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00:03
Disease X is running rampant (no it's not corona despite recent events). There are currently two "vaccines". One vaccine, A, is 90% effective and the other, B, is 85% effective. If you get both vaccines what the chances you don't get disease X? That is, vaccine A works OR vaccine B works OR both work.
make sense?
ahh
yeah, that's different and probably does have an answer
I guess I made an error in my simplification (that looking back isn't really simpler, oops)
heh, happens
the wording is a little ambiguous: is 90% the probability of being cured if you get vaccine A and not vaccine B, or is it the probability of getting cured if you got vaccine A regardless if you got B
Regardless, for both percents
yeah
the way i'd usually tackle this is to consider a hypothetical population
hmm
i'm not sure that's the right way here tho
00:11
It feels like there's enough information, but maybe there isn't?
If A is 0.9 effective and B is 0.85 effective, how to calculate the effectiveness of applying both A and B?
Can't be multiply them. That would result in less than using one.
if I were to guess, the answer will end up being: 1-(1-0.9)(1-.0.85) = 1-(0.1)(0.15) = 0.985
Probably.
there's a 10% probability that vaccine A fails, and a 15% probability that B fails
so there's a 1.5% probability that both fail
and therefore a 98.5% probability that at least one succeeds
00:14
Why does that work for failure but not for success?
Failure= 1- success
@JBis hmm
i'm assuming that the "A fails to cure" and "B fails to cure" events are independent
which is not exactly an innocent assumption
Id agree with that assumption however. Succeeding is also independent I think.
How they can be independent if the fact that both work is included?
hence the
10 mins ago, by JBis
Regardless, for both percents
00:17
Anyways, how to explain truth-table for conditional A->B?
@Ante A working and B working are independent. But you need either one or both to wrk.
the point comes down to 1-(1-p)(1-q) = p+q-pq =/= p+q
bah, i can't algebra tonight
so something something inclusion-exclusion
| A | B | Outcome |
|---|--:|---------|
| 0 | 0 | 0       |
| 0 | 1 | 1       |
| 1 | 0 | 1       |
| 1 | 1 | 1       |  0 indicates failure. 1 indicates success.
The success of A and B are independent but the outcome relies on the value of both or one.
the only way to show that an implication is false is to show that B can be false even though A is true
ok well i'll make that a given
A and B do not effect the effectiveness of each other to cure.
00:21
right
and you definitely can't just add the probabilities
if only because that's bigger than 1 in this case :P
175% chance of me not being infected? I like those odds!
lol
and it being the product of the probabilities also wouldn't make sense, as then you'd be -less- likely to be cured by doing both
the probability of the failures multiplying together, by contrast, is at least not absurd
not much of an endorsement, mind
but that is only coincidental
yeah. it's a necessary test for p+q-pq to make sense, but not a sufficient criterion
Is it worth asking on the main site? I felt it might be easier to work through it in chat.
and i like chat better :D
00:26
what i'd actually do at this point is search the main site
because this seems very likely to have been posed at some point
what would i search for?
ahh, think i found one
2
Q: What is the probability of success when two independent predictors co-occur?

TyroLet's say I have a collection of objects. Each object has a set of predictive attributes where the attribute may be true or false. Each attribute when true predicts "success" with a known probability, and these probabilities are not necessarily the same for different attributes. What is the pr...

...except those answers seem kinda awful
from something online
> Formula for the probability of A and B (independent events): p(A and B) = p(A) * p(B).
so the prob of both working is 0.90 * 0.85 = 0.765
So now we have:
Prob of A: 0.90
Prob of B: 0.85
Prob of Both A and B: 0.765
(1-(1-0.1)(1-0.15) - (1-0.1)(1-0.235) - (1-0.15)(1-0.235)+(1-0.1)(1-0.15)(1-0.235) ?
Is this sort of exclusion-inclusion wanted?
1- that above= solution, is it?
p(A or B) = (0.90 + 0.85) - (0.765) = 0.985
p((A or B) OR (A and B) =( (0.985) + (0.765) ) - (x) )
@Ante No clue what you did. Could you explain?
@Ante thats 1.518525
00:37
Whoops. :(
my math is based on
> p(A or B) = p(A) + p(B) – p(A and B).
Proceed, seems youre on the right track
But how the hell do you find the probability of (A or B) AND (A and B)?
oh you multiply them
p((A or B) OR (A and B) =( (0.985) + (0.765) ) - ( (0.985) * (0.765) ) = 0.996475
so theres a 99.6475% chance you will be cured?
does my math work?
i think we got it?
Well anyway thanks for the help :)
00:57
Is the group ring $\Bbb{C}[\Bbb{Z}^2]$ the same as $\Bbb{C}^2$ (or $*$-isomorphic to it)?
01:10
hey guys, how do you show (roughly) that the indefinite orthogonal group $O(p,q)$ (where $p,q>0$) has (exactly) four connected components?
 
1 hour later…
02:38
0
A: If $f:X\rightarrow Y$ is continuous then $\phi:X\owns x\rightarrow f(x)\in f(X)$ is continuous

topologicalorientablesurfaceYour proof is correct. Your claim can be rephrased in the following manner: Let $X,Y$ be topological spaces. $f:X\rightarrow Y$ is continuous if and only if $f:X\rightarrow f(X)$ is continuous. I will provide two proofs. Proof 1: Forward Direction: If $f: X\rightarrow Y$ is continuous, then...

I would like to hear peoples feedback
@ShaVuklia this is something i should probably know, given that the case of O(3,1) is one I remember seeing in the context of relativity. alas...
 
2 hours later…
04:32
I think I'm being targeting via serial downvoting, but the system hasn't registered it. What should I do?
 
3 hours later…
07:42
I have one doubt!
Anyone is online?
Is anyone online....plz reply
08:00
@AmreshPrasadSinha What topic?
And What level?
It's usually better to just ask the question. Then if a person has the time and knowledge to answer, they can jump straight into answering instead of needing to wait for the question
this was my question :- Prove that (n!)! is divisible by (n!)^{(n-1)!}
This was the question....sorry I got late in posting it as turned off my net!!
Well, I'd start with showing that $(n!)!$ is divisible by $n!$. Perhaps finding that quotient will illuminate how to approach the rest of the problem. Though I am unfamiliar with this result, so that might be as far as I can help without diving into it myself
 
2 hours later…
10:31
Welp
Looks like i won't go to the US after all
I think a lot of conferences have cancelled due to the COVID-19
ABC
ABC
10:46
Hi guys, I need to compute Taylor expansion of $sin(\pi z^2)$ centred in 1 with order 3.

I get:
$\pi z^2=\pi+2\pi(z-1)+\pi(z-1)^2$
$sin(y)=y-\frac{y^3}{3!}+o(y^3)$

When I substitute $y$ with $\pi+2\pi(z-1)+\pi(z-1)^2$ I have to do a lot of calculations. I need to do this expansion during an exam, so I need velocity. This way is good? Exist another faster way?
11:19
$\sin(\pi z^2) = -\sin(\pi (z^2-1))$
11:47
How much integral has been generalized from the time of Riemann? Is it developed so much that integral of every everywhere discontinuous real function of a real variable defined on the segment exists?
12:02
And is unique.
I don't think so
What you're asking about is measure theory I think
A measure on a set is something that tells you how you evaluate integrals
And you can define the integral of functions that are said to be measurable
Yes, but if, for example, function has everywhere discontinuities then the generalization of integral over those functions could be non-trivial.
There are functions that are everywhere discontinuous and that are not measurable
Yes, so the concept of measure itself could (should?) be specially adapted to suit those cases, right?
Hmm I doubt so
I'm not sure I follow what you're saying to be honest
12:08
How would one integrate a nonmeasurable set, for example
that is where most discontinuous everywhere functions are
Well you can't
yup that's what I was thinking
12:37
this question gives me a headache: math.stackexchange.com/q/3578609/137524
............................
(not the underlying question itself---that's garden variety---but the utter laziness of it)
yup
and the fact that it's for a master's dissertation
I have no problem turning that person into a nonmeasurable set
12:39
yuuuuuup
Also hey @Semiclassical and @Secret
nah, they're a set of measure zero, in that for most practical purposes they can be identified with zero
and therefore ignored
well, nonmeasurable sets are a more brutal way to go lol
The way measure theorists plan their murder: Turn the unsuspecting victim into a nonmeasurable set
or measure zero sets
@EdwardEvans This makes me feel good about myself
12:42
@Astyx wait what
If people ask this kind of questions for their master's dissertation, I'm not that bad at maths after all
I see hahaha
...okay, this is just kind of ridiculous:
this question is not very good to start with: math.stackexchange.com/questions/3578591/…
Aren't those the exact same link ?
12:47
oops
second should have been this: math.stackexchange.com/questions/58/…
trying to mine reputation or smth
same question
and same wording
bot?
your guess is as good as mine
i voted to close on other grounds before realizing it was a verbatim copy
so i flagged it as a verbatim copy as well
12:52
good move
Hey that's not fair
I was torn because it's technically an interesting question, if a little half-baked, but showing me the other link made me vote to close lol
He took the time to replace "fascinating" by "interesting and wants to learn more about it". Have some respect for his time and investment.
heh
ah, yes. how shameful of me to miss the author's clear work to make the questions not -exactly- identical
12:55
2
Q: Group Convolution is Associative

user193319Let $G$ be some locally compact group and $\mu$ its associated Haar measure. I am trying to adapt this proof that convolution on the locally compact group $(\Bbb{R},+)$ is associative. Here's what I have so far: $$((f \ast g) \ast h)(u) = \int_{G} (f \ast g)(x)h(x^{-1}u) ~d \mu (x)$$ $$= \int_{...

@user193319 i think you missed an "=" in your first string of equations
Oh, you're right. You're referring to the last line of the first string of equations?
yeah
right after Fubini
also, this is the kind of construction for which the align environment is a great fit:
\begin{align}
((f \ast g) \ast h)(u)
&= \int_{G} (f \ast g)(x)h(x^{-1}u) ~d \mu (x)\\
&= \int_{G} \left[ \int_{G} f(y) g(y^{-x}y) ~d \mu (y) \right] h(x^{-1}u) ~d \mu (x)\\
&= \int_{G} \int_{G} f(y) g(y^{-1}x)h(x^{-1}u) ~d \mu (y)~ d \mu (x)\\
&= \int_{G} \int_{G} f(y) g(y^{-1}x)h(x^{-1}u) ~d \mu (x)~ d \mu (y) &\text{Fubini's theorem}\\
&= \int_{G} f(y) \left[\int_{G} g(y^{-1}x) h(x^{-1}u) ~d \mu (x) \right] ~ d \mu (y)
\end{align}
finally, if I look at the main question:
$$(g \ast h)(y^{-1}u) \underset{?}{=} \int_{G} g(x) h(x^{-1}y^{-1}u) ~ d \mu (x)$$
that took way too many tries to get right
anyways. if we relabel $y^{-1}u\to y$, then that's $$(g\ast h)(y)\underset{?}{=}\int_G g(x)h(x^{-1}y)~d\mu(x)$$
@user193319 isn't the above just the definition of convolution in this context?
@Semiclassical Looks right to me
same, in which case there's nothing left to show
 
1 hour later…
14:10
How do you look at spacings between numbers if you're not working with sequences of natural numbers?
For example, maybe you want to look at the spacing distribtution of some rational numbers
but how do you formalize the notion of a "gap?"
for example, in the naturals you can have a sequence of gaps, g=1,3,5,7,23,4
but in the rational field it's dense so I don't know what to define as a gap because everywhere you look there's another rational number
Well, if you're working in a metric space you can always talk about the distance between two numbers. It's just that the distances become arbitrarily small.
Generally, the traditional metric is defined as $\vert a-b\vert$ where $a$ and $b$ are your points
3
Q: How is it that there are 'gaps' in rational numbers and yet between any two rational numbers, there exists another rational number?

Nishant Pr. DasIf there are gaps in rational numbers then lets assume we have a gap between a and b, both being rational. Then we have $\frac{a+b}{2}$ which is inside the gap which essentially makes it a non-gap. What am I getting wrong?

There's also a set theoretic approach to gaps which basically deals with cofinal set stuff I think...
I was just thinking that permutation gives us the number of choices we have for arranging something but how can we prove that $\frac{n!}{(n-r)!}$ is always a natural number.
@geocalc33 It seems that it can be done only if it can be known where irrationals are and how to measure their distribution in different intervals, I was thinking about such problems.
(Ah, I misunderstood the meaning of "gap." Didn't expect it to be a rigorous term)
14:24
If you're thinking combinatorically, the combinatorial interpretation makes it obvious that $n!/(n-r)!$ is a natural number. Otherwise, convince yourself that $n!/(n-r)!=(n-r+1)\cdot...\cdot n$.
@Knight It is always a natural number, that can be proved, unless the factorial expression is used for some decimal number.
If you're feeling funny, you can apply Lagrange's theorem to the obvious copy of $S_{n-r}$ in $S_n$
@Thorgott Didn’t get that last line
@AbhasKumarSinha See the numerator is an integer and denominator is also an integer but it’s not obvious that there division will also be an integer
14:27
@Knight The numerator must divide denominator all the time
$n! \equiv (n-r)! \mod n$
@AbhasKumarSinha WHAT?
@Knight yes.
@AbhasKumarSinha I think you meant that “numerator is always divisible by denominator “
$$\frac{n!}{(n-r)!}=\frac{1\cdot...\cdot n}{1\cdot...\cdot(n-r)}=\frac{1\cdot...\cdot(n-r)\cdot(n-r+1)\cdot...\cdot n}{1\cdot...\cdot(n-r)}=(n-r+1)\cdot...\cdot n$$
@Knight in case of permutation formula
@Thorgott Yap
14:29
@Thorgott Can you please write that message with double $ as it is very hard to read, sorry!
Thanks
Got you
It's just this thing: $$ P = \prod \limits_{i=1}^{n-i} (n-r+i) $$
@Knight n=(n-r)+r so n!/(n-r)!=((n-r)+r)!/(n-r)! and (a+k)!/k! is a natural number because, k! is product of the first k numbers and (a+k)! is a product of k consecutive numbers and one of them is surely divisible by 1, one surely by 2, etc..., one surely by k. This is a different proof.
@Rithaniel @Ante alright...
I just don't get how to compare the distribution of the rationals over some base field like the reals
sorry wrong formula! :P
@Ante Can you please describe that “(a+k)! is a product of .... “ once more
14:34
For example maybe I can systematically densify a space (0,1) with fibers that eventually cover the real line (in some limit)
@AbhasKumarSinha How’d you doing these days?
@Knight Learning, Tensor, Manifolds, General Relativity
@AbhasKumarSinha Wow! Great
@Knight Thanks :)
@AbhasKumarSinha Then why are you going for engineering?
14:36
@Knight Now. I'm in high school, yes, I think, I'll do a Bachleor's in Engineering, so, yes, B Tech
All right!
@Knight Let's talk in facebook, not here :)
@AbhasKumarSinha How much does the convolution of the constant lines of space and time in a pseudo riemanian manifold equipped with the lorentz metric, constrain the geometry of a class of hyperbolic submanifolds isometrically and densely embedded in a bounded but not compact region?
@geocalc33 The question is how much the answer is very less
That's flat spacetime, (no matter)
@AbhasKumarSinha okay that makes sense
14:41
@geocalc33 unlike your question :P
yeah it's a rough draft lol
@AbhasKumarSinha I guess a better way to phrase it is, first convolute two transveral and isomorphic copies of 2-minkowski space and then understand the geometry of submanifolds
$$ I = \int \limits_{\text{Manifold } \mathcal D} \dfrac{\mathcal L}{\psi^{a\dots d}_{e \dots h}} \nabla \psi_{e \dots h}^{a \dots d} \, d^4 x + \int \limits_{\text{manifold } \mathcal D} \sqrt{-g} R \dfrac{d^4 x}{g_{\mu u}}$$
what's that^?
14:46
How to draw bigger integrals on mathjax?
@geocalc33 let's not take all math seriously :P XD
In my mind I was thinking it was not valid lo
$\psi : \mathbb R^n \rightarrow T$
psi is a map from R^n to T
chart
map
whatever it's same
yeah, go on
14:48
flatspacetime^
yeah, so is the point you're trying to make that I can't convolute two flat spacetimes
yap
@geocalc33 Different techniques are in existance for that.
@AbhasKumarSinha can't you if you make the constant lines of space and time overlap?
@geocalc33 It'z simpel
14:51
it's simple
elbaorate
i t s s i m p l e f l a t M i n k o w s k i s p a c e
7 hours ago, by Amresh Prasad Sinha
this was my question :- Prove that (n!)! is divisible by (n!)^{(n-1)!}
@AmreshPrasadSinha This question is not solvable by me, I have tried it enough
Oh I get it now! @AbhasKumarSinha
@geocalc33 g r e a t t h e n
14:54
@AmreshPrasadSinha I think it’s one of those trick questions, question-maker has something in his mind which he thinks is obvious but it’s not.
@Knight just prove that $n!! \geq (n!)^{(n-1)}!$
@AmreshPrasadSinha yo, we share same surname...
@AbhasKumarSinha 1. That’s not enough, 2. It can’t be proved (at least by me)
That's not fair, you are asking beginner competitive questions!
What!
@AbhasKumarSinha what techniques are used to convolute riemannian manifolds as submanifolds of minkowski then? warped metrics, convolution metrics?
14:56
@Knight 1. that's enough! ;)
@AbhasKumarSinha No, $3 \gt 2$ but 3 is not divisible by 2
@geocalc33 Variation of $g_{\mu u}$ and show that $g_{\mu u} = \eta$
ah sick!
@Knight if both sides of inequality end with factorial, then it'd be true $a! \geq b!$, then $a! | b!$
@geocalc33 I understand
@Knight You get that? if $a! \geq b!$, then $a!$ will divide $b!$

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