\begin{document}

\maketitle

\section{Calcul de lois}

Question 1 :
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X a pour densité sur \mathbb{R} , f\textsubscript{X}(x) = \lambda e \up{$-(\lambda x$ + e\up{- $$\lambda$ x})}
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\Phi (x) = 3x+2
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Z = \Phi (x)
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Pour ~ trouver ~la ~ densité ~ de ~ Z ~ on ~ utilise ~ la ~ fonction ~ de ~
répartition
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F\textsubscript{Z}(x) = P(Z \leq t)
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F\textsubscript{Z}(x) = P(\Phi (x) \leq t)
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F\textsubscript{Z}(x) = P(3X+2 \leq t)
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F\textsubscript{Z}(x) = P(X \leq \frac{t-2}{3})
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F\textsubscript{Z}(x) = $\displaystyle \int_{- \infty}^{t} \lambda e\up{$-(\lambda$ ( $\frac{x-2}{3}$) +e\up{$-\lambda$ ($\frac{x-2}{3}$)}}   \mathrm{d}x$
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F\textsubscript{Z}(x) = \lambda $\displaystyle \int_{- \infty}^{t}  e\up{$-\lambda$ ( $\frac{x-2}{3}$) . e\up{$-\lambda$ ($\frac{x-2}{3}$)}}   \mathrm{d}x$
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F\textsubscript{Z}(x) = 3 $\displaystyle \int_{- \infty}^{t} $\frac{\lambda}{3}$ e\up{$-\lambda$ ( $\frac{x-2}{3}$) . e\up{$-\lambda$ ($\frac{x-2}{3}$)}}   \mathrm{d}x$
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F\textsubscript{Z}(x) = 3 [e\up{-e\up{$-\lambda(\frac{x-2}{3})$}}]_- \infty^t
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F\textsubscript{Z}(x) = 3 (e\up{-e\up{$-\lambda(\frac{t-2}{3})$}})
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F\textsubscript{Z}(x) = 3 (e\up{-e\up{$(\frac{14-7t}{3})$}})
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f\textsubscript{Z}(x) = F\textsubscript{Z}'(x)
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f\textsubscript{Z}(x) = \frac{7}{3}.e\up{$(\frac{14-7t}{3})$}} . 3(e\up{-e\up{$(\frac{14-7t}{3})$}})
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f\textsubscript{Z}(x)= 7e\up{$(\frac{14-7t}{3}$-e\up{$(\frac{14-7t}{3})$}})  \\
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 Question 2 :
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F\textsubscript{Z}(t) = 3 (e\up{-e\up{$-\lambda(\frac{t-2}{3})$}})
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f\textsubscript{Z}(t) = F\textsubscript{Z}'(t) = \lambda e\up{$-\lambda$ ( $\frac{t-2}{3}$) .e\up{-e\up{$-\lambda(\frac{t-2}{3})$}}}
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Donc $\displaystyle \int_{- \infty}^{t} f\textsubscript{Z}(x) ~ \mathrm{d}x$ = 0,5
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\Leftrightarrow  3 (e\up{-e\up{$-\lambda(\frac{t-2}{3})$}}) = 0,5
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\Leftrightarrow -e\up{$-\lambda(\frac{t-2}{3})$}} = ln(\frac{1}{6})
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\Leftrightarrow -\lambda (\frac{t-2}{3}) = ln(-ln(\frac{1}{6}))
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\Leftrightarrow - \frac{\lambda t}{3} + \frac{2 \lambda}{3} = ln(ln(6))
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\Leftrightarrow - \frac{7}{3} t = ln(ln(6)) - \frac{14}{3}
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\Leftrightarrow t = - \frac{3}{7} ln(ln(6)) + 2
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\Leftrightarrow t \simeq 1,750
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Question 3 :
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X \texttildelow U([0,\lambda])

\section{Calcul d'Esperance}

Question 1 :
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X \Rightarrow G(p) ~ avec ~ p $ \in $ ]0;1[
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$ P(X=k) = p(1-p)\up{k-1} $
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 E[e\up{-tx}] = \sum_{k=1}^\infty e\up{-tk} ~ . ~ p(1-p)\up{k-1}
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 E[e\up{-tx}] = \frac{p}{1-p} \sum_{k=1}^\infty e\up{-tk} ~ . ~ p(1-p)\up{k}
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 E[e\up{-tx}] = \frac{p}{1-p} \sum_{k=0}^\infty (e\up{-t} (1-p))\up{k+1}
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 E[e\up{-tx}] = \frac{p}{1-p} e\up{-t} (1-p) \sum_{k=0}^\infty (e\up{-t} (1-p))\up{k}
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 E[e\up{-tx}] = pe\up{-t} . \frac{1}{1-e\up{-t}(1-p)}
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 E[e\up{-tx}] = \frac{p}{e\up{t}-(1-p)}
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 Question 2 :
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E[e\up{-t(x+y)}] = E[e\up{-tx}.e\up{-ty}]
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E[e\up{-t(x+y)}] = E[e\up{-tx}].E[e\up{-ty}] ~ ~ car X et Y sont independants
\end{document}