지수분포

\begin{align*} P~(-\infty \le X \le \infty) &= \int_{-\infty}^{\infty}f_X(x)dx = \int_{-\infty}^{\infty} \lambda e^{-\lambda x}~ I(0 < x)~dx \\[20pt] &=-e^{-\lambda x} \biggr\rvert_{0}^{\infty} = 0+1 = 1 \end{align*}

\begin{align*} E(X) &= \int_{-\infty}^{\infty} x\cdot f_X(x)~dx = \int_{-\infty}^{\infty} x\cdot\lambda e^{-\lambda x} ~I(0 < x)~dx \\[20pt] &= -xe^{-\lambda x}~ \bigg\rvert_{0}^{\infty} ~+~ \int_{-\infty}^{\infty} e^{-\lambda x}~dx \\[20pt] &= \Bigg( \lim\limits_{x \to \infty} \dfrac{x}{e^{\lambda x}} - 0 \Bigg) ~+~ -\dfrac{1}{\lambda}~ e^{-\lambda x}~ \bigg\rvert_{0}^{\infty} \\[20pt] &= \lim\limits_{x \to \infty} \dfrac{1}{\lambda e^{\lambda x}} + \dfrac{1}{\lambda} \\[20pt] &= \dfrac{1}{\lambda} \end{align*}

\begin{align*} E(X^2) &= \int_{-\infty}^{\infty} x^2\cdot f_X(x)~dx = \int_{-\infty}^{\infty} x^2\cdot\lambda e^{-\lambda x} ~I(0 < x)~dx \\[20pt] &= -x^2e^{-\lambda x}~ \bigg\rvert_{0}^{\infty} ~+~ \int_{-\infty}^{\infty} 2xe^{-\lambda x}~dx \\[20pt] &= \lim\limits_{x \to \infty} \dfrac{-x^2} {e^{-\lambda x}} ~+~ \dfrac {2}{\lambda}~ \int_{-\infty}^{\infty} xe^{-\lambda x}~dx \\[20pt] &= \lim\limits_{x \to \infty} \dfrac{-2x} {\lambda e^{-\lambda x}} + \dfrac{2}{\lambda} \cdot \dfrac{1}{\lambda} \\[20pt] &= \dfrac{2}{\lambda^2} \end{align*}

\begin{align*} Var(X) &= E(X^2) - E(X^2) \\[10pt] &= \dfrac{2}{\lambda^2} - \bigg( \dfrac{1}{\lambda} \bigg)^2 \\[15pt] &= \dfrac{1}{\lambda^2} \end{align*}

\begin{align*} M_X(x) &= \int_{-\infty}^{\infty} e^{tx}\cdot f_X(x)~dx = \int_{-\infty}^{\infty} e^{tx}\cdot \lambda e^{-\lambda x} ~dx \\[20pt] &= \lambda\cdot \int_{-\infty}^{\infty}~ e^{-(\lambda-t)x}~dx \\[20pt] &= \dfrac{\lambda}{\lambda-t} \cdot \int_{-\infty}^{\infty}~ (\lambda-t) e^{-(\lambda-t)x}~dx \\[20pt] &= \dfrac{\lambda}{\lambda-t} ~=~\dfrac{\lambda-t}{\lambda} \\[20pt] M_X(x) &= \bigg( 1 - \dfrac{t}{\lambda} \bigg)^{-1} \end{align*}

\begin{align*} M_X(x) &= \bigg( 1 - \dfrac{t}{\lambda} \bigg)^{-1} \\[20pt] \log M_X(x) &= -\log\bigg( 1 - \dfrac{t}{\lambda} \bigg) = -\log(t - \lambda) + \log(\lambda) \\[20pt] \dfrac{d}{dt}~\log M_X(x) &= ~\dfrac{1}{\lambda-t} ~\biggr\rvert_{t=0} = \dfrac{1}{\lambda} \\[20pt] \dfrac{d^2}{dt^2}~\log M_X(x) &= ~\dfrac{1}{(\lambda-t)^2} ~\biggr\rvert_{t=0} = \dfrac{1}{\lambda^2} \end{align*}