From 07b4bec23e8755afa83cc44421caeca10847de00 Mon Sep 17 00:00:00 2001 From: Paul ALNET Date: Sun, 4 Jun 2023 19:37:02 +0200 Subject: [PATCH] tex: NFDBP demo ! --- latex/content.tex | 78 ++++++++++++++++++----------------------------- 1 file changed, 30 insertions(+), 48 deletions(-) diff --git a/latex/content.tex b/latex/content.tex index 0e3e7e3..79e1e28 100644 --- a/latex/content.tex +++ b/latex/content.tex @@ -323,7 +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 @@ -332,61 +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)!} $. @@ -397,15 +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!} $. +