Todai Entrance Exam: Math 2017 – Problem 3

(1)

$\mathbb{E}(T) = \left ( \int_{0}^{\infty} t\lambda e^{-\lambda t}dt + \int_{-\infty}^{0} t0dt\right ) = -e^{-\lambda t}\left ( \frac{1}{\lambda} + t \right ) \Big|^{\infty}_{0} = \frac{1}{\lambda}$

$F(t') = \int_{-\infty}^{t'} f(t)dt \Leftrightarrow \begin{Bmatrix} F(t') = 0 & \text{ if } & t' < 0\\ F(t') = \int_{0}^{t'} \lambda e^{-\lambda t}dt = -e^{-\lambda t} \big |^{t'}_{0} = 1 - e^{-\lambda t'} & \text{ if } & t' \geq 0 \end{matrix}$

(2)

$\mathbb{P}(T > s + t | T > s) = \frac{\mathbb{P}((T > s + t) \cap (T > s))}{\mathbb{P}(T > s)} = \frac{\mathbb{P}(T > s + t)}{\mathbb{P}(T > s)} = \frac{1 - F(s+t)}{1 - F(s)} = \frac{e^{-\lambda (s + t)}}{e^{-\lambda s}} = e^{-\lambda t} = 1 - F(t) = \mathbb{P}(T > t)$

(3)

$F_{\text{earliest}}(t) = \mathbb{P}(T_{\text{earliest}} \leq t)(\mathbb{P}(T_{\text{others}} > t))^{n-1} = F(t)(1 - F(t))^{n-1} = (1 - e^{-\lambda t} )(e^{-\lambda t})^{n-1} = e^{-\lambda (n-1) t} - e^{-\lambda n t}$

$f_{\text{earliest}}(t) = \frac{\mathrm{d} F}{\mathrm{d} t} = \lambda ne^{-\lambda n t} - \lambda (n-1)e^{-\lambda (n-1) t}$

$\mathbb{E}(T_\text{earliest}) = \int_{0}^{\infty} tf_{\text{earliest}}(t)dt = \int_{0}^{\infty} t(\lambda ne^{-\lambda n t} - \lambda (n-1)e^{-\lambda (n-1) t})dt$

$\mathbb{E}(T_\text{earliest}) = e^{-\lambda (n-1) t}\left ( \frac{1}{\lambda (n-1)} + t \right ) \Big|^{\infty}{0} - e^{-\lambda n t}\left ( \frac{1}{\lambda n} + t \right ) \Big|^{\infty}{0} = \frac{1}{\lambda n} - \frac{1}{\lambda (n-1)}$

(4)

$\mathbb{P}(T_{A} < T_{B}) = \mathbb{P}(T_{A} \leq t)\mathbb{P}(T_{B} > t) = F_{\lambda_{A}}(t)(1 - F_{\lambda_{B}}(t)) = (1 - e^{-\lambda_{A} t})e^{-\lambda_{B} t} = e^{-\lambda_{B} t} - e^{-(\lambda_{A} + \lambda_{B})t}$

(5)

From (1), we have:

$10\mathbb{E}(T_\text{Hideo}) = \mathbb{E}(T_\text{others}) \Leftrightarrow \frac{10}{\lambda_{\text{Hideo}}} = \frac{1}{\lambda_{\text{others}}} \Leftrightarrow \lambda_{\text{Hideo}} = 10\lambda_{\text{others}} = 10\lambda$

$\mathbb{P}(\text{Hideo finishes the problem first}) = \mathbb{P}(T_{\text{Hideo}} \leq t)(\mathbb{P}(T_{\text{others}} > t))^{10} = F_{\text{Hideo}}(t)(1 - F_{\text{others}}(t))^{10} = (1 - e^{-10\lambda t} )(e^{-\lambda t})^{10} = e^{-10\lambda t} - e^{-20\lambda t}$

$\mathbb{P}(\text{Hideo finishes the problem fourth}) = (\mathbb{P}(T_{\text{others}} < t))^{3}\mathbb{P}(T_{\text{Hideo}} = t)(\mathbb{P}(T_{\text{others}} > t))^{7} = (F_{\text{others}}(t) - f_{\text{others}}(t))^{3}f_{\text{Hideo}}(t)(1 - F_{\text{others}}(t))^{7}$

$\mathbb{P}(\text{Hideo finishes the problem fourth}) = (1 - e^{-\lambda t} - \lambda e^{-\lambda t})^{3}\lambda e^{-10\lambda t}(e^{-\lambda t})^{7} = (1 - e^{-\lambda t} - \lambda e^{-\lambda t})^{3}\lambda e^{-17\lambda t}$

$\mathbb{P}(\text{Hideo finishes the problem fourth}) = \lambda e^{-17\lambda t} - 3\lambda(1 + \lambda)e^{-18\lambda t} - (1 + \lambda)^{2}e^{-19\lambda t} - \lambda (1 + \lambda)^{3}e^{-20\lambda t}$

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