STAT 400

Wed. March 4th, 2020

Normal (Gaussian) Distribution

Definition. A continuous random variable XX is said to follow a normal distribution with parameter μ\mu and σ2\sigma^2, where σ>0\sigma > 0, if its pdf is f(x;μ,σ)=12πσ2e−(x−μ)22σ2=12πσ2exp(−(x−μ)22σ2).f(x;\mu,\sigma)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}=\frac{1}{\sqrt{2\pi\sigma^2}} \text{exp}(-\frac{(x-\mu)^2}{2\sigma^2}) . In this case, we write XX follows N(μ,σ2)N(\mu,\sigma^2).

Additionally, E(X)=μE(X)=\mu and V(X)=σ2V(X)=\sigma^2. We can prove this ourselves:

E(X)=∫−∞∞x⋅12πσ2e−(x−μ)22σ2dxy=x−μσE(X)=∫−∞∞12π(yσ+μ)e−y22dy=σ2π∫−∞∞ye−y22dy+μ2π∫−∞∞e−y22dy=σ2π⋅0+μ2π⋅2π=μ\begin{aligned} E(X)&=\int_{-\infin}^{\infin}x\cdot \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx\\ y&=\frac{x-\mu}{\sigma}\\ E(X)&=\int_{-\infin}^{\infin}\frac{1}{\sqrt{2\pi}}(y\sigma + \mu)e^{-\frac{y^2}{2}}dy\\ &=\frac{\sigma}{\sqrt{2\pi}}\int_{-\infin}^{\infin}ye^{-\frac{y^2}{2}}dy+\frac{\mu}{\sqrt{2\pi}}\int_{-\infin}^{\infin}e^{-\frac{y^2}{2}}dy\\ &=\frac{\sigma }{\sqrt{2\pi}}\cdot 0+\frac{\mu}{\sqrt{2\pi}}\cdot \sqrt{2\pi}\\ &=\mu \end{aligned} We know that the left integral is 0 because it’s an odd function, and the right integral is 2π\sqrt{2\pi} because it’s the Gaussian integral (look this up on your own). V(X)V(X) can be calculated in a similar way.

Standard Normal Distribution (N(0,1)N(0,1))

Note what happens to the pdf of the normal distribution as the parameters change. Changing μ\mu simply shifts the function left and right, while lowering σ\sigma causes the function to "bunch up" around μ\mu and raising it causes it to "spread out".

Because of this, we can just focus on the simplest case of parameters (μ=0, σ2=1\mu=0,\ \sigma^2=1) and still understand the entire family of normal distributions. N(0,1)N(0,1) is thus called the standard normal distirbution.

We use ZZ to denote an RV of standard normal distribution (ZZ follows N(0,1)N(0,1)). The pdf of ZZ is f(x;0,1)=12πe−x22f(x;0,1)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}. This function is called the standard normal curve.

Properties of N(0,1)N(0,1)

  1. The inflection points of f(x;0,1)f(x;0,1) are at 11 and −1-1. (These are the points where f′′(x)=0f''(x)=0)
  2. The cdf of ZZ, denoted as Φ(x)=P(Z≤x)\Phi(x)=P(Z\le x), is ∫−∞x12πe−y22dy.\int_{-\infin}^x \frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}dy. There is no closed form for Φ\Phi, so we need to use a standard normal table to calculate it.

Important Percentiles of N(0,1)N(0,1)

The (100p)(100p)th percentile of N(0,1)N(0,1) is η(p)\eta(p), which satisfies Φ(η(p))=P(Z≤η(p))\Phi(\eta(p))=P(Z\le \eta(p)). Here’s a table of important values of η(p)\eta(p): p∣0.9∣0.95∣0.975∣0.99∣0.995∣0.999η(p)∣1.28∣1.645∣1.96∣2.33∣2.58∣3.08\begin{aligned} p &\mid&0.9&\mid&0.95&\mid&0.975&\mid&0.99&\mid&0.995&\mid&0.999 \\ \eta(p) &\mid&1.28&\mid&1.645&\mid&1.96&\mid&2.33&\mid&2.58&\mid&3.08 \end{aligned}

Critical Values of ZZ (ZαZ_\alpha)

Given 0<α<10<\alpha<1, we set ZαZ_\alpha to the value satisfying P(Z≥Zα)=1−Φ(Zα)=α.P(Z\ge Z_\alpha)=1-\Phi(Z_\alpha)=\alpha. (This is essentially the same concept as percentiles, but with a different notation.)

α∣0.1∣0.05∣0.025∣0.01∣0.005∣0.001Zα∣1.28∣1.645∣1.96∣2.33∣2.58∣3.08\begin{aligned} \alpha &\mid&0.1&\mid&0.05&\mid&0.025&\mid&0.01&\mid&0.005&\mid&0.001 \\ Z_\alpha &\mid&1.28&\mid&1.645&\mid&1.96&\mid&2.33&\mid&2.58&\mid&3.08 \end{aligned}

Nonstandard Normal Distributions

Proposition. Let XX follow N(μ,σ2)N(\mu,\sigma^2). Then Z=x−μσZ=\frac{x-\mu}{\sigma} follows N(0,1)N(0,1), which we can prove by performing the same change of variable we used to calculate E(X).E(X).

Thus, P(a≤X≤b)=P(a−μσ≤Z≤b−μσ)=Φ(b−μσ)−Φ(a−μσ).P(a\le X\le b)=P(\frac{a-\mu}{\sigma}\le Z\le \frac{b-\mu}{\sigma})=\Phi(\frac{b-\mu}{\sigma})-\Phi(\frac{a-\mu}{\sigma}).

Similarly, P(a≤X)=P(a−μσ≤Z)=1−Φ(a−μσ)P(a\le X)=P(\frac{a-\mu}{\sigma}\le Z)=1-\Phi(\frac{a-\mu}{\sigma}) and P(X≤b)=P(Z≤b−μσ)=Φ(b−μσ).P(X\le b)=P(Z\le \frac{b-\mu}{\sigma})=\Phi(\frac{b-\mu}{\sigma}).