Si \(x > a\)
\[ \text{ on a : } \mathbb{E}[N(a) \mid X_1 = x] = \mathbb{E}[1] = 1. \]
Si \(x \leq a\) \[ \quad (N(a) \mid X_1 = x) = \min\{n \geq 1 \mid X_1 + X_2 + \dots + X_n > a\} \\ \] \[ = \min\{n \geq 1 \mid X_2 + \dots + X_n > a - x\} \\ \] \[ = 1 + \min\{n' \geq 1 \mid X'_1 + \dots + X'_{n'} > a - x\} \\ = 1 + N(a - x). \]
D’où : \[ \mathbb{E}[N(A) \mid X_1 = x] = \mathbb{E}[1] = 1 + \mathbb{1}_{x<a}\mathbb{E}[N(a-x)] \]
\[ \mathbb{E}[N(a)] = \int_{0}^{1} \mathbb{E}[N(a) \mid X_1 = x] f_{X_1}(x) dx \]
\[ = \int_{0}^{a} \mathbb{E}[N(a) \mid X_1 = x] dx + \int_{a}^{1} \mathbb{E}[N(a) \mid X_1 = x] dx = (1-a) + \int_{0}^{a} 1 + \mathbb{E}[N(a-x)] dx \]
\[ = 1 + \int_{0}^{a} \mathbb{E}[N(a-x)] dx = 1 + \int_{0}^{a} \mathbb{E}[N(x)] dx \]
` ### Question 3
\[ \text{On pose } f(a) = \mathbb{E}[N(a)] . \]
\[ \text{Donc, en dérivant : } f'(a) = f(a) \] \[ \text{Alors : } f(a) = \lambda e^{a} \]
\[ \text{Or } \lambda = \mathbb{E}[N(0)] = 1 \]
\[ \text{Ainsi: } f(a) = e^{a} \]
\[ \mathbb{E}[N_{n+1} \mid U_1 = u] = \begin{cases} 1 + f_n(u) & \text{si } u > x, \\ f_n(x) & \text{si } u \leq x. \end{cases} \]
On a: \[ 1 - f_{n+1}(x) = 1 - \int_0^1 \mathbb{E}[N_{n+1} \mid U_1 = u] \, du \]
\[ = 1 - \int_0^x f_n(x) \, du - \int_x^1 1 + f_n(u) \, du \]
\[ = 1 - x f_n(x) - (1-x) - \int_x^1 f_n(u) \, du \]
\[ = x \cdot (1 - f_n(x)) - \int_x^1 f_n(u) \, du \]
En dérivant l’égalité précédente, on trouve:
\[ - g_{n+1}(x) = 1 - f_n(x) - x \cdot g_n(x) + f_n(x) \]
\[ g_{n+1}(x) = x \cdot g_n(x) - 1 \]
Par récurrence, on peut montrer que pour tout \(n \geq 2\) : \[ g_{n}(x) = - \sum_{i=0}^{n-1}x^i \]
D’où \[ f_{n}(x) = C - \sum_{i=1}^{n}\frac{x^i}{i} \]
La valeur de \(f_n(x)\) en \(x = 1\) et \(n = 1\) donne la valeur de \(C\).