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2:40 AM
Can anyone explain how is $\int_0^{\frac 12}-x^n\log x$ a proper integral?
Ofc, $f(x)$ is bounded on $(0,\frac 12].$ However, $f$ is not defined on zero. In order that a function $f$ be integrable in $[a,b]$ , the definition of proper integrals suggests that $f$ is bounded on $[a,b]$ which means that $f$ must be defined on $a.$ But in this case, $f$ is not defined on $0.$
One might say that we may redefine $f$ as $f_R:[0,\frac 12]$ by $f_R(0)=0$ and $f^R(x)=f(x)$ when $x\in (0,\frac 12]$ and then $\int_0^{\frac 12}f_R$ is a proper integral.
But my question is that, we are talking about $f$ not $f_R.$ Also, $f\neq f_R$ as the domains of $f$ and $f_R$ are different. Moreover the given function is $f$ not $f_R$ so, I don't think we can't just 'redefine' $f$ and be done.
3:27 AM
Interesting: If $f:I\to I$ is a $C^\infty$ diffeomorphism satisfying ${d^nf\over dt^n}(\alpha) = 1$ for $n=1$ and $= 0$ for $n>1$ for $\alpha\in\{0,1\}$, then there exists $C^\infty$ diffeomorphisms $c_i,b_i:I\to I$, $i =1,...,n$ satisfying the above conditions so that $f\circ[c_1,b_1]\circ[c_2,b_2]\circ\dots\circ[c_n,b_n] = id$.
surprising? or trivial?
thomas: it is entirely up to you how or whether you define definite integrals for functions with removable discontinuities. the answer you get to your question depends only on that, and not on substance. yes it's defined, or no it isn't depending on setup. [note that you could define f_R in your example by setting f_R(0) to be literally any number, and, assuming you have defined the integral for piecewise continuous functions, the value of the integral of f_R would not depend on that choice]
one runs into this fairly frequently with many setups of the riemann integral - something you might want to consider turns out not quite to be defined, but not for any 'good' reason (e.g. in this example, how you define the function at the endpoint has no affect on the resulting integral at all), but only because your definitional setup doesn't quite let you get there
the good news is, this is also why a lot of the time you can "get away" with not precisely stating the hypotheses of results about integration
as even some calculus books will do
4:08 AM
@leslietownes Ok the definition that I use for a function to Riemann integrable assumes $f$ is defined and is bounded in the closed and bounded interval determined by the upper and lower limits of the integrals.
So this means my definition doesn't take the fact that '$f$ may not defined at some point in the closed and bounded interval of integration' into consideration.
Reading your messages makes the impression that the problem is with my definitional setup.
This is one of those points that it is often best to not get caught up on. The singularity is removable, so remove it
An improper integral is when you have a limit at infinity, or a divergent limit.
4:22 AM
@XanderHenderson by singularity I hope you mean 'discontinuity'? Ok, so the takeaway is just an addition to the definition of Riemann integrability. The part that must be added to the definition of Riemann integrability of a function of it is as follows:
If a function is not defined at some point say $p$, in the (closed and bounded) interval of integration (determined by it's upper and lower limits) and furthermore the limit at that point exists finitely and say it's given by $l,$ then we can consider the function $f_R$ which is precisely the function $f$ at all other points not equal to $p$ and $f_R(p)=l.$
If $f_R$ so defined is Riemann integrable over $[a,b]$ then so is $f$ and $f_R$ is not Riemann integrable in $[a,b]$ then $f$ is not Riemann integrable on $[a,b]$ as well.
@leslietownes and @XanderHenderson Am I good to go with this idea in my mind?
@ThomasFinley No, I mean singularity. It isn't a discontinuity, because the function is continuous on its domain.
You can't be discontinuous where you aren't defined.
But I think that you are overcomplicating things.
i love the game of "i'm too busy noticing things to do any math." i wonder if there is a telephone book somewhere that needs proofreading
@XanderHenderson ah, alright. Stupid me. But is the rest part of the modification, alright?
@XanderHenderson maybe:?)
If two functions differ at a finite number of points, they will have the same integral. So if a function is not defined at a single point in some interval, but you want to integrate over that interval, just define it however you like at that point and get on with life.
Riemann integrability is usually started in terms of limits of sums, anyway.
@leslietownes I didn't get the joke. Maybe I am too stupid to understand it :?)
4:31 AM
In any event, my phone just turned black and white, which means that it is past my bedtime.
xander: or you're traveling back in time to the 1950s.
you know, when phones were in black and white
@leslietownes before they invented "color".
But I'm pretty sure phones were only black back then.
@XanderHenderson though we can't use the result you mentioned in this case as the two functions here can't be said different at $0$ as one isn't defined at $0$ firstly. But ok, I will move on.
It was only in the 70s or 80s that you could get colors.
due to woke
4:33 AM
@ThomasFinley Which functions?
@XanderHenderson $f(x)=x^{n-1}logx$ where $n-1\gt 0$ and $f_R(x)=f(x),\forall x\neq 0$ and $f_R(x)=0.$
Take your f, and define it at the singularity to get the function g. Then any function g' which differs from g at that one point will have the same integral as g.
Those are the two functions I have in mind.
What this says is that no matter how you extend f to the extra point, the integral is the same.
So who cares about that point?
@XanderHenderson yes, we can say that surely. But I think we just have to consider the integral of f is equal to that of integral of $g$ in this case as a fact or some kind of "necessary axiom"
The value of a function at a single point just doesn't matter.
Since $f$ is not defined there
4:37 AM
@ThomasFinley I have just given you a definition of the integral of f.
And that theorem is valid for all functions that are defined everywhere
Your original definition doesn't apply to functions with singularities. But extending the definition of the Riemann integral to such functions doesn't require a run of machinery.
If a function is not defined at a finite number of points, just define it however you like at those points. The integral will be the same for any choices, su just call that the integral of f.
@XanderHenderson If I get you correctly: You wanted to mean that if $f$ is not defined at a point say $c.$ Then we may extend $f$ to $f_R$(say) by defining it at $c.$ If by this extension the resulting function $f_R$ is integrable then so is $f.$ And If $f_R$ is not integrable then $f$ is not integrable.
If you are an overachiever, show that this definition is consistent with the original definition.
@XanderHenderson I might not be an overachiever:P But I just want to know, if I got you correctly as to whether you wanted me to use this definition chat.stackexchange.com/transcript/message/65815445#65815445
Maybe for the consistency part, I will keep it as a homework.
4:42 AM
I am giving a new, extended definition of what the integral of a function is, when that function has a finite number of singularities: it is the integral of any extension of that function to the whole interval.
And then don't overthink it.
The mantra is "integrals don't care about single points".
@XanderHenderson perfect. I take this as the definition. It makes things more rigorous and consistent.
In any event, my phone is black and white, the mobile interface sucks anyway, and I need to sleep
@XanderHenderson or say, "Integrals don't care about a finite number of singularities" to be more specific :D
@XanderHenderson Have sweet dreams...
@ThomasFinley No, points. Because you can redefine things at points, whether they be singularities or not.
@XanderHenderson oh...so perfect!
You have broadened my horizons in my humble opinion.
I remain thankful to you.
To end this great chat, I will write this mantra: Integrals don't care about a finite number of points
4:53 AM
or more generally a set of measure zero
@SoumikMukherjee haha...that's true... But I think that's the highest possible generalization one can make in an undergraduate course imho
there is something to be said here about when $\lim_{x\to 0^+} \int_x^a f$ and $\int_0^a f$ coincide
everyone should learn Henstock integration, really...
5:12 AM
Is there a function which is not Riemann integrable but is Henstock integrable where the gauge is continuous?
 
3 hours later…
7:48 AM
0
Q: Show that $\int_0^{\frac{\pi}{2}}\frac{x^m}{\sin^n x}dx$ is convergent if and only if $n < 1 + m.$

Thomas FinleyShow that $\int_0^{\frac{\pi}{2}}\frac{x^m}{\sin^n x}dx$ is convergent if and only if $n < 1 + m.$ The solution given in the book is as follows: Let the given integral be $\int_0^{\frac{\pi}{2}}f(x)dx.$ If $m - n\geq 0,$ it is a proper integral since $\lim ( \frac{\sin x}{x})^n = 1.$ If $m-n < 0...

I need some help with this.
 
2 hours later…
9:19 AM
Consider the conditional expectation of a r.v. $X$ given an event $B$, $$E(X|B):=\frac{E(X\mathbf1_{B})}{P(B)}.$$ We know the expectation $E(X)$ is $$E(X)=\int_\Omega X\,dP.$$It is claimed that $E(X|B)$ is simply the expectation of $X$ under the measure $P(\cdot |B):=P(\cdot\cap B)/P(B)$. I don't see how. I'd be grateful for any clarification. Also, is $X\mathbf1_B$ simply a product?
9:43 AM
ok, never mind what I wrote. Let's just write $\nu(A)=P(A|B)$, and so $$E(X|B)=\int_\Omega X\,d\nu.$$
And to see that $$\int_\Omega X\,d\nu=E(X\mathbf1_B)/P(B),$$we use measure theoretic induction.
 
2 hours later…
11:34 AM
@SoumikMukherjee this is a Riemann integral. Sets of measure zero can mess it up.
11:54 AM
we take real numbers and define divisibility as "p is divisible by q if p/q is rational"
is this an interesting thing to define
12:24 PM
@RyderRude That is, essentially, what it means for two real numbers to be commensurable.
In mathematics, two non-zero real numbers a and b are said to be commensurable if their ratio a/b is a rational number; otherwise a and b are called incommensurable. (Recall that a rational number is one that is equivalent to the ratio of two integers.) There is a more general notion of commensurability in group theory. For example, the numbers 3 and 2 are commensurable because their ratio, 3/2, is a rational number. The numbers 3 {\displaystyle {\sqrt {3}}} and 2...
It is a notion which pops up in number theory and (weirdly enough) fractal analysis.
Wow! everybody left and we're the only ones left.
I'm here
@SoumikMukherjee what do you mean by the gauge?
There is no canonical gauge attached to each Henstock integrable function
Gauges are just generalized deltas
Do you mean definition where we only allow for continuous gauges? When we integrate a function on a closed interval, this interval is compact so such gauge has a positive minimum. We can replace the gauge by this constant and obtain Riemann integrals
12:45 PM
@Jakobian Meh. The Henstock integral allows one to integrate more functions, but this isn't necessarily a good thing. And the space of Henstock integrable functions isn't all that useful. The Riemann integral is worth knowing because it is fairly simple to understand, but measure theoretic notions of integration are far more useful (in particular, $L^p$-spaces are very useful, and are generally best understood in terms of the Lebesgue integral).
Note that I say this as someone who went down an HK-integral rabbit hole in grad school.
What helped you the most to climb out?
Learning Henstock integration helps you to understand integration in general better. That's what I think
1:01 PM
Space of Henstock integrable functions is horrible? Sure, there is couple complicated ways to assign structure here but why would you want it to be nice anyway? It just shows these functions are badly behaved as topological vector spaces, in the big picture, so what though. That's not what you study when studying Henstock integration
Just because the way you want to apply it in is not useful doesn't make the concept
You're working with one of the purest forms of analysis, its gnarly. That's what analysis should be
@Jakobian I didn't say "horrible". I said "not all that useful".
I'm saying its horrible, because it is
@Jakobian Yes, in the definition, gauges can be any function, so I am asking if we restrict gauges to be a continuous function only. that is, if there is a function which is not Riemann integrable but is Henstock integrable w.r.t a continuous gauge.
@Jakobian That feels like a toxic view of mathematics. "IT'S NOT WORTH DOING IF IT'S NOT GNARLY! RWAR!"
@XanderHenderson aren't you the one saying that
1:14 PM
@Jakobian It is very much how you come across.
"...its gnarly. That's what analysis should be."
Why?
Why should it be gnarly?
It often is gnarly, but that's a bug, not a feature.
I say thats a feature
(And I say that as someone who works with fractals, which are, by definition, gnarly).
@Jakobian Yes, I know. You've been clear about that. I disagree with you.
I'm not saying that because I think its not worth doing. You assumed my opinion. I think that its worth doing because its "gnarly" as I said
so completely the opposite of what you said
stop putting words in my mouth
I have only paraphrased what you seem to have said. If the representation is incorrect, I would argue that it is your job to make your position more clear. As it is, this is exactly how you come across to me.
How else am I supposed to interpret "...its gnarly. That's what analysis should be." You seem to be making the value judgment that mathematics (or analysis, at least) should be hard. It is good that it is "gnarly". This is a feature, not a bug.
I disagree.
That's all.
@XanderHenderson yeah, i missed the context
1:24 PM
@XanderHenderson not at all. You went and assumed things
its not my job to "come across" in a certain subjective way to you
just pay attention to what I said and what I said not, its your fault for not paying attention
1:55 PM
@Jakobian Communication requires two parties. If I have misunderstood you, I apologize, but you have done nothing to clear up the confusion. You are constantly blaming other people in this room for misunderstanding you---at some point, you have to accept some responsibility for the misunderstandings.
You are the person who wrote "...its gnarly. That's what analysis should be." The only way I can see to interpret this is as a value judgement that analysis "should be" gnarly. It is good for things to be gnarly. All I have said in response to that is that I think it is toxic to believe that any part of mathematics should be difficult (difficulty is a bug, not a feature---I think that it is elitist to assume that things should be hard).
@Jakobian Would it be possible to give a further hint here? I spent yesterday evening and I can't see where I gave up tightness in the bound
2:11 PM
@EE18 there is a prime between n and 2n
@XanderHenderson oh
@SoumikMukherjee Just to be sure, we're talking about the sequence question with decreasing distances?
I ask because I can't understand how primes are relevant here lol
And I know I ask lots of diff questions so want to be sure
this is like an equivalence relation on rationals
Just to be sure, this was the question
And my work was:
for $n > m$ we have
$$d(x_{n},x_m) \leq d(x_m,x_{m-1})\sum_{k=1}^{n-m} \alpha^k \leq \alpha^{m-1}d(x_{1},x_{0})\sum_{k=1}^{n-m} \alpha^k = \alpha^{m-1}d(x_{1},x_{0})\sum_{k=1}^{n-m} \alpha^k = d(x_{1},x_{0})\sum_{k=1}^{n-m} \alpha^{k+ m -1} $$
$$\leq d(x_{1},x_{0})\sum_{k=1}^{n-m} \alpha^{m} = \alpha^m d(x_{1},x_{0}) (n-m)$$
But this wasn't a tight enough bound since $n-m$ unbounded
2:31 PM
@EE18 no its a joke
Ooh I thought you were asking about the sum1/k^2+k question
@EE18 the inequality in the second line here is too strong
No worries Soumik. Just wanted to check cause I figured some wires were getting crossed lol
That other one I got eventually )
by bounding $\alpha^{k+m-1}$ by $\alpha^m$ you lose too much information
this is unnecessary, this is just a geometric sum
OK I see now. I should have figured this out when you said geometric sum before. Will work on this now
2:37 PM
@RyderRude equivalence relation on (non-zero?) real numbers
@Jakobian yes... i mis typed
i wondered about this idea becuz if we take a matrix $A$, then $e^{iAt}$ hits identity at some $t$ if the eigenvalues of $A$ are commensurable
@XanderHenderson its funny because I think its you thats the one avoiding responsibility
I've shown before I can admit when I'm wrong
and A needs to be Hermitian too
and we need non zero t
and yes I did clear up the confusion, see messages above. Maybe you should get another pair of glasses, Xander
so when we have commensurable eigenvalues, the exponential is a periodic function
2:43 PM
Definition Let $X$ and $Y$ be discrete, jointly distributed random variables. For $P(X=x)>0$ the conditional probability function of $Y$ given that $X=x$ is $$p_{Y|X=x}(y)=P(Y=y|X=x)=\frac{p_{X,Y}(x,y)}{p_X(x)}.$$
Exercise Show that $p_{Y|X=x}(y)$ is a probability function of a true probability distribution.
I don't understand the exercise. What is it I need to show?
By probability function, the author means probability mass function, i.e. the discrete analog of probability density function.
@psie that it consists of positive numbers that sum up to $1$
there is also a question of, does there exists a discrete random variable with such probability mass function, but if you don't know such result already then you might learn it later
ok, so basically $\sum_y p_{Y|X=x}(y)=\frac{\sum_y p_{X,Y}(x,y)}{p_X(x)}=\frac{p_X(x)}{p_X(x)}=1$
yes. Another thing is to construct a random variable $Z$ such that $P(Z = y) = p_{Y|X = x}(y)$. This might be harder to do on your own
@Jakobian I would really love to know what it is that I have done to offend you. Every time I disagree with something you say, it devolves into you accusing me of misunderstanding you or putting words in your mouth, yet you never actually attempt to clarify whatever it is that I've misunderstood. Do you actually want to clear up the miscommunication, or are you just looking to argue? Like, what is your goal in constantly attacking me?
2:58 PM
but if you have some function $p:\mathbb{R}\to \mathbb{R}_+$ and $\sum_x p(x) = 1$ (where we interpret this as supremum over all finite sums), then there is always a random variable $Z$ with $P(Z = x) = p(x)$
ok
@XanderHenderson I try to look at each case individually, and from how you phrased it, it seems that you are attacking me rather than the opposite
So, basically "I know you are, but what am I"?
Got that question in the end
Thanks again Jakobian
4:06 PM
I'm saying that your methodology seems to be that of "focus on the person" rather than "focus on whats being discussed"
What is a cure for laziness?
Excercise?
@Jakobian I have given you multiple opportunities in this discussion to focus on the math. I have asked you, repeatedly, to explain what part of your statement I have misunderstood. I have invited you to clarify.
I will restate my question again: you very much seem to have said that analysis should be gnarly. I disagree, and feel that this is a toxic and elitist attitude towards mathematics---mathematics should be accessible. The fact that it isn't is a bug, not a feature. Perhaps this is an unavoidable bug, but that doesn't make it what should be.
2
("Should" is a value judgement).
So where did I misunderstand you?
4:28 PM
> mathematics should be accessible
that couldn't be farther from truth, mathematics just is what it is
you can't influence this by your feelings of what it should be
@Jakobian Nothing I've said contradicts that statement.
and being gnarly is a good thing, because it gives people research opportunities
you don't want to study a field that is "done" with the intention of proving things in it
There is a world of difference between "accessible" and "done".
analysis should be gnarly, yes, because this how functions behave
its not gatekeeping or elitist attitude
its how things are
You are making a virtue out of necessity.
4:35 PM
that's to say its expected of it to be gnarly, its not something surprising
@Jakobian Again, I haven't contradicted this point.
if we want to take it all without losing on a lot of information to make it as simple of a concept as possible to be able to envelop it into this definitions such as Banach spaces, without doing such sacrifices, if you want to look at analysis in its more raw form, thats what Henstock integrable functions are to me, and this doesn't make them not useful because this is just the nature of the problem
the problem of understanding them
just because you would like to apply methods that you are used to, doesn't mean its not useful
@Jakobian When did I ever say "I would like to apply method [x], HK integrals don't let you use method [x], therefore HK integrals are not all that useful"?
@XanderHenderson here
I have made two different statements: (1) HK integrals don't have a lot of the nice properties of the Lebesgue integral (i.e. the space of HK integrable functions doesn't have a lot of the nice structures of, say, the space of $L^p$ functions), and (2) HK integrals are not all that useful (as in, they don't have a lot of applications outside of themselves).
There is overlap in those statements, but there is no implication that HK integrals are not useful because they lack these properties.
4:43 PM
that HK functions don't form such structures like Banach or Frechet spaces only indicates to me that this is a wrong way to think about them
as for what is useful and what isn't, well, that's really relative to what we are talking about
Okay. But you are fighting an uphill battle on that one, since those tools have already proven to be really useful in terms of understanding the world. Given the current state of mathematical knowledge, the burden is on the proponent of HK integrals to explain why they are more useful.
Aside from giving a value to certain integrals which don't have a value with respect to the Riemann integral or the Lebesgue integral, what can you do with an HK integral that you can't do with the Lebesgue integral?
What interesting theorems do you get?
What applications do you have to other areas of mathematics? or to the sciences?
I don't assign value to something based on what is useful, but I think that learning HK integration is useful
that's not to say HK integrals will be useful to you, but it will help you to learn the picture
Also, to repeat myself, I went down the HK integral rabbithole when I was in graduate school. I spent a lot of time thinking about HK integrals, and working through the theory of them. I get the attraction. But I ultimately found that there just wasn't much there. There is no real advantage of the HK integral over other versions of integration.
again this is not about utility
I'm not saying "HK integrals are useful" I'm saying "learning HK integrals is useful"
In what way? Can you be specific?
Note: if learning something is useful, then I would argue that the thing itself is useful, if simply for the reason that learning it is useful. An example may have no utility beyond explicating some principal, and that makes it useful. But I am not convinced that this is the case with the HK integral.
4:50 PM
because it helps you to think about integrals in ways that Riemann or Lebesgue integration doesn't
its useful to be able to think "this improper Riemann integral is actually just like a normal integral, without limit process being necessary"
But how is that useful? Just "thinking different" isn't, as far as I am concerned, useful (in and of itself). What insight do you get by thinking differently about integrals?
if Thomas knew about Henstock integration, say Hake's theorem, he'd have no problem accepting the integral of $x\ln(x)$ from $0$ to $1$
@Jakobian Possibly, but that very much depends on the pedagogical goal of assigning that problem.
I guarantee that no one actually cares about the value of that integral. Rather, they care about the integral as a way of demonstrating certain principles.
I'm talking about how to understand that integral, an extension by letting $0\cdot \ln(0) = 0$ or limit process? Which one is valid? Both are
And there is the "law of conservation of difficulty": in mathematics, problems generally have some level of difficulty. You can hide that difficulty in different places, but you ultimately have to deal with it. For example, you can prove the fundamental theorem of algebra using relatively elementary techniques, but the proof is a technical mess.
Alternatively, you can hit it over the head with Liouville, and prove it in two lines. But you have to learn a lot of complex analysis, first.
4:56 PM
besides that, there is also a trap people might come up with, where there is this hidden assumption that $f'$ needs to be integrable in order for some theorems to hold
this gets fixed when talking about Henstock integrable functions, and you begin to understand why the assumption is there
so, yes, it does help because it makes you think about all those problems with derivatives being integrable
even if HK integrals aren't inherently useful for someone, it still makes them consider things they normally wouldn't
just by learning the theory
ultimately I'm not even advocating for HK integration, but benefits that come from learning about them
Okay, you have, at long last, made a tangible claim: learning HK integration helps to explain the relation between integrals and derivatives. I'm not sure that I agree that this is the best way to highlight this relation, but that's all I've been asking for!
5:58 PM
Consider a fair die thrown twice. Let $U_1$ be the dots on the first throw and $U_2$ on the second. Let $X=U_1+U_2$ and $Y=\min(U_1,U_2)$. I'm trying to find the function $$p_{Y|X=x}(y)=\frac{p_{X,Y}(x,y)}{p_X}=\frac{p_{X,Y}(x,y)}{\sum_z p_{X,Y}(x,z)}.$$ It looks as if I need to find the joint pmf. How do I do that?
6:09 PM
I guess $p_X(x)$ is just the convolution, i.e. $$p_X(x)=\sum_u p_{U_1}(u)p_{U_2}(x-u).$$ I wonder though how I can simplify this further.
6:24 PM
@psie Of course, its not stated, but I assume the observations $U_1$ and $U_2$ are independent
yeah
the other day I worked on finding the pdf of the sum of two independent (continuous) uniform distributed r.v.s., but here we have discrete ones
$P(X = x, Y = y)$ is what you need to find. It will be useful to identify what the two conditions are equivalent to in terms of $U_1$ and $U_2$
ok
($U_1 = y$, $U_2 = x-y$ or $U_2 = y, U_1 = x-y$) and $x\geq 2y$
is what it looks like to me
i.e.
$$P(X = x, Y = y) = \begin{cases} P(U_1 = y, U_2 = x-y\text{ or }U_2 = y, U_1 = x-y) & x\geq 2y \\ 0 & x < 2y\end{cases}$$
now I'd split it up into $x = 2y$ and $x > 2y$
hmm, where does the $x\geq 2y$ condition come from?
6:35 PM
@psie yes this will be convolution as seen by an explicit calculation. You can further try to simplify it by plugging in $u = 1, ..., 6$
@psie rephrasing of $U_1\geq U_2$
or $U_2\geq U_1$, depending on which is larger
ok, makes sense
@psie using the same strategies like before for integrals, this should be $\frac{\min(6, x-1)-\max(1, x-6)}{6}$ (for $x$ in appropriate interval)
$2\leq x\leq 12$, $x\in\mathbb{N}$
@Jakobian ok, the "or", is this simply a union? In other words, is $$\{U_1=y,U_2 = x-y \text{ or }U_2 = y, U_1 = x-y\}=\left(\{U_1=y\}\cap\{U_2=x-y\}\right)\bigcup\left(\{U_2 = y\}\cap \{U_1 = x-y\}\right)?$$
@psie I use $\bigcup$ for unions of families, here I'd just use $\cup$. Here this is a logical connective. It means the same as set of $\omega$ for which the thing in the brackets holds. But this would be an union, yes
ok
6:47 PM
$A\cup B$ but $\bigcup\mathcal{U}$ or $\bigcup_{i\in I} A_i$
note the difference in how it looks
I know this is petty, but \bigcup for union of two sets just doesn't look right
it's alright :)
7:49 PM
@Jakobian how did you get this? that is, $\frac{\min(6, x-1)-\max(1, x-6)}{6}$? We have $$p_X(x)=\sum_u p_{U_1}(u)p_{U_2}(x-u),$$ and this can be viewed as an integral with respect to counting measure, or? I see how the terms in this sum will only be non-zero if $\max(1, x-6)< u <\min(x-1,6)$, but from here I'm lost.
@psie I'm not using such advanced things as integrals with respect to the counting measure
you just observe $p_{U_1}(u)p_{U_2}(x-u)$ is simply an indicator function on a certain set
oh sorry I realized I made an error
it should be $6^2$ in the denominator
8:29 PM
@Jakobian ok, I think I understand, but there's an additional error I think, we want $$\frac{\min(6, x-1)-\max(1, x-6)+1}{6^2}.$$ Just plug in $x=12$ in your old expression, and you'd get $0$, but the probability is actually $1/36$.
@psie yeah, good catch. There is $b-a+1$ natural numbers in $[a, b]$, not $b-a$
common mistake I make
 
3 hours later…
11:05 PM
@XanderHenderson by the way, the definition of gnarly I had in mind was "unpleasant or unattractive."
your remarks about elitism were unnecessary
this is actually, the only definition I was aware of at the time, you most likely understood it to mean "difficult"
I think that's why you were raising those, weird to me, objections

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