tex: NFDBP demo !
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Paul ALNET
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1 changed files with 30 additions and 48 deletions
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@ -323,7 +323,7 @@ bin. We have that
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T_i = k \iff U_1 + U_2 + \ldots + U_{k-1} < 1 \text{ and } U_1 + U_2 + \ldots + U_{k} \geq 1
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\end{equation}
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Let $ A_k = \{ U_1 + U_2 + \ldots + U_{k-1} < 1 \}$. Hence,
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Let $ A_k = \{ U_1 + U_2 + \ldots + U_{k} < 1 \}$. Hence,
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\begin{align*}
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% TODO = k
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@ -332,61 +332,33 @@ Let $ A_k = \{ U_1 + U_2 + \ldots + U_{k-1} < 1 \}$. Hence,
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& = P(A_{k-1}) - P(A_k) \qquad \text{ (as $ A_k \subset A_{k-1} $)} \\
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\end{align*}
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We will try to show that $ \forall k \geq 2 $, $ P(A_k) = \frac{1}{k!} $. To do
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We will try to show that $ \forall k \geq 1 $, $ P(A_k) = \frac{1}{k!} $. To do
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so, we will use induction to prove the following proposition \eqref{eq:induction},
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$ \forall k \geq 2 $:
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$ \forall k \geq 1 $:
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\begin{equation}
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\label{eq:induction}
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\tag{$ \mathcal{H}_k $}
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P(U_1 + U_2 + \ldots + U_{k-1} < a) = \frac{a^k}{k!} \qquad \forall a \in [0, 1],
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P(U_1 + U_2 + \ldots + U_{k} < a) = \frac{a^k}{k!} \qquad \forall a \in [0, 1],
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\end{equation}
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Let us denote $ S_k = U_1 + U_2 + \ldots + U_{k-1} \qquad \forall k \geq 2 $.
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Let us denote $ S_k = U_1 + U_2 + \ldots + U_{k} \qquad \forall k \geq 1 $.
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\paragraph{Base cases} $ k = 2 $ : $ P(U_1 < a) = a \neq \frac{a^2}{2}$ supposedly proving $ (\mathcal{H}_2) $.
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\paragraph{Base case} $ k = 1 $ : $ P(U_1 < a) = a = \frac{a^1}{1!}$, proving $ (\mathcal{H}_1) $.
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$ k = 2 $ : \[ P(U_1 + U_2 < a) = \iint_{\cal{D}} f_{U_1, U_2}(x, y) \cdot (x + y) dxdy \]
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Where $ \mathcal{D} = \{ (x, y) \in [0, 1]^2 \mid x + y < a \} $.
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$ U_1 $ and $ U_2 $ are independent, so
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\begin{align*}
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f_{U_1, U_2}(x, y) & = f_{U_1}(x) \cdot f_{U_2}(y) \\
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& = \begin{cases}
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1 & \text{if } x \in [0, 1] \text{ and } y \in [0, 1] \\
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0 & \text{otherwise} \\
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\end{cases} \\
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\end{align*}
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Hence,
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\paragraph{Induction step} Let $ k \geq 2 $. We assume $ (\mathcal{H}_{k-1}) $ is
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true. We will show that $ (\mathcal{H}_{k}) $ is true.
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\begin{align*}
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P(U_1 + U_2 < a)
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& = \iint_{\cal{D}} (x + y)dxdy \\
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& = \int_{0}^{a} \int_{0}^{a - x} (x + y) dy dx \\
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& = \int_{0}^{a} \left[ xy + \frac{y^2}{2} \right]_{y=0}^{y=a - x} dx \\
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& = \int_{0}^{a} \left( ax - x^2 + \frac{a^2}{2} - ax + \frac{x^2}{2} \right) dx \\
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& = \int_{0}^{a} \left( \frac{a^2}{2} - \frac{x^2}{2} \right) dx \\
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& = \left[ \frac{a^2 x}{2} - \frac{x^3}{6} \right]_{0}^{a} \\
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& = \frac{a^3}{2} - \frac{a^3}{6} \\
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P(S_k < a) & = P(S_{k-1} + U_k < a) \\
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& = \iint_{\cal{D}} f_{S_{k-1}, U_k}(x, y) dxdy \\
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\text{Where } \mathcal{D} & = \{ (x, y) \in [0, 1]^2 \mid x + y < a \} \\
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& = \{ (x, y) \in [0, 1]^2 \mid 0 < x < a \text{ and } 0 < y < a - x \} \\
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P(S_k < a) & = \iint_{\cal{D}} f_{S_{k-1}}(x) \cdot f_{U_k}(y) dxdy \qquad
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\text{because $ S_{k-1} $ and $ U_k $ are independent} \\
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& = \int_{0}^{a} f_{S_{k-1}}(x) \cdot \left( \int_{0}^{a-x} f_{U_k}(y) dy \right) dx \\
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\end{align*}
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\paragraph{Induction step} For a fixed $ k > 2 $, we assume that $
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(\mathcal{H}_{k-1}) $ is true. We will try to prove $ (\mathcal{H}_{k}) $.
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\[
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P(S_{k-1} + U_{k-1} < a)
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= \iint_{\cal{D}} f_{S_{k-1}, U_{k-1}}(x, y) \cdot (x + y) dxdy \\
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\]
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where $ \mathcal{D} = \{ (x, y) \in [0, 1]^2 \mid x + y < a \} $.
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As $ S_{k-1} $ and $ U_{k-1} $ are independent,
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\[
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P(S_{k-1} + U_{k-1} < a)
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= \iint_{\cal{D}} f_{S_{k-1}}(x) \cdot f_{U_{k-1}}(y) \cdot (x + y) dxdy \qquad \\
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\]
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$ (\mathcal{H}_{k-1}) $ gives us that $ \forall x \in [0, 1] $,
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$ F_{S_{k-1}}(x) = P(S_{k-1} < x) = \frac{x^{k-1}}{(k-1)!} $.
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@ -397,15 +369,25 @@ By differentiating, we get that $ \forall x \in [0, 1] $,
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\]
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Furthermore, $ U_{k-1} $ is uniformly distributed on $ [0, 1] $, so
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$ f_{U_{k-1}}(y) = 1 $.
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$ f_{U_{k-1}}(y) = 1 $. We can then integrate by parts :
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\begin{align*}
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\text{Hence, }
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P(S_{k-1} + U_{k-1} < a)
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& =
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& = \frac{a^{k}}{k!}
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P(S_k < a)
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& = \int_{0}^{a} f_{S_{k-1}}(x) \cdot \left( \int_{0}^{a-x} 1 dy \right) dx \\
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& = \int_{0}^{a} f_{S_{k-1}}(x) \cdot (a - x) dx \\
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& = a \int_{0}^{a} f_{S_{k-1}}(x) dx - \int_{0}^{a} x f_{S_{k-1}}(x) dx \\
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& = a \int_0^a F'_{S_{k-1}}(x) dx - \left[ x F_{S_{k-1}}(x) \right]_0^a
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+ \int_{0}^{a} x F_{S_{k-1}}(x) dx \qquad \text{(IPP)} \\
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& = a \left[ F_{S_{k-1}}(x) \right]_0^a - \left[ x F_{S_{k-1}}(x) \right]_0^a
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+ \int_{0}^{a} \frac{x^{k-1}}{(k-1)!} dx \\
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& = \left[ \frac{x^k}{k!} \right]_0^a \\
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& = \frac{a^k}{k!} \\
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\end{align*}
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We have shown that $ (\mathcal{H}_{k}) $ is true, so by induction, $ \forall k \geq 1 $,
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$ \forall a \in [0, 1] $, $ P(U_1 + U_2 + \ldots + U_{k} < a) = \frac{a^k}{k!} $. Take
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$ a = 1 $ to get $ P(U_1 + U_2 + \ldots + U_{k} < 1) = \frac{1}{k!} $.
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