PDF & CDF
The probability density function is $$f(x; \mu, \sigma) = {1\over\sqrt{2\pi}\sigma}e^{-{1\over2}{(x-\mu)^2\over\sigma^2}}$$ The cumulative distribution function is defined by $$F(x; \mu, \sigma) = \Phi\left({x-\mu\over\sigma}\right)$$ where $$\Phi(z) = {1\over\sqrt{2\pi}} \int_{-\infty}^{z}e^{-{1\over2}x^2}\ dx$$
Proof:
$$ \begin{align*} \int_{-\infty}^{\infty}f(x; \mu, \sigma) &= \int_{-\infty}^{\infty}{1\over\sqrt{2\pi}\sigma}e^{-{1\over2}{(x-\mu)^2\over\sigma^2}}\ dx\\ &= {1\over\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}e^{-{1\over2}{(x-\mu)^2\over\sigma^2}}\ dx\\ &= {1\over\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-{1\over2}y^2}\ dy\quad\quad\quad\quad\quad(\mbox{setting}\ y={x-\mu\over\sigma} \Rightarrow dx = \sigma dy)\\ \end{align*} $$ Let $I = \int_{-\infty}^{\infty}e^{-{1\over2}y^2}\ dy$, then $$ \begin{eqnarray*} I^2 &=& \int_{-\infty}^{\infty}e^{-{1\over2}y^2}\ dy\int_{-\infty}^{\infty}e^{-{1\over2}x^2}\ dx\\ &=& \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-{1\over2}(y^2+x^2)}\ dydx\quad\quad\quad\quad(\mbox{setting}\ x=r\cos\theta, y=r\sin\theta)\\ &=& \int_{0}^{\infty}\int_{0}^{2\pi}e^{-{1\over2}r^2}\ rd\theta dr \\ & & (\mbox{double integral}\ \iint\limits_{D}f(x, y)\ dxdy = \iint\limits_{D^*}f(r\cos\theta, r\sin\theta)r\ drd\theta) \\ &=& 2\pi\int_{0}^{\infty}re^{-{1\over2}r^2}\ dr\\ &=& -2\pi e^{-{1\over2}r^2}\Big|_{0}^{\infty}\\ &=& 2\pi \end{eqnarray*} $$ Hence $$\int_{-\infty}^{\infty}f(x; \mu, \sigma) = {1\over\sqrt{2\pi}} \cdot\sqrt{2\pi} = 1$$
Standard Normal Distribution
If $X$ is normally distributed with parameters $\mu$ and $\sigma^2$, then $$Z = {X-\mu\over\sigma}$$ is normally distributed with parameters 0 and 1.
Proof:
An important conclusion is that if $X$ is normally distributed with parameters $\mu$ and $\sigma^2$, then $Y = aX + b$ is normally distributed with parameters $a\mu + b$ and $a^2\sigma^2$. Denote $F_{Y}$ as the cumulative distribution function of $Y$: $$ \begin{align*} F_{Y}(x) &= P(Y \leq x)\\ &= P(aX + b \leq x)\\ &= P(X \leq {x-b\over a})\\ &= F_{X}\left({x-b\over a}\right) \end{align*} $$ where $F_{X}(x)$ is the cumulative distribution function of $X$. By differentiation, the probability density function of $Y$ is $$ \begin{align*} f_{Y}(x) &= {1\over a}f_{X}\left({x-b\over a}\right)\\ &= {1\over\sqrt{2\pi}a\sigma}e^{-{1\over2}{({x-b\over a} - \mu)^2\over \sigma^2}}\\ &= {1\over\sqrt{2\pi}(a\sigma)}e^{-{1\over2}{(x-b - a\mu)^2\over a^2\sigma^2}}\\ &= {1\over\sqrt{2\pi}(a\sigma)}e^{-{1\over2}{(x-(b + a\mu))^2\over (a\sigma)^2}} \end{align*} $$ which shows that $Y$ is normally distributed with parameters $a\mu + b$ and $a^2\sigma^2$. According to the above result, we can easily deduce that $Z = {X-\mu\over\sigma}$ follows the normally distributed with parameters 0 and 1.
Mean
The expected value is $$E[X] = \mu$$
Proof:
$$ \begin{align*} E[Z] &= \int_{-\infty}^{\infty}xf_{Z}(x)\ dx\quad\quad\quad \quad\quad \quad\quad (\mbox{setting}\ Z={X-\mu\over\sigma})\\ &= {1\over\sqrt{2\pi}}\int_{-\infty}^{\infty}xe^{-{1\over2}x^2}\ dx\\ &= -{1\over\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-{1\over2}x^2}\ d\left(-{1\over2}x^2\right)\\ &= -{1\over\sqrt{2\pi}}e^{-{1\over2}x^2}\Big|_{-\infty}^{\infty}\\ &= 0 \end{align*} $$ Hence $$ \begin{align*} E[X] &= E\left[\sigma Z+\mu\right]\\ &= \sigma E[Z] + \mu\\ &= \mu \end{align*} $$
Variance
The variance is $$\mbox{Var}(X) = \sigma^2$$
Proof:
$$ \begin{align*} E\left[Z^2\right] &= {1\over\sqrt{2\pi}}\int_{-\infty}^{\infty}x^2e^{-{1\over2}x^2}\ dx\quad\quad\quad \quad\quad \quad\quad\quad\quad\quad (\mbox{setting}\ Z={X-\mu\over\sigma})\\ &= {1\over\sqrt{2\pi}}\left(-xe^{-{1\over2}x^2}\Big|_{-\infty}^{\infty} +\int_{-\infty}^{\infty}e^{-{1\over2}x^2}\ dx\right)\quad\quad\quad(\mbox{integrating by parts})\\ &= {1\over\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-{1\over2}x^2}\ dx \quad\quad\quad\quad\quad\quad\quad(\mbox{standard normal distribution})\\ &= 1 \end{align*} $$ the integral by parts: $$u= x,\ dv = xe^{-{1\over2}x^2}\ dx$$ $$\implies du = dx,\ v = \int xe^{-{1\over2}x^2}\ dx = -e^{-{1\over2}x^2}$$ $$\implies \int x^2e^{-{1\over2}x^2}\ dx =-xe^{-{1\over2}x^2} +\int e^{-{1\over2}x^2}\ dx$$ Hence $$\mbox{Var}(X) = \mbox{Var}(\sigma Z + \mu)= \sigma^2\mbox{Var}(Z) = \sigma^2$$
Examples
1. If $X$ is a normal random variable with parameters $\mu = 3$ and $\sigma^2 = 9$, find (a) $P(2 < X <5 data-blogger-escaped-b=""> 0)$; (c) $P(|X - 3| > 6)$.
Solution:
(a) $$ \begin{align*} P(2 < X < 5) &= P\left({2-3\over3} < {X - 3\over 3} < {5-3\over 3}\right)\\ &= P\left(-{1\over3} < Z < {2\over3}\right)\\ &= \Phi\left({2\over3}\right) - \Phi\left(-{1\over3}\right) = 0.3780661 \end{align*} $$ R code:
pnorm(2/3) - pnorm(-1/3)
# [1] 0.3780661
(b) $$ \begin{align*} P(X > 0) &= P\left({X-3\over3} > {0-3\over3}\right)\\ &= P\left(Z > -1\right)\\ &= 1 - \Phi(-1) = 0.8413447 \end{align*} $$ R code:
1 - pnorm(-1)
# [1] 0.8413447
(c) $$ \begin{align*} P(|X - 3| > 6) &= P(X > 9) + P(X < -3)\\ &= P\left({X-3\over3} > {9-3\over3}\right) + P\left({X-3\over3} < {-3-3\over3}\right)\\ &= P(Z > 2) + P(Z < -2)\\ &= 1-\Phi(2) + \Phi(-2) = 0.04550026 \end{align*} $$ R code:
1 - pnorm(2) + pnorm(-2)
# [1] 0.04550026
2. Let $X$ be normally distributed with standard deviation $\sigma$. Determine $P\left(|X-\mu| \geq 2\sigma\right)$. Compare with Chebyshev's Inequality.
Solution:
$$ \begin{align*} P\left(|X-\mu| \geq 2\sigma\right) &= P\left({X-\mu\over\sigma} \geq 2\right) + P\left({X-\mu\over\sigma} \leq -2\right)\\ &=2\cdot P\left({X-\mu\over\sigma} \leq -2\right) = 2\Phi(-2) \end{align*} $$ R code:
2 * pnorm(-2)
# [1] 0.04550026
By Chebyshev's Inequality, the probability is $$P\left(|X-\mu| \geq 2\sigma\right) \leq {1\over2^2}=0.25$$ which is a weaker estimation.
3. Let $X$ be a normally distributed random variable with expected value $\mu=5$. Assume $P(X \leq 0) = 0.1$. What is the variance of $X$?
Solution:
$$ \begin{align*} P(X \leq 0) &= P\left({X - 5\over\sigma} \leq {0-5\over\sigma}\right)\\ &= P\left(Z \leq -{5\over\sigma}\right) = 0.1 \end{align*} $$ Hence by using R:
z = qnorm(0.1)
var = (-5/z)^2
# [1] -1.281552
# [1] 15.22186
$$-{5\over\sigma} = -1.281552\Rightarrow \sigma^2 = 15.22186$$
4. A normally distributed random variable $X$ satisfies $P(X \leq 0) = 0.4$ and $P(X \geq 10) = 0.1$. What is the expected value $\mu$ and the standard deviation $\sigma$?
Solution:
$$P(X \leq 0) = 0.4\Rightarrow \Phi\left({-\mu\over\sigma}\right) = 0.4$$ and $$P(X \geq 10) = 0.1\Rightarrow\Phi\left({10-\mu\over \sigma}\right) = 0.9$$ Thus $$\begin{cases}{-\mu\over\sigma}=-0.2533471\\ {10-\mu\over \sigma}=1.281552 \end{cases}\Rightarrow \begin{cases}\mu = 1.650579\\ \sigma= 6.515088 \end{cases}$$ R code:
z1 = qnorm(0.4); z2 = qnorm(0.9)
s = 10 / (z2 - z1)
mu = -s * z1
z1; z2
# [1] -0.2533471
# [1] 1.281552
mu; s
# [1] 1.650579
# [1] 6.515088
5. Consider independent random variables $X\sim N(1, 3)$ and $Y\sim N(2, 4)$. What is $P(X + Y \leq 5)$?
Solution:
$X +Y$ is still normally distributed with parameters $$\mu = \mu_1 + \mu_2 = 3$$ and $$\sigma^2 = \sigma_1^2 + \sigma_2^2 = 7$$ Hence $$ \begin{align*} P(X + Y \leq 5) &= P\left(Z \leq {5-3 \over\sqrt{7}}\right)\\ &= \Phi\left({2 \over\sqrt{7}}\right) = 0.7751541 \end{align*} $$ R code:
pnorm(2 / sqrt(7))
# [1] 0.7751541
Reference
- Ross, S. (2010). A First Course in Probability (8th Edition). Chapter 5. Pearson. ISBN: 978-0-13-603313-4.
- Brink, D. (2010). Essentials of Statistics: Exercises. Chapter 5 & 15. ISBN: 978-87-7681-409-0.