Normal Random Variable

Definition We say a continuous random variable $\mathbb{X}$ is has a normal distribution with parameter $\mu, \sigma^2$, denoted $\mathbb{X} \sim normal(\mu, \sigma^2)$ if $$\mathbb{X} \sim f(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}} \text{ for all } x\in\mathbb{R}$$

Property If $\mathbb{X} \sim normal(\mu, \sigma^2)$, then $\mu$ is the mean and $\sigma^2$ is the variance

Proof 1 By definition

First note that if $\mathbb{X} \sim normal(\mu, \sigma^2)$, then $\mathbb{X} = \sigma\mathbb{Z}+\mu$, where $\mathbb{Z}$ is standard normal distribution, with $\mu = 0, \sigma = 1$. So we only need to prove that the mean and variance of $\mathbb{Z}$ is $0$ and $1$, (to see why $\mathbb{X} = \sigma\mathbb{Z}+\mu$, look at their c.d.f).

$\mathbb{E}[\mathbb{Z}] = \int_{-\infty}^\infty \frac{x}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}\mathrm{d}x = 0 \text{ because } \frac{x}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \text{ is odd and the integration exists}$ $\text{variance of } \mathbb{Z} = \int_{-\infty}^\infty \frac{x^2}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}\mathrm{d}x = \int_{-\infty}^\infty x * \frac{x}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}\mathrm{d}x$ $\text{(integration by part) } = 0 + \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-\frac{x^2}{2}}\mathrm{d}x = 1$

Here we used the identity $\int_{-\infty}^\infty e^{-\frac{x^2}{2}}\mathrm{d}x = \sqrt{2\pi}$

Proof 2 By moment generating function

Since $D_t (f(x)e^{xt}) = xe^{xt}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$ is continuous on $\mathbb{R}^2$, we can use m.g.f method.

The moment generating function $M_x(t) = \mathbb{E}[e^{\mathbb{X}t}] = \int_{-\infty}^\infty e^{xt} \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}} \mathrm{d}x = \frac{1}{\sqrt{2\pi}\sigma} \int_{-\infty}^\infty e^{-\frac{(x-\mu)^2-2 \sigma^2 t x}{2\sigma^2}} \mathrm{d}x$

$= \frac{1}{\sqrt{2\pi}\sigma} e^{\frac{(\mu+\sigma^2 t)^2 - \mu^2}{2\sigma^2}} \int_{-\infty}^\infty e^{-\frac{(x-(\mu+\sigma^2 t))^2}{2\sigma^2}} \mathrm{d}x = e^{\frac{2\mu\sigma^2 t+\sigma^4 t^2 }{2\sigma^2}} = e^{\frac{\sigma^2 t^2}{2}+\mu t}$ So the mean $\mu = \mathbb{E}[\mathbb{X}] = M_x'(0) = (\sigma^2 t + \mu) e^{\frac{\sigma^2 t^2}{2}+\mu t} |_{t=0} = \mu$ and variance $\sigma^2 = M_X''(0) - \mu^2 = (\sigma^2 t + \mu)^2 e^{\frac{\sigma^2 t^2}{2}+\mu t} + \sigma^2 |_{t=0} - \mu^2 = \sigma^2$