variables?

Julio

On 16/12/2021 08:10, Julio Di Egidio wrote:
> On Thursday, 16 December 2021 at 03:48:05 UTC+1, Mike Terry wrote:
>> On 16/12/2021 01:38, RichD wrote:
>>> On December 15, Mike Terry 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 -
>>>>
>>>> When you divide by a length that can be arbitrarily small (as we must do
>>>> here) there's no upper bound on the result so there is the possibility
>>>> that the expectation may not exist (or put another way, it might be
>>>> infinite).
>>>> In your case, let the breakage point be distance x from one end - so the
>>>> ratio of longer to shorter is going to be greater than (C/x) for a
>>>> suitable constant C
>>>> This integral is infinite, as the indefinite integral is C log(x)
>>>
>>> hmmm... my first reaction is, no problem, the integral is weighted by the
>>> distribution. But that's uniform, which doesn't ameliorate the problem.
>>>
>>> Surprising, isn't it? The ratio of short piece to long is well defined, but the
>>> converse blows up. I tricked myself! Good catch.
>>>
>>> Re-formulate: What's the expected ratio of the shorter to longer piece?
>
> One problem is insoluble and the reciprocal is trivial? Doesn't sound right.
>
>> I'll take your rewording as asking for
>> E( short length / long length)
>> which won't have any divergence issues as the random variable is bounded.
>>
>> But I'll have to look at it tomorrow, and I'm out most of the day so
>> probably someone else will have got the answer by then...
>>
>> Quick note: the expectation can't be derived by dividing
>> E(short length) / E(long length).
>> (even if they were independent, which they're not..)
>
> The expectation of a quotient is not the quotient of the expectations.

Exactly!
>
> But can't we write our problem as E(x/(L-x))? Why is that two variables?

Yes, that's what we want. We can calculate it as an integral. Both x
and (L-x) are (obviously dependent) "random variables", as is x/(L-x).
That's a different usage of the word "variable" compared with familiar
equations in algebra.

>
> Julio
>

On 16/12/2021 01:38, RichD wrote:
> On December 15, Mike Terry 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 -
>>
>> When you divide by a length that can be arbitrarily small (as we must do
>> here) there's no upper bound on the result so there is the possibility
>> that the expectation may not exist (or put another way, it might be
>> infinite).
>> In your case, let the breakage point be distance x from one end - so the
>> ratio of longer to shorter is going to be greater than (C/x) for a
>> suitable constant C
>> This integral is infinite, as the indefinite integral is C log(x)
>
> hmmm... my first reaction is, no problem, the integral is weighted by the
> distribution. But that's uniform, which doesn't ameliorate the problem.
>
> Surprising, isn't it? The ratio of short piece to long is well defined, but the
> converse blows up. I tricked myself! Good catch.
>
> Re-formulate: What's the expected ratio of the shorter to longer piece?
>
> It appears you're well on the way to the solution. A mere schoolboy task,
> as promised.
>
> --
> Rich
>

I get 2*ln(2)-1, but I think I'm missing the trick.

Duncan

On December 16, ju...@diegidio.name 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?
>>> Re-formulate: What's the expected ratio of the shorter to longer piece?
>
> But can't we write our problem as E(x/(L-x))? Why is that two variables?

Informationally, they're equivalent. So it looks like one unknown.

Let x be the length of the short piece, y is the long piece.
Draw a histogram for each. The have different domains, hence
distinct random variables.

We want E[x/y]
y = L - x
etc.

--
Rich

On Friday, 17 December 2021 at 02:00:31 UTC+1, RichD wrote:
> On December 16, ju...@diegidio.name 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?
> >>> Re-formulate: What's the expected ratio of the shorter to longer piece?
> >
> > But can't we write our problem as E(x/(L-x))? Why is that two variables?
> Informationally, they're equivalent. So it looks like one unknown.
>
> Let x be the length of the short piece, y is the long piece.
> Draw a histogram for each. The have different domains, hence
> distinct random variables.
>
> We want E[x/y]
> y = L - x
> etc.

I am not the expert here and if you guys keep not answering I'll have to go back reading definitions: indeed, you should show that it's two (or even more, according to Mike) random variables starting *from definitions*, not just by repeating yourself in different words and/or by restricting the problem, which changes nothing in that sense...

Julio

Julio Di Egidio was thinking very hard :
> On Friday, 17 December 2021 at 02:00:31 UTC+1, RichD wrote:
>> On December 16, ju...@diegidio.name 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?
>>>>> Re-formulate: What's the expected ratio of the shorter to longer piece?
>>>
>>> But can't we write our problem as E(x/(L-x))? Why is that two variables?
>> Informationally, they're equivalent. So it looks like one unknown.
>>
>> Let x be the length of the short piece, y is the long piece.
>> Draw a histogram for each. The have different domains, hence
>> distinct random variables.
>>
>> We want E[x/y]
>> y = L - x
>> etc.
>
> I am not the expert here and if you guys keep not answering I'll have to go
> back reading definitions: indeed, you should show that it's two (or even
> more, according to Mike) random variables starting *from definitions*, not
> just by repeating yourself in different words and/or by restricting the
> problem, which changes nothing in that sense...

That could be difficult when dealing with 'random' without any useable
distribution. It seems to me that asserting the left-right to be long-
short or short-long misses the fact that improper ratios are ratios
too.

On Friday, 17 December 2021 at 12:06:41 UTC+1, FromTheRafters wrote:
> Julio Di Egidio was thinking very hard :
> > On Friday, 17 December 2021 at 02:00:31 UTC+1, RichD wrote:
> >> On December 16, ju...@diegidio.name 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?
> >>>>> Re-formulate: What's the expected ratio of the shorter to longer piece?
> >>>
> >>> But can't we write our problem as E(x/(L-x))? Why is that two variables?
> >> Informationally, they're equivalent. So it looks like one unknown.
> >>
> >> Let x be the length of the short piece, y is the long piece.
> >> Draw a histogram for each. The have different domains, hence
> >> distinct random variables.
> >>
> >> We want E[x/y]
> >> y = L - x
> >> etc.
> >
> > I am not the expert here and if you guys keep not answering I'll have to go
> > back reading definitions: indeed, you should show that it's two (or even
> > more, according to Mike) random variables starting *from definitions*, not
> > just by repeating yourself in different words and/or by restricting the
> > problem, which changes nothing in that sense...
>
> That could be difficult when dealing with 'random' without any useable
> distribution.

Which specific distribution it is (and we should assume uniform here, since no specific information was given) is irrelevant to my question, which is a formal question. That said, I think I have found the information I was looking for:

<https://en.wikipedia.org/wiki/Random_variable#Functions_of_random_variables>

Now all that is left is do the calculations... ;)

> It seems to me that asserting the left-right to be long-
> short or short-long misses the fact that improper ratios are ratios
> too.

1/0 is simply undefined in standard mathematics, but I'd grant you that "divergent" is an answer...

Julio

On 17/12/2021 08:16, Julio Di Egidio wrote:
> On Friday, 17 December 2021 at 02:00:31 UTC+1, RichD wrote:
>> On December 16, ju...@diegidio.name 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?
>>>>> Re-formulate: What's the expected ratio of the shorter to longer piece?
>>>
>>> But can't we write our problem as E(x/(L-x))? Why is that two variables?
>> Informationally, they're equivalent. So it looks like one unknown.
>>
>> Let x be the length of the short piece, y is the long piece.
>> Draw a histogram for each. The have different domains, hence
>> distinct random variables.
>>
>> We want E[x/y]
>> y = L - x
>> etc.
>
> I am not the expert here and if you guys keep not answering I'll have to go back reading definitions: indeed, you should show that it's two (or even more, according to Mike) random variables starting *from definitions*, not just by repeating yourself in different words and/or by restricting the problem, which changes nothing in that sense...
>
> Julio
>

I'll go back to definitions for you (I'll follow with a couple of
examples which hopefully will make it clearer, as there's a fair bit to
take in just by reading definitions):

First, we start with an underlying set which we can think of as
containing the outcomes for a hypothetical experiment we might perform.
(The experiment is not essential for theoretical development but guides
our intuition.) I forget what this is normally called - the outcome
space? The underlying probability space? Something with "space" in the
name :)

Then we have a set of "events" whose probability we measure, and each
event has it's own probability. The events are just subsets of the
outcome space. There are rules for how the probabilities of different
events must relate, e.g. if events A, B are disjoint, then P(A U B) =
P(A) + P(B) etc.. [If you are familiar with measure theory I'd just say
what we have here corresponds exactly to a measure space where the total
measure of the space is 1. The events are the measurable sets.]

Then we define "random variables" to be functions from the underlying
outcome space to the real numbers. [The function must be consistent
with the probability measure, in the sense that it's required to be a
"measurable" function. If you don't know what that is you'll need to
get a background in measure theory, or I suggest for now just forget
about the requirement since it's likely all functions you might think up
or be interested in are going to be measurable. In particular, for a
discrete space [see first example below] all subsets are typically
measureable and all functions are measurable.]

Two examples:

-----------------------
1) Modelling rolling of two fair dice: [a "discreet" probability]

The outcome space X will have 36 elements which will be ordered pairs of
(1st die outcome, 2nd die outcome)
i.e. X = {1,2,3,4,5,6} x {1,2,3,4,5,6}
= {(1,1), (1,2), (1,6), (2,1),..... (6,6)}

Events are all the subsets of X, and the probability measure assigns
1/36 to each singleton event, and for larger events we just add up the
probabilities for all the individual outcomes that make up that event,
in the natural way. So let event
E = "roll a double"
= { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) }
and then we have
P(E) = 1/6 [.. = P({(1,1)}) + ... + P({(6,6)}) ]

You can see that individual outcomes are not numbers (they're ordered
pairs in our case) so it would not make sense to ask what is the
"expected value" of rolling two dice, with this model we've adopted.

That's why we have an added layer of "random variables" which map
outcomes to real numbers. We can talk about expected values of random
variables.

So here's a selection of some random variables:
D1 = "result of 1st die"
D2 = "result of 2nd die"
S = "sum of both dice"
Remember D1,D2,S are FUNCTIONS X :--> R, e.g.
D1( (2,6) ) = 2
S ( (3,1) ) = 4 and so on.

We can add, subtract, multiply etc. RVs in the obvious way. E.g. we have
S = D1 + D2.

We also can talk about probabilities of random variables doing things,
like e.g. P( D1 = 6 ). The D1=6 must be translated back to it's
corresponding event in the underlying probability space. I.e.
P( D1 = 6 ) = P( {(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} )
= 1/6

And we can talk of expectations of random variables. This corresponds
to an integral over the underlying probability space, but since this is
discreet in our case, it just ammounts to the corresponding sum:
E( S ) = 1.P(S=1) + 2.P(S=2) ... + 12.P(S=12)
= 7
There is a general result (needs proof of course, but it's correct) that
for any RVs A,B : E(A+B) = E(A) + E(B).

Question: S = D1 + D2, so with E(S) = E(D1+D2) = E(D1) + E(D2) do we
have one or two random variables?

[answer: obviously S, D1, D2 are 3 separate RVs, but in asking what an
expectation value is, the RV you're asking it of is 1 RV. Knowing what
RVs are, the question probably seems a bit misplaced... If you think an
RV is like an algebra question variable the question sounds sensible.]

-----------------------
2) A "continuous probability" example: Modelling RichD's breaking rod.

This is more sophisticated, mathematically, because there are infinitely
many outcomes forming a continuum. But the overall procedure is the same:

Outcome space: X = the interval [0,1], where an outcome x represents
the rod breaking at distance x from left hand end. [I've fixed the rod
length as 1 here.]

Events: subsets of X which are "measurable". Without going into detail
thats likely to be all subsets you would think up plus lots more :)
E.g. event "rod breaks in left half of rod" = [0, 1/2) etc.

Probabilities: Remember these are defined for /events/ rather than for
individual outcomes. The probability measure is built up from the
measure of intervals [a,b] which naturally are given measure (b-a), e.g.
P("rod breaks in left side of rod") = P([0,1/2)) = 1/2

Hmm, ok I should say there are plenty of other measures that can be
defined on [0,1]. The one here is the "uniform distribution" we've
assumed for this problem, reflecting the idea that the rod is equally
likely to break at any point. Also when I said probability is defined
for /events/ rather than /outcomes/, I should add that in this example
singleton events like { x } /do/ have a probability, which is zero. But
that's less useful to us than in a discrete probability, because we
can't use this to get the probability of events containing uncountably
many outcomes like the event [0, 1/2).

Random variables: they're the (measurable) functions on interval [0,1].
Examples:
K = length of left side of rod after breaking
= x [in this example, x is a real number, so this makes
sense. In the dice example, an outcome like (3,4)
is not an RV]
L = length of right side of rod after breaking
= 1-x
= 1 - K
M = length of shortest piece of rod after break
= minimum of x, 1-x
= minimum of K, L
N = length of longest piece of rod after break
= maximum of K,L
R = ratio (M/N)

Those are all RVs. RichD asks for

E( R ) = E( M/N )
= E( min(K,L) / max(K,L) )
= E( min(K,L) / max(K,L) )

How many RVs are in there? :) Well obviously R is one RV in and of
itself, so we could say 1. [By definition, an expectation is always
applied to one particular RV, in our case to R]. Also K, L, M and N are
obviously RVs in their own right, and R can be expressed in terms of
those RVs. (Like in the dice example where S = D1 + D2, where S,D1,D2
are all RVs.)

For this example, because the probability space is continuous the
expectation needs an integral to evaluate it, unlike in the discreet case.

Hope that clarifies things, at least enough to satisfy your curiousity
so you won't have to aquire a text book to learn everything thoroughly.

Regards,
Mike.

On 16/12/2021 19:49, duncan smith wrote:
> On 16/12/2021 01:38, RichD wrote:
>> On December 15, Mike Terry 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 -
>>>
>>> When you divide by a length that can be arbitrarily small (as we must do
>>> here) there's no upper bound on the result so there is the possibility
>>> that the expectation may not exist (or put another way, it might be
>>> infinite).
>>> In your case, let the breakage point be distance x from one end - so the
>>> ratio of longer to shorter is going to be greater than (C/x) for a
>>> suitable constant C
>>> This integral is infinite, as the indefinite integral is C log(x)
>>
>> hmmm... my first reaction is, no problem, the integral is weighted by the
>> distribution. But that's uniform, which doesn't ameliorate the problem.
>>
>> Surprising, isn't it? The ratio of short piece to long is well defined, but the
>> converse blows up. I tricked myself! Good catch.
>>
>> Re-formulate: What's the expected ratio of the shorter to longer piece?
>>
>> It appears you're well on the way to the solution. A mere schoolboy task,
>> as promised.
>>
>> --
>> Rich
>>
>
>
> I get 2*ln(2)-1, but I think I'm missing the trick.
>
> Duncan
>

I get the same.
Mike.

On Friday, 17 December 2021 at 18:39:42 UTC+1, Mike Terry wrote:
> On 17/12/2021 08:16, Julio Di Egidio wrote:
> > On Friday, 17 December 2021 at 02:00:31 UTC+1, RichD wrote:
> >> On December 16, ju...@diegidio.name 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?
> >>>>> Re-formulate: What's the expected ratio of the shorter to longer piece?
> >>>
> >>> But can't we write our problem as E(x/(L-x))? Why is that two variables?
> >> Informationally, they're equivalent. So it looks like one unknown.
> >>
> >> Let x be the length of the short piece, y is the long piece.
> >> Draw a histogram for each. The have different domains, hence
> >> distinct random variables.
> >>
> >> We want E[x/y]
> >> y = L - x
> >> etc.
> >
> > I am not the expert here and if you guys keep not answering I'll have to go back reading definitions: indeed, you should show that it's two (or even more, according to Mike) random variables starting *from definitions*, not just by repeating yourself in different words and/or by restricting the problem, which changes nothing in that sense...
>
> I'll go back to definitions for you (I'll follow with a couple of
> examples which hopefully will make it clearer, as there's a fair bit to
> take in just by reading definitions):
<snipped pink elephants>

That was quite uncalled for, as usual with you: instead of a straight answering of what you said, you transform it in a pre-pre-amateurish lesson... great stuff indeed. Next time, either don't bother and stick to it, or quoting the relevant mathematical definitions is enough (see my previous post for a link), no need to restart from Adam and Eve and even less if you make it up approximately, as if an actual and straight mathematical answer is not even possible...

Julio

On Wed, 15 Dec 2021 13:52:30 -0600, Serg io wrote:

>
> 1
>
> next...
>

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.

--
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