tex: NFDBP demo !

latex/content.tex

@@ -180,14 +180,75 @@ Mathematically, the NFBP algorithm imposes the following constraint on the first
 We implemented the NFBP algorithm in Python \footnotemark, for its ease of use
 and broad recommendation. We used the \texttt{random} library to generate
 random numbers between $ 0 $ and $ 1 $ and \texttt{matplotlib} to plot the
-results in the form of histograms. We ran $ R = 10^6 $ simulations with
-$ N = 10 $ different items each.
+results in the form of histograms.
 
 \footnotetext{The code is available in Annex \ref{annex:probabilistic}}
 
+We will try to approximate $ \mathbb{E}[X] $ and $ \mathbb{E}[V] $ with $
+	\overline{X_N} $ using $ {S_n}^2 $. This operation will be done for both $ R =
+	2 $ and $ R = 10^6 $ simulations.
+
+\[
+	\overline{X_N} = \frac{1}{N} \sum_{i=1}^{N} X_i
+\]
+
+As the variance value is unknown, we will use $ {S_n}^2 $ to estimate the
+variance and further determine the Confidence Interval (95 \% certainty).
+
+\begin{align*}
+	{S_N}^2      & = \frac{1}{N-1} \sum_{i=1}^{N} (X_i - \overline{X_N})^2                                      \\
+	IC_{95\%}(m) & = \left[ \overline{X_N} \pm \frac{S_N}{\sqrt{N}} \cdot t_{1 - \frac{\alpha}{2}, N-1} \right] \\
+\end{align*}
+
+
+
+
+\paragraph{2 simulations} We first ran $ R = 2 $ simulations to observe the
+behavior of the algorithm and the low precision of the results.
+
+% TODO graph T_i 2 sim
+
+On this graph, we can see each value of $ T_i $. Our calculations have yielded
+that $ \overline{T_1} = 1.0 $ and $ {S_N}^2 = 2.7 $. Our student coefficient is
+$ t_{0.95, 2} = 4.303 $.
+
+\begin{align*}
+	\overline{T_1} = \sum_{k=1}^{2} {T_1}_k & = 1.0                                                                 \\
+	IC_{95\%}(T_1)                          & = \left[ 1.0 \pm 1.96 \frac{\sqrt{2.7}}{\sqrt{2}} \cdot 4.303 \right] \\
+	                                        & = \left[ 1 \pm 9.8 \right]                                            \\
+\end{align*}
+
+With two simulations, we obtain $ \overline{T_1} = 1.0 $.
+
+
+
+IC observed
+
+We then ran $ R = 10^6 $ simulations with $ N = 50 $ different items each.
+With 10 6 simulations, we obtain Xn barre = cf graphe
+Calcul Sn carre
+IC observed
+
+
+Same for V.
+
+
+Graphe H
+
 \paragraph{Distribution of $ T_i $} We first studied how many items were
 present per bin.
 
+% TODO sim of T_i
+
+We determined the empirical mean to be
+
+\[
+	\overline{T_i} = \frac{1}{20} \sum_{k=1}^{20} T_k = 1.5 \qquad \forall 1 \leq i \leq 20
+\]
+
+
+We can show
+
 \paragraph{Distribution of $ V_i $} We then looked at the size of the first
 item in each bin.
 
@@ -262,7 +323,7 @@ bin. We have that
 	T_i = k \iff U_1 + U_2 + \ldots + U_{k-1} < 1 \text{ and } U_1 + U_2 + \ldots + U_{k} \geq 1
 \end{equation}
 
-Let $ A_k = \{ U_1 + U_2 + \ldots + U_{k-1} < 1 \}$. Hence,
+Let $ A_k = \{ U_1 + U_2 + \ldots + U_{k} < 1 \}$. Hence,
 
 \begin{align*}
 	% TODO = k
@@ -271,61 +332,33 @@ Let $ A_k = \{ U_1 + U_2 + \ldots + U_{k-1} < 1 \}$. Hence,
 	 & = P(A_{k-1}) - P(A_k) \qquad \text{ (as $ A_k \subset A_{k-1} $)} \\
 \end{align*}
 
-We will try to show that $ \forall k \geq 2 $, $ P(A_k) = \frac{1}{k!} $. To do
+We will try to show that $ \forall k \geq 1 $, $ P(A_k) = \frac{1}{k!} $. To do
 so, we will use induction to prove the following proposition \eqref{eq:induction},
-$ \forall k \geq 2 $:
+$ \forall k \geq 1 $:
 
 \begin{equation}
 	\label{eq:induction}
 	\tag{$ \mathcal{H}_k $}
-	P(U_1 + U_2 + \ldots + U_{k-1} < a) = \frac{a^k}{k!} \qquad \forall a \in [0, 1],
+	P(U_1 + U_2 + \ldots + U_{k} < a) = \frac{a^k}{k!} \qquad \forall a \in [0, 1],
 \end{equation}
 
-Let us denote $ S_k = U_1 + U_2 + \ldots + U_{k-1} \qquad \forall k \geq 2 $.
+Let us denote $ S_k = U_1 + U_2 + \ldots + U_{k} \qquad \forall k \geq 1 $.
 
-\paragraph{Base cases} $ k = 2 $ : $ P(U_1 < a) = a \neq \frac{a^2}{2}$ supposedly proving $ (\mathcal{H}_2) $.
+\paragraph{Base case} $ k = 1 $ : $ P(U_1 < a) = a = \frac{a^1}{1!}$, proving $ (\mathcal{H}_1) $.
 
-$ k = 2 $ : \[ P(U_1 + U_2 < a) = \iint_{\cal{D}} f_{U_1, U_2}(x, y) \cdot (x + y) dxdy \]
-
-Where $ \mathcal{D} = \{ (x, y) \in [0, 1]^2 \mid x + y < a \} $.
-
-$ U_1 $ and $ U_2 $ are independent, so
-\begin{align*}
-	f_{U_1, U_2}(x, y) & = f_{U_1}(x) \cdot f_{U_2}(y) \\
-	                   & = \begin{cases}
-		1 & \text{if } x \in [0, 1] \text{ and } y \in [0, 1] \\
-		0 & \text{otherwise}                                  \\
-	\end{cases}  \\
-\end{align*}
-
-Hence,
+\paragraph{Induction step} Let $ k \geq 2 $. We assume $ (\mathcal{H}_{k-1}) $ is
+true. We will show that $ (\mathcal{H}_{k}) $ is true.
 
 \begin{align*}
-	P(U_1 + U_2 < a)
-	 & = \iint_{\cal{D}} (x + y)dxdy                                                  \\
-	 & = \int_{0}^{a} \int_{0}^{a - x} (x + y) dy dx                                  \\
-	 & = \int_{0}^{a} \left[ xy + \frac{y^2}{2} \right]_{y=0}^{y=a - x} dx            \\
-	 & = \int_{0}^{a} \left( ax - x^2 + \frac{a^2}{2} - ax + \frac{x^2}{2} \right) dx \\
-	 & = \int_{0}^{a} \left( \frac{a^2}{2} - \frac{x^2}{2} \right) dx                 \\
-	 & = \left[ \frac{a^2 x}{2} - \frac{x^3}{6} \right]_{0}^{a}                       \\
-	 & = \frac{a^3}{2} - \frac{a^3}{6}                                                \\
+	P(S_k < a)                & = P(S_{k-1} + U_k < a)                                                             \\
+	                          & = \iint_{\cal{D}} f_{S_{k-1}, U_k}(x, y)  dxdy                                     \\
+	\text{Where } \mathcal{D} & = \{ (x, y) \in [0, 1]^2 \mid x + y < a \}                                         \\
+	                          & = \{ (x, y) \in [0, 1]^2 \mid 0 < x < a \text{ and } 0 < y < a - x \}              \\
+	P(S_k < a)                & = \iint_{\cal{D}} f_{S_{k-1}}(x) \cdot f_{U_k}(y) dxdy \qquad
+	\text{because $ S_{k-1} $ and $ U_k $ are independent}                                                         \\
+	                          & = \int_{0}^{a} f_{S_{k-1}}(x) \cdot \left( \int_{0}^{a-x} f_{U_k}(y) dy \right) dx \\
 \end{align*}
 
-
-\paragraph{Induction step} For a fixed $ k > 2 $, we assume that $
-	(\mathcal{H}_{k-1}) $ is true. We will try to prove $ (\mathcal{H}_{k}) $.
-
-\[
-	P(S_{k-1} + U_{k-1} < a)
-	= \iint_{\cal{D}} f_{S_{k-1}, U_{k-1}}(x, y) \cdot (x + y) dxdy                 \\
-\]
-where $ \mathcal{D} = \{ (x, y) \in [0, 1]^2 \mid x + y < a \} $.
-As $ S_{k-1} $ and $ U_{k-1} $ are independent,
-\[
-	P(S_{k-1} + U_{k-1} < a)
-	= \iint_{\cal{D}} f_{S_{k-1}}(x) \cdot f_{U_{k-1}}(y) \cdot (x + y) dxdy \qquad \\
-\]
-
 $ (\mathcal{H}_{k-1}) $ gives us that $ \forall x \in [0, 1] $,
 $ F_{S_{k-1}}(x) = P(S_{k-1} < x) = \frac{x^{k-1}}{(k-1)!} $.
 
@@ -336,15 +369,25 @@ By differentiating, we get that $ \forall x \in [0, 1] $,
 \]
 
 Furthermore, $ U_{k-1} $ is uniformly distributed on $ [0, 1] $, so
-$ f_{U_{k-1}}(y) = 1 $.
+$ f_{U_{k-1}}(y) = 1 $. We can then integrate by parts :
 
 \begin{align*}
-	\text{Hence, }
-	P(S_{k-1} + U_{k-1} < a)
-	 & =
-	 & = \frac{a^{k}}{k!}
+	P(S_k < a)
+	 & = \int_{0}^{a} f_{S_{k-1}}(x) \cdot \left( \int_{0}^{a-x} 1 dy \right) dx   \\
+	 & = \int_{0}^{a} f_{S_{k-1}}(x) \cdot (a - x) dx                              \\
+	 & = a \int_{0}^{a} f_{S_{k-1}}(x) dx - \int_{0}^{a} x f_{S_{k-1}}(x) dx       \\
+	 & = a \int_0^a F'_{S_{k-1}}(x) dx - \left[ x F_{S_{k-1}}(x) \right]_0^a
+	+ \int_{0}^{a} x F_{S_{k-1}}(x) dx \qquad \text{(IPP)}                         \\
+	 & = a \left[ F_{S_{k-1}}(x) \right]_0^a - \left[ x F_{S_{k-1}}(x) \right]_0^a
+	+ \int_{0}^{a} \frac{x^{k-1}}{(k-1)!} dx                                       \\
+	 & = \left[ \frac{x^k}{k!} \right]_0^a                                         \\
+	 & = \frac{a^k}{k!}                                                            \\
 \end{align*}
 
+We have shown that $ (\mathcal{H}_{k}) $ is true, so by induction, $ \forall k \geq 1 $,
+$ \forall a \in [0, 1] $, $ P(U_1 + U_2 + \ldots + U_{k} < a) = \frac{a^k}{k!} $. Take
+$ a = 1 $ to get $ P(U_1 + U_2 + \ldots + U_{k} < 1) = \frac{1}{k!} $.
+