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tex: finalize NFDBP demo

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Paul ALNET 2023-06-04 21:20:38 +02:00
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@ -312,25 +312,25 @@ new constraints on the first bin can be expressed as follows :
\subsection{La giga demo} \subsection{La giga demo}
Let $ k \in \mathbb{N} $. Let $ (U_n)_{n \in \mathbb{N}} $ be a sequence of Let $ k \geq 2 $. Let $ (U_n)_{n \in \mathbb{N}^*} $ be a sequence of
independent random variables with uniform distribution on $ [0, 1] $, representing independent random variables with uniform distribution on $ [0, 1] $, representing
the size of the $ n $-th item. the size of the $ n $-th item.
Let $ i \in \mathbb{N} $. $ T_i $ denotes the number of items in the $ i $-th Let $ i \in \mathbb{N} $. $ T_i $ denotes the number of items in the $ i $-th
bin. We have that bin. We have that
\begin{equation} \begin{equation*}
T_i = k \iff U_1 + U_2 + \ldots + U_{k-1} < 1 \text{ and } U_1 + U_2 + \ldots + U_{k} \geq 1 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} \end{equation*}
Let $ A_k = \{ U_1 + U_2 + \ldots + U_{k} < 1 \}$. Hence, Let $ A_k = \{ U_1 + U_2 + \ldots + U_{k} < 1 \}$. Hence,
\begin{align*} \begin{align}
% TODO = k \label{eq:prob}
P(T_i = k) P(T_i = k)
& = P(A_{k-1} \cap A_k^c) \\ & = P(A_{k-1} \cap A_k^c) \\
& = P(A_{k-1}) - P(A_k) \qquad \text{ (as $ A_k \subset A_{k-1} $)} \\ & = P(A_{k-1}) - P(A_k) \qquad \text{ (as $ A_k \subset A_{k-1} $)} \\
\end{align*} \end{align}
We will try to show that $ \forall k \geq 1 $, $ 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}, so, we will use induction to prove the following proposition \eqref{eq:induction},
@ -377,16 +377,25 @@ $ f_{U_{k-1}}(y) = 1 $. We can then integrate by parts :
& = \int_{0}^{a} f_{S_{k-1}}(x) \cdot (a - x) 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 - \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 & = 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)} \\ + \int_{0}^{a} F_{S_{k-1}}(x) dx \qquad \text{(IPP : }x, F_{S_{k-1}} \in C^1([0,1]) \\
& = a \left[ F_{S_{k-1}}(x) \right]_0^a - \left[ x F_{S_{k-1}}(x) \right]_0^a & = 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 \\ + \int_{0}^{a} \frac{x^{k-1}}{(k-1)!} dx \\
& = \left[ \frac{x^k}{k!} \right]_0^a \\ & = \left[ \frac{x^k}{k!} \right]_0^a \\
& = \frac{a^k}{k!} \\ & = \frac{a^k}{k!} \\
\end{align*} \end{align*}
We have shown that $ (\mathcal{H}_{k}) $ is true, so by induction, $ \forall k \geq 1 $, \paragraph{Conclusion} 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 $ \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!} $. $ a = 1 $ to get
\[ P(U_1 + U_2 + \ldots + U_{k} < 1) = \frac{1}{k!} \]
Finally, plugging this into \eqref{eq:prob} gives us
\[
P(T_i = k) = P(A_{k-1}) - P(A_{k}) = \frac{1}{(k-1)!} - \frac{1}{k!} \qquad \forall k \geq 2
\]