How Moment generation function of Gaussian R.V. can be divided into this one?

I know the moment generating function of Gaussian random variable is $$E\{e^{rx}\}=\int^{+\infty}_{-\infty}e^{rx}f(x)dx=e^{mr+r^2\sigma^2/2}$$ where $f(x)$ is PDF of Gaussian R.V., $m$ is mean value and $\sigma$ is standard deviation.

But in my question, some papers refer that that MGF of Gaussian R.V. can be divided into $$\int^0_{-\infty}e^{rx}f(x)dx=\left(e^{mr+r^2\sigma^2/2}\right)\Phi\left(\frac{-m-\sigma^2r}{\sigma}\right)$$ where $\Phi$ is CDF of Gaussian R.V., and I don't understand why that MGF can be divided into above equation.

I think it just have basic principle.

Thank you.

1 Answer

By definition, we have

$$f(x) = \frac{1}{\sqrt{2\pi \sigma^2}} \exp \left(- \frac{(x-m)^2}{2\sigma^2} \right).$$

Hence,

$$\int_{-\infty}^0 e^{rx} \cdot f(x) \, dx = \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^0 \exp \left(rx - \frac{(x-m)^2}{2\sigma^2} \right) \, dx.$$

Now write

\begin{align*} rx - \frac{(x-m)^2}{2\sigma^2} &= \frac{1}{2\sigma^2} \bigg[ -(x-(m+\sigma^2 r))^2 + (\sigma^2 r)^2 + 2m\sigma^2 r \bigg]. \end{align*}

Setting $\tilde{m} := m+\sigma^2 r$, we see that

$$\int_{-\infty}^0 e^{rx} \cdot f(x) \, dx = \frac{1}{\sqrt{2\pi \sigma^2}} e^{rm +\frac{1}{2} \sigma^2 r^2} \int_{-\infty}^0 \exp \left( -\frac{(x-\tilde{m})^2}{2\sigma^2} \right) \, dx.$$

By a standard substitution this yields

\begin{align*} \int_{-\infty}^0 e^{rx} \cdot f(x) \, dx &=e^{rm + \frac{1}{2} \sigma^2 r^2} \int_{-\infty}^{-\frac{\tilde{m}}{\sigma}} \frac{1}{\sqrt{2\pi}} \exp \left( -\frac{z^2}{2} \right) \, dz \\ &=e^{rm + \frac12 \sigma^2 r^2} \cdot \Phi \left( - \frac{\tilde{m}}{\sigma}\right) \\&= e^{rm + \frac{1}{2} \sigma^2 r^2} \cdot \Phi \left( - \frac{m+\sigma^2 r}{\sigma} \right). \end{align*}