Discussion between clarinetist and OneRaynyDay

OneRaynyDay

23:25

Oh neat-o!

Clarinetist

Okay, so let's see what we have here

OneRaynyDay

Okie, let me just copy paste from previously :D

$P(X)$ :

$P(A\cap B^c) = P(A) + P(B^c) - P(A \cup B^c)$

$P(Y)$:

$P(A^c \cap B) = P(A^c) + P(B) - P(A^c \cup B)$

The proof in question is:

prove: $P((A\cap B^c) \cup (A^c\cap B)) = P(A) + P(B) - 2P(A \cap B)$

Clarinetist

where $X = A \cap B^c$ and $Y = A^c \cap B$

OneRaynyDay

Yupp

Clarinetist

Okay, I agree with everything you have

OneRaynyDay

23:28

And $P(X) + P(Y)$ is equal to the LHS of the proof statement

Clarinetist

@OneRaynyDay Yep.

OneRaynyDay

awesome, okay so after adding both and rearranging I get:

$P(A) + P(A^c) + P(B) + P(B^c) - P(A \cup B^c) - P(A^c \cup B)$

Whereas we can see it's equal to $ 2 - P(A \cup B^c) - P(A^c \cup B)$

Clarinetist

Yeah, and then you have $P(X) + P(Y) = $ that stuff ^

Now here's the problem

OneRaynyDay

yup :D

Clarinetist

The only thing you can do from here is use the addition rule again... chances are, if you're having to use the addition rule twice on the same union, you're probably doing something wrong

OneRaynyDay

23:30

yeah... So that was my first attempt

I figured I was in a dead-end in this respect

Clarinetist

K, let's see your 2nd one

OneRaynyDay

Cool, so for the 2nd one I tackled it straight in the statement $(A \cap B^c) \cup (A^c \cap B)$

(the statement within the LHS probability function)

Clarinetist

Yep

OneRaynyDay

This was where I asked for the foil-y statement confirmation haha

So following that I got:

$((A\cap B^c)\cup A^c) \cap ((A \cap B^c)\cup B)$

And upon rearranging:

$((A \cup A^c) \cap B^c) \cap ((B^c \cup B) \cap A)$

and as we know that means it's

$B^c \cap A$

Clarinetist

Unfortunately, what you did there isn't true

OneRaynyDay

23:33

Oh nos

Clarinetist

Actually, $(A \cap B^c) \cup A^c = (A \cap A^c) \cup (B^c \cap A^c)$

OneRaynyDay

OH.

Ohmy- okay I see what you mean

Then the correct version would be...

Clarinetist

@OneRaynyDay OOOPS

I did that wrong

OneRaynyDay

did what wrong?

Clarinetist

Wait, nvm

My work is right

I always screw up those set properties

OneRaynyDay

23:36

yeah same haha

Clarinetist

@OneRaynyDay Wait

Lol

Let's look here

Algebra of sets

The algebra of sets defines the properties and laws of sets, the set-theoretic operations of union, intersection, and complementation and the relations of set equality and set inclusion. It also provides systematic procedures for evaluating expressions, and performing calculations, involving these operations and relations. Any set of sets closed under the set-theoretic operations forms a Boolean algebra with the join operator being union, the meet operator being intersection, and the complement operator being set complement. == Fundamentals == The algebra of sets is the set-theoretic analogue of...

$(A \cap B^c) \cup A^c = A^c \cup (A \cap B^c)$

so that would be

$(A^c \cup A) \cap (A^c \cup B^c)$

Whew, there we go

OneRaynyDay

wait... but if we apply the associative principle

$(X \cap Y) \cup Z$

Clarinetist

@OneRaynyDay No, notice that everything faces the same way for associative

They're either all $\cup$ or $\cap$

OneRaynyDay

Ahhhhh

Clarinetist

Not the case here

OneRaynyDay

23:38

Wow I would've gotten that wrong so hard on an exam

thanks for the save haha

Okie, then following the previous agreed answer, we'd have something like...

$P(A^c \cup B^c)$ right?

woopsies not so fast

Clarinetist

@OneRaynyDay Yeah, it's some strange compound set

OneRaynyDay

Yeah - it would be

Clarinetist

Let's just put it this way... you don't want to work with $(A^c \cup A) \cap (A^c \cup B^c)$.

OneRaynyDay

but $(A^c \cup A) = \omega$

Clarinetist

If you'd like, I can guide you through an approach

OneRaynyDay

23:41

oops the capital omega

oh yeah sure!

Clarinetist

Yes, but the thing is, you want to work with disjoint unions. $P$ works well with those, not so well with intersections

OneRaynyDay

ohh I see

got it

Clarinetist

So we want to prove $P((A \cap B^c) \cup (A^c \cap B)) = P(A) + P(B) - 2P(A \cap B)$

To start, it should be easy to see that $A \cap B^c$ and $A^c \cap B$ are a disjoint union

so on the LHS, we can do $P(A \cap B^c) + P(A^c \cap B)$

OneRaynyDay

right

ohhhh I see it LOL

the moment you stated the additiveness, I see it pretty straight

Clarinetist

Now let's see how this works

OneRaynyDay

23:43

$P(A \cap B^c) = P(A) - P(A \cup B)$?

Clarinetist

Yeah, then apply that same equation to the other one, since $A^c \cap B = B \cap A^c$

If you've already proven that equation, this is easy

OneRaynyDay

omg I see it LOL

Clarinetist

and you're done

OneRaynyDay

ahhhh -scratches head- how did you see that so easily

Clarinetist

I've had a lot of practice with these

To see it visually though

Quick sec

Here's $A^c \cap B$:

OneRaynyDay

23:46

ahh right

Clarinetist

and here's $A \cap B^c$:

This isn't a "rigorous" proof that they're disjoint, but it's easy to see

OneRaynyDay

right, intuitively that makes sense

Clarinetist

and what's nice about $P$ is that when you take a union of two sets that are disjoint, you can add those two probabilities together instead

I.e., if $X$ and $Y$ are disjoint, $P(X \cup Y) = P(X) + P(Y)$

OneRaynyDay

Also I didn't have a "rigorous" proof that $P(A \cap B^c) = P(A) - P(A \cap B)$ but rather visualized it

Mmm, gotcha

I should look out for the disjointness of sets when doing these problems

Clarinetist

@OneRaynyDay Well, let's make it rigorous

@OneRaynyDay Oh wait

@OneRaynyDay It should be $A \cap B$, not $A \cup B$

OneRaynyDay

23:49

ohh woops

Clarinetist

@OneRaynyDay Anyway, it's pretty easy to prove rigorously. Show $A = (A \cap B^c) \cup (A \cap B)$ is a disjoint union

@OneRaynyDay Thus, $P(A) = P(A \cap B^c) + P(A \cap B)$. Done.

OneRaynyDay

ah yeah - after I made it into a $\cap$ it made a ton of sense LOL

silly mistakes oh man ;-;

Clarinetist

Better to make them now than on the exam, I'd say

It'll take some practice

OneRaynyDay

Gotcha. Thank you sooooo much man :) This was an enlightening discussion

Clarinetist

I've seen this stuff for like 4-5 years, so I've had a lot of practice

No problem

OneRaynyDay

23:51

yep! I was just reading this book: athenasc.com/Prob-2nd-Ch1.pdf

Oh I see :D Do you do a lot of probability theory/statistics?

Clarinetist

@OneRaynyDay Yeah, I'm doing a masters in statistics

OneRaynyDay

Ahhh - inference based?

Clarinetist

This is the kinda stuff I've been doing lately

OneRaynyDay

Sounds fun nonetheless!

Clarinetist

For now, yes; I'm bringing in machine learning into my toolset as well

OneRaynyDay

23:52

Ah wow - one day I hope to be like you haha

I'm a computer science major, but looking to specialize in machine learning

Clarinetist

Haha, I'm flattered. Probability's so much fun

I wish I had a better CS background

OneRaynyDay

unlike everyone who follows the latest trends without understanding it I'd much rather learn the material rigorously

so I'm starting from the basics of probability and building up from there :) so I'd say I envy your background!

Clarinetist

My math/stats background is fine, but I'm having to self-teach the CS stuff. Looking into learning about algorithmic efficiency soon, maybe some C++ too

OneRaynyDay

Ooh - if you'd like, I could teach you C++ and we can trade skills?

We can even trade contacts that way

Clarinetist

Hmm, if you're serious about that, I'd like to pursue it

OneRaynyDay

23:54

I'm pretty proficient in C++, enough to get me into Microsoft and Bloomberg's internship program

Clarinetist

Dang

Congrats on that

OneRaynyDay

So yeah! PLEASE let me know if you'd like any help haha

Clarinetist

Yeah, here's my programming background

OneRaynyDay

Thank you! It's nothing special though, when compared to your statistics background LOL

Clarinetist

VBA, LaTeX (if you even consider that programming), Python, and I use SQL and R on a daily basis

OneRaynyDay

23:55

Ah yeah, R is a statistician's language

Clarinetist

Well, what would you prefer: FB or e-mail?

OneRaynyDay

both is fine! :D

Clarinetist

facebook.com/BbClarinetist

I look forward to working with you more :)

OneRaynyDay

Sent you a request (I'm Ray Zhang)

Email is [email protected]

Awesome :) thank you man!

Clarinetist

Well, Ray, nice to meet you. I'm starting to learn C++ since it underlies R

OneRaynyDay

23:56

I'll be taking a machine learning class next quarter and I'll definitely need some help haha, so don't hesitate to message me whenever you want to know something about C++, it's my most proficient language second to java/python

because if you don't, I'd feel like I'm bothering you too much ^_^

Clarinetist

Yeah, I'm no machine learning expert lol. I'm still learning. Stats helps some, but not much

My forte is the math/stats side

OneRaynyDay

yeah - but it's no problem! The class is very probability-theory heavy and so as a result the class is more theoretical

The programming aspect is easy if you understand the math :)

Clarinetist

K, well, feel free to just chat me on FB whenever you feel like it. Got any other questions?

OneRaynyDay

no questions for now! :D thanks!

Clarinetist

@OneRaynyDay K, then I have one quick question for you

OneRaynyDay

23:59

and same for you; I'm pretty active on messenger :)

Clarinetist

If I know basic programming in R/Python, you know - variables, loops, conditional statements, data structures and the like, how do I become a better programmer? @OneRaynyDay

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$Mathematics$ Discussion between clarinetist and On…