W06 Notes

Normal distribution

01 Theory

Theory 1

Normal distribution

A variable $X$ has a normal distribution, written $X\sim 𝒩(\mu ,{\sigma }^2)$ or “$X$ is Gaussian $(\mu ,\sigma )$,” when it has PDF given by:

$$f_X(x)=\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-{(x-\mu )}^2/2{\sigma }^2}$$

The standard normal is $Z\sim 𝒩(\mathrm{0,1})$ and its PDF is usually denoted by $\phi (x)$:

$$\phi (x)=\frac{1}{\sqrt{2\pi }}{e}^{-x^2/2}$$

The standard normal CDF is usually denoted by $\mathrm{\Phi }(z)$:

$$\mathrm{\Phi }(z)={\int }_{-\infty }^z\frac{1}{\sqrt{2\pi }}{e}^{-u^2/2}du$$

• To show that $\phi (x)$ is a valid probability density, we must show that ${\int }_{-\infty }^{+\infty }\phi (x)dx=1$.
  • This calculation is not trivial; it requires a double integral in polar coordinates!
• There is no explicit antiderivative of $\phi$
  • A computer is needed for numerical calculations.
  • A chart of approximate values of $\mathrm{\Phi }$ is provided for exams.

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• To check that $E[Z]=0$:
  • Observe that $x\phi (x)$ is an odd function, i.e. symmetric about the $y$-axis.
  • One must then simply verify that the improper integral converges.
• To check that $\mathrm{Var}[Z]=1$:
  • Since $\mu =E[Z]=0$, we find:

$$\mathrm{Var}[Z]=E[Z^2]\gg \gg \frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{x}^2{e}^{-x^2/2}dx=:I$$

• Use integration by parts to compute that $I=1$. (Select $u=x$ and $dv=x{e}^{-x^2/2}dx$.)

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General and standard normals

Assume that $Z\sim 𝒩(\mathrm{0,1})$ and $\sigma ,\mu$ are constants. Define $X=\sigma Z+\mu$. Then:

$$f_X=\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-{(x-\mu )}^2/2{\sigma }^2}$$

That is, $\sigma Z+\mu$ has the distribution type $𝒩(\mu ,{\sigma }^2)$.

Derivation of PDF of $\sigma Z+\mu$

Suppose that $X=\sigma Z+\mu$. Then:

$$\begin{array}{cc}{F}_X(x)& =P[X\le x]\\ & \\ & =P[\sigma Z+\mu \le x]\\ & \\ & =P[Z\le \frac{x-\mu }{\sigma }]\\ & \\ & =\mathrm{\Phi }\left(\frac{x-\mu }{\sigma }\right)\end{array}$$

Differentiate to find $f_X$:

$$\begin{array}{cc}{f}_X(x)& =\frac{d}{dx}{F}_X(x)\\ & \\ & =\frac{d}{dx}\mathrm{\Phi }\left(\frac{x-\mu }{\sigma }\right)\\ & \\ & =\frac{1}{\sigma }\phi \left(\frac{x-\mu }{\sigma }\right)\\ & \\ & =\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-{(x-\mu )}^2/2{\sigma }^2}\end{array}$$

From this fact we can infer that $E[X]=\mu$ and $\mathrm{Var}[X]={\sigma }^2$ whenever $X\sim 𝒩(\mu ,{\sigma }^2)$.

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02 Illustration

Example - Basic generalized normal calculation

Basic generalized normal calculation

Suppose $X\sim 𝒩(-\mathrm{3,4})$. Find $P[X\ge -1.7]$.

Solution

First write $X$ as a linear transformation of $Z$:

$$X\sim 2Z-3$$

Then:

$$X\ge -1.7⟺Z\ge 0.65$$

Look in a table to find that $\mathrm{\Phi }(0.65)\approx 0.74$ and therefore:

$$\begin{array}{c}P[Z\ge 0.65]\gg \gg 1-P[Z\le 0.65]\\ \\ \gg \gg \approx 1-0.74\gg \gg 0.26\end{array}$$ Link to original

Example - Gaussian: interval of $2/3$

Gaussian: interval of 2/3

Find the number $a$ such that $P[-a\le Z\le +a]=2/3$.

Solution

First convert the question:

$$\begin{array}{cc}P[-a\le Z\le +a]& \gg \gg F_Z(a)-F_Z(-a)\\ & \\ & \gg \gg \mathrm{\Phi }(a)-\mathrm{\Phi }(-a)\\ & \\ & \gg \gg 2\mathrm{\Phi }(a)-1\end{array}$$

Solve for $a$ so that this value is $2/3$:

$$2\mathrm{\Phi }(a)-1=2/3\gg \gg \mathrm{\Phi }(a)=5/6\gg \gg a={\mathrm{\Phi }}^{-1}(5/6)$$

Use a $\mathrm{\Phi }$ table to conclude $a\approx 0.97$.

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Example - Heights of American males

Heights of American males

Suppose that the height of an American male in inches follows the normal distribution $𝒩(\mathrm{71,6.25})$.

(a) What percent of American males are over 6 feet, 2 inches tall?

(b) What percent of those over 6 feet tall are also over 6 feet, 5 inches tall?

Solution

(a) Let $H$ be a random variable measuring the height of American males in inches, so $H\sim 𝒩({\mathrm{71,2.5}}^2)$. Thus $H\sim 2.5Z+71$, and:

$$\begin{array}{cc}P[H>74]& \gg \gg 1-P[H\le 74]\\ & \\ & \gg \gg 1-P[2.5Z+71\le 74]\\ & \\ & \gg \gg 1-P[Z\le 1.20]\\ & \\ & \gg \gg 1-0.8849\approx 11.5\%\end{array}$$

(b) We seek $P[H>77|H>72]$ as the answer. Compute as follows:

$$\begin{array}{cc}P[H>77|H>72]& =\frac{P[H>77]}{P[H>72]}\\ & \\ & \gg \gg \frac{P[2.5Z+71>77]}{P[2.5Z+71>72]}\\ & \\ & \gg \gg \frac{1-P[Z\le 2.4]}{1-P[Z\le 0.4]}=\frac{1-0.9918}{1-0.6554}\approx 2.38\%\end{array}$$ Link to original

Example - Variance of normal from CDF table

Variance of normal from CDF table

Suppose $X\sim 𝒩(5,{\sigma }^2)$, and suppose you know $P[X>9]=0.2$.

Find the approximate value of $\sigma$ using a $\mathrm{\Phi }$ table.

Solution

$$X\sim 𝒩(5,{\sigma }^2)⟹X\sim \sigma Z+5$$

So $1-P[X\le 9]=0.2$ and thus $P[\sigma Z+5\le 9]=0.8$. Then:

$$P[\sigma Z+5\le 9]=P[Z\le 4/\sigma ]$$

so $P[Z\le 4/\sigma ]=0.8$.

Looking in the chart of $\mathrm{\Phi }$ for the nearest inverse of 0.8, we obtain $4/\sigma =0.842$, hence $\sigma =4.75$.

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