解答
1
$$
\int_0^{\infty}f_X(x)dx = \int_0^{\infty}\lambda e^{-\lambda x}dx = \left[-e^{-\lambda x}\right]_0^{\infty} = 1
$$
2
$$\begin{aligned}
\mathbb{E}\left(X\right)
&= \int_0^{\infty}xf_X(x)dx = \int_0^{\infty}x\lambda e^{-\lambda x}dx\\
&= \int_0^{\infty}x\left(-e^{-\lambda x}\right)^{\prime}dx\\
&= \left[-xe^{-\lambda x}\right]_0^{\infty} - \int_0^{\infty}-e^{-\lambda x} dx\\
&= \left[-\frac{1}{\lambda}e^{-\lambda x}\right]_0^{\infty}\\
&= \frac{1}{\lambda}
\end{aligned}$$
3
$$
\mathbb{P}\left(X>t\right) = \int_t^{\infty}f(x)dx = \left[-e^{-\lambda x}\right]_t^{\infty} = e^{-\lambda t}
$$
4
$$\begin{aligned}
\mathbb{P}\left(X>s+t|X>s\right) &= \frac{\mathbb{P}\left(X>s+t\right)}{\mathbb{P}\left(X>s\right)}\\
&= \frac{\int_{s+t}^{\infty}f_X(x)dx}{\int_s^{\infty}f_X(x)dx}\\
&= \frac{e^{-\lambda\left(s+t\right)}}{e^{-\lambda s}}\\
&= e^{-\lambda t} = \mathbb{P}\left(X>t\right)
\end{aligned}$$