Open thyecust opened 2 years ago
对于标准正态分布 $Z\sim N(0, 1)$
$$ \begin{align} M_Z(t) &= E[e^{tZ}] = \int e^{tz} \frac 1 {\sqrt{2\pi}} e^{- \frac12z^2} dz \ &= \frac 1 {\sqrt{2\pi}} \int e^{tz - \frac12 z^2}dz \ &= \frac 1 {\sqrt{2\pi}} \int e^{\frac 12 t^2 - \frac12 (z-t)^2}dz \ &= \frac {e^{\frac12 t^2}} {\sqrt{2\pi}} \int e^{- \frac12 (z-t)^2}dz \ &= \exp(\frac12 t^2) \end{align} $$
对于正态分布 $X\sim N(\mu, \sigma^2)$
$$ M_X(t) = E[e^{tX}] = \exp(\mu t+\frac 12 \sigma^2t^2) $$
对于混合高斯分布 $I\sim \mathrm{Ber}(p), S=IX+(1-I)Y$
$$ \begin{align} M_S(t) &= E[\exp(tS)] = E[\exp(tIX+t(1-I)Y)] \ &= E[\exp(tIX+t(1-I)Y) | I=1]P(I=1)+E[\exp(tIX+t(1-I)Y) | I=0]P(I=0) \ &= pE[\exp(tX)]+(1-p)E[\exp(tY)] \ &= pM_X(t)+(1-p)M_Y(t) \end{align} $$
Normal Distribution MGF
对于标准正态分布 $Z\sim N(0, 1)$
$$ \begin{align} M_Z(t) &= E[e^{tZ}] = \int e^{tz} \frac 1 {\sqrt{2\pi}} e^{- \frac12z^2} dz \ &= \frac 1 {\sqrt{2\pi}} \int e^{tz - \frac12 z^2}dz \ &= \frac 1 {\sqrt{2\pi}} \int e^{\frac 12 t^2 - \frac12 (z-t)^2}dz \ &= \frac {e^{\frac12 t^2}} {\sqrt{2\pi}} \int e^{- \frac12 (z-t)^2}dz \ &= \exp(\frac12 t^2) \end{align} $$
对于正态分布 $X\sim N(\mu, \sigma^2)$
$$ M_X(t) = E[e^{tX}] = \exp(\mu t+\frac 12 \sigma^2t^2) $$
对于混合高斯分布 $I\sim \mathrm{Ber}(p), S=IX+(1-I)Y$
$$ \begin{align} M_S(t) &= E[\exp(tS)] = E[\exp(tIX+t(1-I)Y)] \ &= E[\exp(tIX+t(1-I)Y) | I=1]P(I=1)+E[\exp(tIX+t(1-I)Y) | I=0]P(I=0) \ &= pE[\exp(tX)]+(1-p)E[\exp(tY)] \ &= pM_X(t)+(1-p)M_Y(t) \end{align} $$