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\begin_body

\begin_layout Title
Introduction to Probability Theory
\begin_inset Newline newline
\end_inset

Problem set #1
\end_layout

\begin_layout Author
Student: Sergi Blanco Cuaresma
\end_layout

\begin_layout Abstract
Solutions to the problem set number one of the subject Statistics, Monte
 Carlo Methods and Data Processing - Master in Astrophysics, Particle Physics
 and Cosmology (Universitat de Barcelona).
\end_layout

\begin_layout Section
Problem
\end_layout

\begin_layout Subsection
Individual cumulative function
\end_layout

\begin_layout Standard
Suppose we have samples for each random variable X, Y and Z (i.i.d):
\end_layout

\begin_layout Standard
\align center
\begin_inset Formula $\left\{ X_{1},...,X_{n}\right\} $
\end_inset

 
\begin_inset Formula $\left\{ Y_{1},...,Y_{n}\right\} $
\end_inset

 
\begin_inset Formula $\left\{ Z_{1},...,Z_{n}\right\} $
\end_inset


\end_layout

\begin_layout Standard
Let define the maximum of each sample as:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
W_{x}=max\left(X_{1},...,X_{n}\right)\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
W_{y}=max\left(Y_{1},...,Y_{n}\right)\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
W_{z}=max\left(Z_{1},...,Z_{n}\right)\]

\end_inset


\end_layout

\begin_layout Standard
And let's consider that these maximums are ordered:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
W_{x}<W_{y}<W_{z}\]

\end_inset


\end_layout

\begin_layout Standard
The probability distribution function of 
\begin_inset Formula $X_{n}$
\end_inset

, 
\begin_inset Formula $Y_{n}$
\end_inset

 and 
\begin_inset Formula $Z_{n}$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
f\left(x\right)=\begin{cases}
6x^{5} & if\:0\leq x\leq1\\
0 & elsewhere\end{cases}\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Graphics
	filename Problem 1 - Density graph.png
	scale 50

\end_inset


\end_layout

\begin_layout Standard
The cumulative distribution function of individual 
\begin_inset Formula $X_{n}$
\end_inset

, 
\begin_inset Formula $Y_{n}$
\end_inset

 and 
\begin_inset Formula $Z_{n}$
\end_inset

:
\end_layout

\begin_layout Standard
\align center
\begin_inset Formula $F\left(x\right)=P\left(X\leq x\right)=\intop_{0}^{x}6x^{5}dx=x^{6}$
\end_inset

 where 
\begin_inset Formula $0\leq x\leq1$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Graphics
	filename Problem 1 - Cumulative distribution.png
	scale 50

\end_inset


\end_layout

\begin_layout Subsection
Cumulative distribution for X, Y and Z
\end_layout

\begin_layout Standard
We seek the cumulative distribution for the maximum of X, Y and Z:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
F_{x}\left(x\right)=P\left(W_{x}<x\right)=x^{6}\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
F_{x,y}\left(x\right)=P\left(W_{x}<x,W_{y}<x\right)=x^{6}\cdot x^{6}=x^{12}\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
F_{x,y,z}\left(x\right)=P\left(W_{x}<x,W_{y}<x\right)=x^{6}\cdot x^{6}\cdot x^{6}=x^{18}\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Graphics
	filename Problem 1 - Cumulative distribution - Final version.png
	scale 50

\end_inset


\end_layout

\begin_layout Subsection
Probability density of X, Y and Z
\end_layout

\begin_layout Standard
The probability density is derived from the cumulative distribution:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
f_{x,y,z}\left(x\right)=\frac{d}{dx}F_{x,y,z}\left(x\right)=18x^{17}\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Graphics
	filename Problem 1 - Density function X Y Z - Final version.png
	scale 50

\end_inset


\end_layout

\begin_layout Section
Problem
\end_layout

\begin_layout Subsection
Probability of containing only two queens in the set A
\end_layout

\begin_layout Standard
Recall (combinations without repetition where order does not matter):
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\binom{n}{k}=\frac{n!}{k!\cdot\left(n-k\right)!}\]

\end_inset


\end_layout

\begin_layout Standard
Number of ways we can pick 26 cards to form set A:
\end_layout

\begin_layout Itemize
Population n: 52 cards
\end_layout

\begin_layout Itemize
Sample k: 26 cards
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\binom{n}{k}=\frac{n!}{k!\cdot\left(n-k\right)!}=\frac{52!}{26!\cdot\left(52-26\right)!}=495.918.532.948.104\]

\end_inset


\end_layout

\begin_layout Standard
Number of ways we can pick 2 queens from 4:
\end_layout

\begin_layout Itemize
Population n: 50 cards
\end_layout

\begin_layout Itemize
Sample k: 24 cards
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\binom{4}{2}=\frac{n!}{k!\cdot\left(n-k\right)!}=\frac{4!}{2!\cdot\left(4-2\right)!}=6\]

\end_inset


\end_layout

\begin_layout Standard
Number of ways we can pick 24 card from the remaining:
\end_layout

\begin_layout Itemize
Population n: 48 cards
\end_layout

\begin_layout Itemize
Sample k: 24 cards
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\binom{48}{24}=\frac{n!}{k!\cdot\left(n-k\right)!}=\frac{48!}{24!\cdot\left(48-24\right)!}=32.247.603.683.100\]

\end_inset


\end_layout

\begin_layout Standard
Probability of containing two queens in the set A:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
P\left("Two\, queens\, in\, set\, A"\right)=6\cdot\frac{32.247.603.683.100}{495.918.532.948.104}=\frac{325}{833}=0.39016\]

\end_inset


\end_layout

\begin_layout Subsection
Probability of containing the queen of spades and the queen of diamonds
 in the set A
\end_layout

\begin_layout Standard
Consider we have already picked the two queens and there are 50 remaining
 cards.
 Number of ways of completing the set A with 24 more cards:
\end_layout

\begin_layout Itemize
Population n: 50 cards
\end_layout

\begin_layout Itemize
Sample k: 24 cards
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\binom{n}{k}=\frac{n!}{k!\cdot\left(n-k\right)!}=\frac{50!}{24!\cdot\left(50-24\right)!}=121.548.660.036.300\]

\end_inset


\end_layout

\begin_layout Standard
Probability of containing the two queens in the set A:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
P\left("Two\, concrete\, queens\, in\, set\, A"\right)=\frac{121.548.660.036.300}{495.918.532.948.104}=\frac{25}{102}=0.2451\]

\end_inset


\end_layout

\begin_layout Section
Problem
\end_layout

\begin_layout Subsection
Probability density for Z = X / Y (ratio distribution)
\end_layout

\begin_layout Standard
Given:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
X=e^{\frac{-x^{2}}{2\sigma_{x}^{2}}}\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
Y=e^{\frac{-x^{2}}{2\sigma_{y}^{2}}}\]

\end_inset


\end_layout

\begin_layout Standard
The probability density for Z:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
Z=\frac{X}{Y}=\frac{e^{\frac{-x^{2}}{2\sigma_{x}^{2}}}}{e^{\frac{-x^{2}}{2\sigma_{y}^{2}}}}=e^{\frac{-x^{2}}{2\sigma_{x}^{2}}-\frac{-x^{2}}{2\sigma_{y}^{2}}}=e^{-\left(\frac{x^{2}}{2\sigma_{x}^{2}}-\frac{x^{2}}{2\sigma_{y}^{2}}\right)}=e^{-\left(\alpha\right)}\]

\end_inset


\end_layout

\begin_layout Standard
Detailed calculation of the exponent:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\alpha=\frac{x^{2}}{2\sigma_{x}}-\frac{x^{2}}{2\sigma_{y}}=\frac{x^{2}\cdot\sigma_{y}-x^{2}\cdot\sigma_{x}}{2\sigma_{x}^{2}\sigma_{y}^{2}}=\frac{x^{2}\cdot\left(\sigma_{y}-\sigma_{x}\right)}{2\sigma_{x}^{2}\sigma_{y}^{2}}=\frac{x^{2}}{2\cdot\frac{\sigma_{x}^{2}\sigma_{y}^{2}}{\sigma_{y}-\sigma_{x}}}\]

\end_inset


\end_layout

\begin_layout Standard
Therefore:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
Z=\frac{X}{Y}=e^{-\left(\alpha\right)}=e^{-\left(\frac{x^{2}}{2\cdot\frac{\sigma_{x}^{2}\sigma_{y}^{2}}{\sigma_{y}-\sigma_{x}}}\right)}\]

\end_inset


\end_layout

\begin_layout Standard
And it can be considered that the standard deviation of the new probability
 density is:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\sigma_{z}=\sqrt{\frac{\sigma_{x}^{2}\sigma_{y}^{2}}{\sigma_{y}-\sigma_{x}}}\]

\end_inset


\end_layout

\begin_layout Standard
In consequence, the probability density for Z can be written as:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
Z=\frac{X}{Y}=e^{-\frac{x^{2}}{2\sigma_{z}^{2}}}\]

\end_inset


\end_layout

\begin_layout Section
Problem
\end_layout

\begin_layout Standard
We've got a limited alphabet of L elements:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
X=\left\{ x_{1},x_{2},\ldots,x_{L}\right\} \]

\end_inset


\end_layout

\begin_layout Standard
If we choose m letters from this alphabet, each new letter is independent
 from the past picked letters:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
P\left(x_{n}|x_{n-1},...,x_{n-n}\right)=P\left(x_{n}\right)=\frac{1}{L}\]

\end_inset


\end_layout

\begin_layout Standard
Therefore, individual probability for one pick (m=1 and n=1):
\end_layout

\begin_layout Standard
\begin_inset Formula \[
P\left(A\, letter\, being\, picked\right)=\frac{1}{L}\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
P\left(A\, letter\, NOT\, being\, picked\right)=1-\frac{1}{L}\]

\end_inset


\end_layout

\begin_layout Standard
And the probability of picking only 1 concrete letter (n=1) for m picks:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
P\left(A\, letter\, being\, picked\, only\,1\, time\right)=\left(\frac{1}{L}\right)^{1}\cdot\left(1-\frac{1}{L}\right)^{m-1}\]

\end_inset


\end_layout

\begin_layout Standard
Finally, the probability of picking n times a letter for m picks:
\end_layout

\begin_layout Standard
\align center
\begin_inset Formula $P\left(A\, letter\, being\, picked\, n\, times\right)=\left(\frac{1}{L}\right)^{n}\cdot\left(1-\frac{1}{L}\right)^{m-n}$
\end_inset

 where 
\begin_inset Formula $n\leq m$
\end_inset


\end_layout

\begin_layout Section
Problem
\end_layout

\begin_layout Standard
Let's consider the following random variables (numbers correspond to the
 different boxes):
\end_layout

\begin_layout Standard
\noindent
\align center
\begin_inset Box Frameless
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
width "75col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout
\begin_inset Formula $SelectedBox\in\left\{ 1,2,3\right\} $
\end_inset

My box selection
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $ObjectBox\in\left\{ 1,2,3\right\} $
\end_inset

 Box where the desired object is located
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $OpenBox\in\left\{ 1,2,3\right\} $
\end_inset

 Box opened by the Oracle
\end_layout

\end_inset


\end_layout

\begin_layout Standard
My box selection and the box where the desired object is located are independent
 variables:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
P\left(SelectedBox\right)=P\left(SelectedBox|ObjectBox\right)=\frac{1}{3}\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
P\left(ObjectBox\right)=P\left(ObjectBox|SelectedBox\right)=\frac{1}{3}\]

\end_inset


\end_layout

\begin_layout Standard
Otherwise, the Oracle will open one box or another depending on my selection
 and on the location of the desired object:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
P\left(OpenBox|SelectedBox,ObjectBox\right)=\begin{cases}
\frac{1}{2} & if\, SelectedBox=ObjectBox\\
1 & if\, OpenBox\neq ObjectBox\neq SelectedBox\\
0 & if\, OpenBox=SelectedBox\\
0 & if\, OpenBox=ObjectBox\end{cases}\]

\end_inset


\end_layout

\begin_layout Standard
And we wan to know, what is the probability of a box containing the desired
 object if we already have made an initial selection and the Oracles has
 open a box:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
P\left(ObjectBox|SelectedBox,OpenBox\right)\]

\end_inset


\end_layout

\begin_layout Standard
To make the calculation we can use the Bayes' theorem:
\end_layout

\begin_layout Standard
\noindent
\align center
\begin_inset Box Boxed
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
width "50col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout

\emph on
Abbreviations:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $SelectedBox=Select$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $ObjectBox=Obj$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $OpenBox=Open$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
P\left(Obj|Select,Open\right)=\frac{P\left(Open|Select,Obj\right)P\left(Obj|Select\right)}{P(Open|Select)}\]

\end_inset


\end_layout

\begin_layout Standard
where 
\begin_inset Formula \[
P\left(Open|Select\right)=\sum_{Obj=1}^{3}P\left(Open,Obj|Select\right)=\sum_{ObjBox=1}^{3}P\left(Open|Select,Obj\right)P\left(Obj|Select\right)\]

\end_inset


\end_layout

\begin_layout Standard
For example, once I have selected a box (number 2) and the Oracle has opened
 number 3:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
SelectedBox=2\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
OpenBox=3\]

\end_inset


\end_layout

\begin_layout Standard
The probability of the object being in the remaining box (number 1):
\end_layout

\begin_layout Standard
\begin_inset Formula \[
ObjectBox=1\neq SelectedBox\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
P(ObjectBox|SelectedBox,OpenBox)=\frac{1\cdot\frac{1}{3}}{\frac{1}{2}\cdot\frac{1}{3}+1\cdot\frac{1}{3}+0\cdot\frac{1}{3}}=\frac{2}{3}\]

\end_inset


\end_layout

\begin_layout Standard
And the probability that the object is on my selected box (number 2):
\end_layout

\begin_layout Standard
\begin_inset Formula \[
ObjectBox=SelectedBox\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
P(Obj=Select|Select,Open)=1-P(Obj|Select,Open)=\frac{1}{3}\]

\end_inset


\end_layout

\begin_layout Standard
In conclusion, it is better to choose the remaining box and change my initial
 choice.
\end_layout

\begin_layout Section
Problem
\end_layout

\begin_layout Subsection
Inequality
\end_layout

\begin_layout Standard
We have four random variables (from Alice and Bob's perspective):
\end_layout

\begin_layout Standard
\noindent
\align center
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height_special "totalheight"
status open

\begin_layout Plain Layout
\begin_inset Formula $\left\{ A_{1},A_{2}\right\} $
\end_inset

 Properties that Alice is interested in
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\left\{ B{}_{1},B_{2}\right\} $
\end_inset

 Properties that Bob is interested in
\end_layout

\end_inset


\end_layout

\begin_layout Standard
This random variables can take values 
\begin_inset Formula $\left\{ -1,+1\right\} $
\end_inset

.
 Oscar, the oracle, draws a random variable 
\begin_inset Formula $\Lambda$
\end_inset

 according to the probability distribution 
\begin_inset Formula $\pi$
\end_inset

.
 In this context, this variable acts as a hidden variable that determines
 the outcome (
\begin_inset Formula $a_{i}$
\end_inset

 for Alice, 
\begin_inset Formula $b_{i}$
\end_inset

 for Bob) of any measurement:
\end_layout

\begin_layout Standard
\noindent
\align center
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\begin_layout Plain Layout
\begin_inset Formula $a_{i}=f_{A}\left(A_{i},\lambda\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $b_{i}=f_{B}\left(B_{i},\lambda\right)$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Where
\begin_inset Formula $\lambda$
\end_inset

 is a value of the random variable 
\begin_inset Formula $\Lambda$
\end_inset

.
 Given these premises, it is clear that one of the following equations must
 be true:
\end_layout

\begin_layout Standard
\noindent
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\begin_layout Plain Layout
Subtraction: 
\begin_inset Formula $f_{A}\left(A_{1},\lambda\right)-f_{A}\left(A_{2},\lambda\right)=0$
\end_inset


\end_layout

\begin_layout Plain Layout
Addition: 
\begin_inset Formula $f_{A}\left(A_{1},\lambda\right)+f_{A}\left(A_{2},\lambda\right)=0$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
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False
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\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
True
\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\end_layout

\begin_layout Standard
Therefore:
\end_layout

\begin_layout Standard
\noindent
\align center
\begin_inset Box Frameless
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout
\begin_inset Tabular
<lyxtabular version="3" rows="3" columns="2">
<features>
<column alignment="center" valignment="top" width="0">
<column alignment="center" valignment="top" width="0">
<row>
<cell alignment="center" valignment="top" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\noindent
\align left
\begin_inset Formula $f_{A}\left(A_{1},\lambda\right)f_{B}\left(B_{1},\lambda\right)+f_{A}\left(A_{2},\lambda\right)f_{B}\left(B_{1},\lambda\right)+f_{A}\left(A_{1},\lambda\right)f_{B}\left(B_{2},\lambda\right)$
\end_inset


\end_layout

\end_inset
</cell>
<cell multicolumn="1" alignment="center" valignment="top" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\noindent
\align left

\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\noindent
\align left
\begin_inset Formula $-f_{A}\left(A_{2},\lambda\right)f_{B}\left(B_{2},\lambda\right)=f_{B}\left(B_{1},\lambda\right)\left(f_{A}\left(A_{1},\lambda\right)+f_{A}\left(A_{2},\lambda\right)\right)$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $+f_{B}\left(B_{2},\lambda\right)\left(f_{A}\left(A_{1},\lambda\right)-f_{A}\left(A_{2},\lambda\right)\right)\leq2$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\noindent
\align left
(6.1.1)
\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
If addition
\begin_inset Formula $=0$
\end_inset

, the first term is equal to zero and the second one cannot be greater (less)
 than +2 (-2) (i.e.
 maximum: 
\begin_inset Formula $f_{A}\left(A_{1},\lambda\right)=+1$
\end_inset

 and 
\begin_inset Formula $f_{A}\left(A_{2},\lambda\right)=+1$
\end_inset

).
 Otherwise, if subtraction
\begin_inset Formula $=0$
\end_inset

, the second term is equal to zero and the first one cannot be also greater
 (lesser) than +2 (-2) (i.e.
 maximum: 
\begin_inset Formula $f_{A}\left(A_{1},\lambda\right)=+1$
\end_inset

 and 
\begin_inset Formula $f_{A}\left(A_{2},\lambda\right)=-1$
\end_inset

).
\end_layout

\begin_layout Standard
After a large number of experiments, Alice and Bob can measure the correlators
 
\begin_inset Formula $\left\langle A_{i}B_{j}\right\rangle $
\end_inset

 (the mean value of the product of their outcomes):
\end_layout

\begin_layout Standard
\noindent
\align center
\begin_inset Box Frameless
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
width "100col%"
special "none"
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height_special "totalheight"
status open

\begin_layout Plain Layout
\begin_inset Formula $\left\langle A_{1}B_{1}\right\rangle =\int_{\Lambda}f_{A}\left(A_{1},\lambda\right)f_{B}\left(B_{1},\lambda\right)\pi\left(\lambda\right)d\lambda$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\left\langle A_{1}B_{2}\right\rangle =\int_{\Lambda}f_{A}\left(A_{1},\lambda\right)f_{B}\left(B_{2},\lambda\right)\pi\left(\lambda\right)d\lambda$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\left\langle A_{2}B_{1}\right\rangle =\int_{\Lambda}f_{A}\left(A_{2},\lambda\right)f_{B}\left(B_{1},\lambda\right)\pi\left(\lambda\right)d\lambda$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\left\langle A_{2}B_{2}\right\rangle =\int_{\Lambda}f_{A}\left(A_{2},\lambda\right)f_{B}\left(B_{2},\lambda\right)\pi\left(\lambda\right)d\lambda$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Now, let's consider:
\end_layout

\begin_layout Standard
\noindent
\align center
\begin_inset Box Frameless
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout
\begin_inset Formula $\left|\left\langle A_{1}B_{1}\right\rangle +\left\langle A_{1}B_{2}\right\rangle +\left\langle A_{2}B_{1}\right\rangle -\left\langle A_{2}B_{2}\right\rangle \right|=$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\int_{\Lambda}f_{A}\left(A_{1},\lambda\right)f_{B}\left(B_{1},\lambda\right)\pi\left(\lambda\right)d\lambda+\int_{\Lambda}f_{A}\left(A_{2},\lambda\right)f_{B}\left(B_{1},\lambda\right)\pi\left(\lambda\right)d\lambda+\int_{\Lambda}f_{A}\left(A_{1},\lambda\right)f_{B}\left(B_{2},\lambda\right)\pi\left(\lambda\right)d\lambda-\int_{\Lambda}f_{A}\left(A_{2},\lambda\right)f_{B}\left(B_{2},\lambda\right)\pi\left(\lambda\right)d\lambda$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $=\int_{\Lambda}\{f_{A}\left(A_{1},\lambda\right)f_{B}\left(B_{1},\lambda\right)+f_{A}\left(A_{2},\lambda\right)f_{B}\left(B_{1},\lambda\right)+f_{A}\left(A_{1},\lambda\right)f_{B}\left(B_{2},\lambda\right)-f_{A}\left(A_{2},\lambda\right)f_{B}\left(B_{2},\lambda\right)\}\pi\left(\lambda\right)d\lambda$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Inside the integral we have the term 
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
(6.1.1) which we know it is less (greater) than +2 (-2), therefore:
\end_layout

\begin_layout Standard
\begin_inset Formula $\left|\left\langle A_{1}B_{1}\right\rangle +\left\langle A_{1}B_{2}\right\rangle +\left\langle A_{2}B_{1}\right\rangle -\left\langle A_{2}B_{2}\right\rangle \right|\leq2$
\end_inset


\end_layout

\begin_layout Subsection
Joint probability distribution
\end_layout

\begin_layout Standard
Alice and Bob choose what measurement to do (
\begin_inset Formula $A_{i}$
\end_inset

and 
\begin_inset Formula $B_{i}$
\end_inset

 respectively) at random from some joint distribution 
\begin_inset Formula $\psi\left(A_{i},B_{j}\right)$
\end_inset

, and they get their outcomes 
\begin_inset Formula $a_{i}$
\end_inset

 and 
\begin_inset Formula $b_{j}$
\end_inset

.
 So, the whole experiment's outcome has the following probability distribution:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
P\left(A_{i},B_{j};a_{i},b_{j}\right)=\psi\left(A_{i},B_{j}\right)P\left(a_{i},b_{j}|A_{i},B_{j}\right)\]

\end_inset


\end_layout

\begin_layout Standard
And outcomes 
\begin_inset Formula $a_{i}$
\end_inset

 and 
\begin_inset Formula $b_{j}$
\end_inset

 are determined by a hidden random variable 
\begin_inset Formula $\Lambda$
\end_inset

 (according to the probability distribution 
\begin_inset Formula $\pi$
\end_inset

) with value 
\begin_inset Formula $\lambda$
\end_inset

 and the functions 
\begin_inset Formula $f_{A}\left(A_{i},\lambda\right)$
\end_inset

 and 
\begin_inset Formula $f_{B}\left(B_{j},\lambda\right)$
\end_inset

.
 So it can be written as a convex combination of independent probability
 distributions and the joint probability distribution can be considered
 local.
\end_layout

\begin_layout Subsection
Quantum mechanical experiments
\end_layout

\begin_layout Standard
As the problem stated, Alice can use two detector setting from which will
 measure a spin singlet property (i.e.
 
\begin_inset Formula $x_{A}$
\end_inset

 and 
\begin_inset Formula $z_{A}$
\end_inset

 axis), and Bob can also use its own detector for measuring other two properties
 (i.e.
 axis 
\begin_inset Formula $x_{B}$
\end_inset

 and 
\begin_inset Formula $z_{B}$
\end_inset

, which are rotated 135º relative to Alice's axis).
\end_layout

\begin_layout Standard
If we consider the Pauli matrices:
\end_layout

\begin_layout Standard
\noindent
\align center
\begin_inset Box Frameless
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout
\begin_inset Formula $\sigma^{x}=\left(\begin{array}{cc}
0 & 1\\
1 & 0\end{array}\right)$
\end_inset

 
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\sigma^{y}=\left(\begin{array}{cc}
0 & -i\\
i & 0\end{array}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\sigma^{z}=\left(\begin{array}{cc}
1 & 0\\
0 & -1\end{array}\right)$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Let's use 
\begin_inset Formula $\sigma^{x}$
\end_inset

 and 
\begin_inset Formula $\sigma^{x}$
\end_inset

 as the observables and the following eigenvectors:
\end_layout

\begin_layout Standard
\noindent
\align center
\begin_inset Box Frameless
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout
\begin_inset Formula $\left|\frac{1}{2},+\frac{1}{2}\right\rangle $
\end_inset

 eigenvector of 
\begin_inset Formula $\sigma^{z}$
\end_inset

 with eigenvalue 
\begin_inset Formula $+1$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\left|\frac{1}{2},-\frac{1}{2}\right\rangle $
\end_inset

 eigenvector of 
\begin_inset Formula $\sigma^{z}$
\end_inset

 with eigenvalue 
\begin_inset Formula $-1$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Then, the spin singlet state is represented by:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\frac{1}{\sqrt{2}}\left(\left|\frac{1}{2},+\frac{1}{2}\right\rangle _{A}\left|\frac{1}{2},-\frac{1}{2}\right\rangle _{B}-\left|\frac{1}{2},-\frac{1}{2}\right\rangle _{A}\left|\frac{1}{2},+\frac{1}{2}\right\rangle _{B}\right)\]

\end_inset


\end_layout

\begin_layout Standard
Note that products among eigenvectors correspond to tensor products 
\begin_inset Formula $\otimes$
\end_inset

.
\end_layout

\begin_layout Standard
Therefore, the measurements that Alice and Bob can perform:
\end_layout

\begin_layout Standard
\noindent
\align center
\begin_inset Box Frameless
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
width "100col%"
special "none"
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height_special "totalheight"
status open

\begin_layout Plain Layout
\begin_inset Formula $f_{A}\left(A_{1}\right)=\sigma^{z}\otimes I$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $f_{A}\left(A_{2}\right)=\sigma^{x}\otimes I$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $f_{B}\left(B_{1}\right)=-\frac{1}{\sqrt{2}}\cdot I\otimes\left(\sigma^{z}+\sigma^{x}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $f_{B}\left(B_{2}\right)=\frac{1}{\sqrt{2}}\cdot I\otimes\left(\sigma^{z}-\sigma^{x}\right)$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Where I is the identity matrix:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
I=\left(\sigma^{x}\right)^{2}\cdot\left(\sigma^{y}\right)^{2}\cdot\left(\sigma^{z}\right)^{2}=-i\sigma^{x}\sigma^{y}\sigma^{z}=\left(\begin{array}{cc}
1 & 0\\
0 & 1\end{array}\right)\]

\end_inset


\end_layout

\begin_layout Standard
Then we can state that:
\end_layout

\begin_layout Standard
\begin_inset Formula \[
\left\langle f_{A}\left(A_{1}\right)f_{A}\left(B_{1}\right)\right\rangle =\left\langle f_{A}\left(A_{2}\right)f_{A}\left(B_{1}\right)\right\rangle =\left\langle f_{A}\left(A_{2}\right)f_{A}\left(B_{2}\right)\right\rangle =\frac{1}{\sqrt{2}}\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula \[
\left\langle f_{A}\left(A_{1}\right)f_{A}\left(B_{2}\right)\right\rangle =-\frac{1}{\sqrt{2}}\]

\end_inset


\end_layout

\begin_layout Standard
Therefore, the inequality established at the beginning of this problem is
 violated:
\end_layout

\begin_layout Standard
\begin_inset Formula $\left\langle f_{A}\left(A_{1}\right)f_{A}\left(B_{1}\right)\right\rangle +\left\langle f_{A}\left(A_{2}\right)f_{A}\left(B_{1}\right)\right\rangle +\left\langle f_{A}\left(A_{2}\right)f_{A}\left(B_{2}\right)\right\rangle -\left\langle f_{A}\left(A_{1}\right)f_{A}\left(B_{2}\right)\right\rangle =\frac{4}{\sqrt{2}}=2\sqrt{2}>2$
\end_inset


\end_layout

\begin_layout Subsection
Conclusion
\end_layout

\begin_layout Standard
The initial theoretical inequality corresponds to 
\emph on
Bell's theorem
\emph default
 and it is deduced considering that:
\end_layout

\begin_layout Itemize
Realism: Hidden variables exists (although nowadays we cannot measure them)
 and determine the measurements that can do Bob and Alice
\end_layout

\begin_layout Itemize
The principle of locality is respected: an object is influenced only by
 its near surroundings
\end_layout

\begin_layout Standard
But as quantum experiments show, this inequality is violated and therefore
 at least one of the above statements is false.
 It's implication are huge for science and physics, although it does not
 demonstrate that quantum physics is a complete theory, it invalidates any
 real-local theory which consider the above statements true.
\end_layout

\begin_layout Standard
Either instantaneous large distance influence can be possible or our universe
 does not have fixed properties until we measure them, which seems to be
 counter-intuitive ideas with a great philosophical impact about our concept
 of reality.
\end_layout

\begin_layout Bibliography
\begin_inset CommandInset bibitem
LatexCommand bibitem
key "key-1"

\end_inset

Iblisdir, S.
 (2010).
 
\emph on
An Introduction to Probability Theory.

\emph default
 Universitat de Barcelona.
\end_layout

\begin_layout Bibliography
\begin_inset CommandInset bibitem
LatexCommand bibitem
key "key-10"

\end_inset

Pérez Olano, J.
 
\emph on
Capitulo 1: Combinatoria.

\emph default
 Retrieved October 30, 2010, from the World Wide Web: http://sauce.pntic.mec.es/~jp
eo0002/Archivos/PDF/T01.pdf
\end_layout

\begin_layout Bibliography
\begin_inset CommandInset bibitem
LatexCommand bibitem
key "key-4"

\end_inset

Math Forum (2001).
 
\emph on
Random Variables and Order Statistics.

\emph default
 Retrieved October 30, 2010, from the World Wide Web: http://mathforum.org/librar
y/drmath/view/52220.html
\end_layout

\begin_layout Bibliography
\begin_inset CommandInset bibitem
LatexCommand bibitem
key "key-5"

\end_inset

Saenz, R (2009).
 
\emph on
Muestreo de las letras del alfabeto.

\emph default
 Retrieved October 30, 2010, from the World Wide Web: http://seriesdivergentes.wo
rdpress.com/2009/03/09/muestreo-de-las-letras-del-alfabeto/
\end_layout

\begin_layout Bibliography
\begin_inset CommandInset bibitem
LatexCommand bibitem
key "key-11"

\end_inset

Bromiley, P.A.
 (2003).
 
\emph on
Products and Convolutions of Gaussian Distributions.

\emph default
 Medical School, University of Manchester.
\end_layout

\begin_layout Bibliography
\begin_inset CommandInset bibitem
LatexCommand bibitem
key "key-3"

\end_inset

Wikipedia (2010).
 
\emph on
Monty Hall problem.

\emph default
 Retrieved October 30, 2010, from the World Wide Web: http://en.wikipedia.org/wiki
/Monty_Hall_problem
\end_layout

\begin_layout Bibliography
\begin_inset CommandInset bibitem
LatexCommand bibitem
key "key-8"

\end_inset

Gómez-Esteban, P.
 (2010).
 
\emph on
Cuántica sin fórmulas – El Teorema de Bell.
 
\emph default
Retrieved October 30, 2010, from the World Wide Web:
\emph on
 
\emph default
http://eltamiz.com/2010/10/27/cuantica-sin-formulas-el-teorema-de-bell/
\end_layout

\begin_layout Bibliography
\begin_inset CommandInset bibitem
LatexCommand bibitem
key "key-6"

\end_inset

Wikipedia (2010).
 
\emph on
Bell's theorem.

\emph default
 Retrieved October 30, 2010, from the World Wide Web: http://en.wikipedia.org/wiki
/Bell's_theorem
\end_layout

\begin_layout Bibliography
\begin_inset CommandInset bibitem
LatexCommand bibitem
key "key-7"

\end_inset

Gill, R.D.
 (2007).
 
\emph on
Better Bell inequalities.

\emph default
 Mathematical Institute, Leiden University and EURANDOM, NWO.
\end_layout

\end_body
\end_document
