227 lines
6.9 KiB
TeX
227 lines
6.9 KiB
TeX
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\documentclass[main.tex]{subfiles}
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\begin{document}
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\begin{enumerate}
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\item À l'extérieur du carré, $f_{XY}(x,y) = 0$. À l'intérieur, le couple ($X,Y$) est uniformément réparti donc :
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\[ f_{XY}(x,y) =
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\left\{
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\begin{array}{ll}
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\frac{1}{a^2} & \si x\in[0,a[, y\in[0,a[ \\
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0 & \sinon
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\end{array}
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\right.
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\]
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$f_{XY}(x,y) = g_X(x)g_Y(x)$, $f_{XY}$ est séparable donc $X$ et $Y$ sont indépendantes.
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\item On considère la VA $Z = X + Y$. Comme $X\in[0,a[$ et $Y\in[0,a[$, $Z\in[0,2a[$
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Calculons la fonction de répartition de la VA Z.
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\begin{align*}
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F_Z(z) & =
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\left\{
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\begin{array}{ll}
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1 & \si z > 2a \\
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? & \si z \in [0,2a] \\
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0 & \si z < 0
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\end{array}
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\right.\\
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\intertext{Si $z\in[0,2a[$, l'expression de la fonction de répartition n'est pas immédiate :}
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F_Z(z) & = P[Z<z] \\
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& = P[(X,Y) \in D_z] \\
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& =
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\left\{
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\begin{array}{ll}
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\frac{z^2/2}{a^2} & \si z\in[0,a] \\
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\frac{a^2-\frac{(2a-z)^2}{2}}{a^2} & \si z\in[a,2a]
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\end{array}
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\right.
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\end{align*}
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\begin{center}
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\begin{tikzpicture}[scale=0.5]
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\draw [>=latex,->] (-1,0) -- (7,0) node[right]{$x$} ;
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\draw [>=latex,->] (0,-1) -- (0,7) node[left]{$y$};
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\draw (3,0) node[below]{$a$} -- (3,3) -- (0,3) node[left]{$a$};
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\draw [dashed,blue] (-1,2) node[left]{$D_z, z\in[0,a[$} -- (2,-1);
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\draw (0,6) node[left]{$2a$} -- (6,0) node[below]{$2a$};
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\fill [color=blue] (0,1) -- (1,0) -- (0,0);
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\draw [>=latex,->] (9,0) -- (17,0) node[right]{$x$} ;
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\draw [>=latex,->] (10,-1) -- (10,7) node[left]{$y$};
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\draw (13,0) node[below]{$a$} -- (13,3) -- (10,3) node[left]{$a$};
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\draw [dashed,red] (9,5) node[left]{$D_z, z\in[a,2a[$} -- (15,-1);
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\draw (10,6) node[left]{$2a$} -- (16,0) node[below]{$2a$};
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\fill [color=red] (10,3) -- (11,3) -- (13,1) -- (13,0) -- (10,0);
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\end{tikzpicture}
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\end{center}
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On peut donc résumer les résultats comme suit :
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\begin{multicols}{2}
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\[F_Z(z) =
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\left\{
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\begin{array}{ll}
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0 & \si z < 0 \\
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\frac{z^2/2}{a^2} & \si z\in[0,a] \\
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\frac{a^2-\frac{(2a-z)^2}{2}}{a^2} & \si z\in[a,2a] \\
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1 & \si z > 2a
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\end{array}
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\right.
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\]
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\[f_Z(z) =
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\left\{
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\begin{array}{ll}
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0 & \si z < 0 \text{ ou } z > 2a \\
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\frac{z}{a^2} & \si z\in[0,a] \\
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\frac{2a-z}{a^2} & \si z\in[a,2a]
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\end{array}
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\right.
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\]
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\end{multicols}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\draw [>=latex,<->] (5,0) node[right]{$z$} -- (0,0) -- (0,2) node[left]{$f_Z(z)$};
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\draw (0,0) node[below]{$0$} -- (2,1) -- (4,0) node[below]{$2a$};
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\draw [dashed] (0,1) node[left]{$1/a$} -- (2,1) -- (2,0) node[below]{$a$};
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\end{tikzpicture}
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\end{center}
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\item On commence par expliciter $F_Z(z)$ en fonction de $f_{XY}(x,y)$ :
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\begin{align*}
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F_Z(z) & = P[Z<z] = \int_{-\infty}^z f_Z(w)dw \\
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& = P[(X,Y) \in D_z] \\
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& = \int \int_{D_z} f_{XY}(x,y)dxdy \\
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F_Z(z) & = \int_ {-\infty}^{+\infty} ( \int_{-\infty}^{z-x} f_{XY}(x,y)dy)dx
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\end{align*}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\fill [color=red!10] (4.5,-2.5) -- (-2,4) -- (-2,-2.5);
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\draw [>=latex,->] (-2,0) -- (5,0) node[right]{$x$} ;
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\draw [>=latex,->] (0,-2) -- (0,3) node[left]{$y$};
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\draw [dashed] (-1,3) -- (4,-2);
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\draw [>=latex,->,red] (3,-2.5) -- (3,-1.5);
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\draw [red] (3,-1.5) -- (3,-1) ;
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\draw [dashed, red] (0,-1) node[left]{$y=z-x$} -- (3,-1) -- (3,0) node[above]{$x$};
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\end{tikzpicture}
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\end{center}
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On en déduit $f_Z(z)$
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\begin{align*}
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f_Z(z) & = \frac{dF_Z(z)}{dz} = \frac{d}{dz} \int_ {-\infty}^{+\infty} ( \int_{-\infty}^{z-x} f_{XY}(x,y)dy)dx \\
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& = \int_ {-\infty}^{+\infty} \frac{\partial}{\partial z} ( \int_{-\infty}^{z-x} f_{XY}(x,y)dy)dx \\
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f_Z(z) & = \int_ {-\infty}^{+\infty} f_{XY}(x,z-x)dx
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\end{align*}
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\item Les VA $X$ et $Y$ indépendantes donc la ddp $f_{XY}(x,y)$ est séparable :
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\begin{align*}
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f_Z(z) & = \int_ {-\infty}^{+\infty} f_{XY}(x,z-x)dx \\
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& = \int_{-\infty}^{+\infty} f_X(x)f_Y(z-x)dx \\
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f_Z(z) & = (f_X * f_Y)(z)
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\end{align*}
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\item Par définition de la fonction caractéristique de la VA $Z$ :
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\begin{align*}
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\phi_Z(u) & = E[e^{juZ}] \\
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& = \int_{\mathbb{R}} f_Z(z) e^{juz} dz \\
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\intertext{Ainsi, on peut réécrire }
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\phi_Z(u) & = TF[f_Z(z)]_{f=-\frac{u}{2\pi}} \\
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& = E[e^{ju(X+Y)}] = E[e^{juX}e^{juY}] \\
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\intertext{Et par indépendance de $X$ et $Y$,}
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\phi_Z(u) & = \phi_X(u)\phi_Y(u) \\
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f_Z(z) & = TF^{-1}[\phi_Z(u)](z) \\
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& = (TF^{-1}[\phi_X(u)] * TF^{-1}[\phi_Y(u)])(z) \\
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f_Z(z) & =(f_X * f_Y)(z)
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\end{align*}
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\item $X_1,...X_n$ indépendantes dans leur ensemble
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\begin{eqnarray*}
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Y_{12} = & X_1 + X_2 & \rightarrow f_{Y_{12}}(y) = (f_{X_1}*f_{X_2})(y) \\
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Y_{123} = & X_1 + X_2 + X_3 = Y_{12} + X_3 & \rightarrow f_{Y_{123}}(y) = (f_{Y_{12}} * f_{X_3})(y) = (f_{X_1}*f_{X_2}*f_{X_3})(y)
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\end{eqnarray*}
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Par récurrence, on montre alors que pour $Y = X_1 + ... + X_n$,
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\[f_Y(y) = (f_{X_1}* ... * f_{X_n})(y) \]
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On montre que pour $X_n, n=1,...,N$ VA réelles et scalaires indépendantes et identiquement distribuées, centrées et d'écart-type $\sigma$,
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\[Z_N = \frac{\sum_{n=1}^N X_n}{\sqrt{N}} \text{ tend vers une VA gaussienne quand N tend vers } +\infty \]
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\item Par linéarité de l'espérance, et comme les variables $X_N$ sont centrées ($E[X_N] = 0$),
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\[ E[Z_N] = E[\frac{\sum_{n=1}^N X_n}{\sqrt{N}}] = \frac{\sum_{n=1}^N E[X_n]}{\sqrt{N}} = 0 \]
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De plus,
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\begin{align*}
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\sigma_Z^2 & = E[(Z_N - m_{Z_N})^2] = E[Z_N^2] \\
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& = \frac{1}{N} E[(\sum_{n=1}^N X_n)^2] \\
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& = \frac{1}{N} E[ \sum_{n=1}^N X_n^2 + \sum_{i\neq j} X_iX_j] \\
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& = \frac{1}{N} (\sum_ {n=1}^N E[X_n^2] + \sum_{i\neq j} E[X_iX_j]) \\
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& = \frac{1}{N} \sum_{n=1}^N \sigma^2 \\
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& = \sigma^2
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\end{align*}
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\newpage
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\item Deux personnes se donnent rendez-vous entre 18h et 19h. On associe aux deux instants d'arrivées deux VA X et Y indépendantes, de ddp uniforme sur l'intervalle [18,19].
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On introduit la VA $\Delta = |Y-X|$. Calculons sa fonction de répartition.
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\begin{align*}
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F_{\Delta}(\delta) & = P[\Delta \leq \delta] \\
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& = P[ |Y-X| \leq \delta ] \\
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& = P[ Y-X \leq \delta \et X-Y \leq \delta ] \\
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& = P[ Y \leq X + \delta \et Y \geq X - \delta ]
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\end{align*}
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\begin{figure}[h!]
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\centering
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\begin{tikzpicture}[scale=0.8]
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\draw [>=latex,->] (-1,0) -- (7,0) node[right]{$x$} ;
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\draw [>=latex,->] (0,-1) -- (0,7) node[left]{$y$};
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\draw [dashed] (1,0) node[below]{$18$} -- (1,7);
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\draw [dashed] (5,0) node[below]{$19$} -- (5,7);
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\draw [dashed] (0,1) node[left]{$18$} -- (7,1);
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\draw [dashed] (0,5) node[left]{$19$} -- (7,5) ;
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\draw [blue] (0,1.5) -- (6,7.5) node[right]{$y=x+\delta$};
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\draw [blue] (1.5,0) -- (7.5,6) node[right]{$y=x-\delta$};
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\fill [color=blue!20] (1,1) -- (1,2.5) -- (3.5,5) -- (5,5) -- (5,3.5) -- (2.5,1) -- (1,1);
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\draw [dashed] (2.5,-1) -- (2.5,1);
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\draw [<->] (1,-1) -- (2.5,-1);
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\draw (1.75,-1) node[below]{$\delta$};
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\draw [<->] (5,-1) -- (2.5,-1);
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\draw (3.75,-1) node[below]{$1-\delta$};
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\end{tikzpicture}
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\end{figure}
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Ainsi, \[F_{\Delta}(\delta) =
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\left\{
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\begin{array}{ll}
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0 & \si \delta < 0 \\
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1 - (1-\delta)^2 & \si 0 \leq \delta < 1\\
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1 & \si \delta \geq 1
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\end{array}
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\right.
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\]
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Donc \[f_{\Delta}(\delta) =
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\left\{
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\begin{array}{ll}
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2 - 2 \delta & \si 0 \leq \delta < 1\\
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0 & \sinon
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\end{array}
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\right.
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\]
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\end{enumerate}
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\end{document}
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