<< Chapter < Page | Chapter >> Page > |
$n!=n(n-1)(n-2)...\left(1\right)\text{}$
$0!=1\text{}$
Formula$\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{(n-r)!r!}\text{}$
Formula$X~B(n,p)$
$P(X=x)=\left(\genfrac{}{}{0ex}{}{n}{x}\right){p}^{x}{q}^{n-x}$ , for $x=0,1,2,...,n$
Formula$X~G\left(p\right)$
$P(X=x)={q}^{x-1}p$ , for $x=1,2,3,...$
Formula$X~H(r,b,n)$
$P(X=x)=\left(\frac{\left(\genfrac{}{}{0ex}{}{r}{x}\right)\left(\genfrac{}{}{0ex}{}{b}{n-x}\right)}{\left(\genfrac{}{}{0ex}{}{r+b}{n}\right)}\right)$
Formula$X~P\left(\mu \right)$
$P(X=x)=\frac{{\mu}^{x}{e}^{-\mu}}{x!}$
Formula$X~U(a,b)$
$f\left(X\right)=\frac{1}{b-a}$ , $a<x<b$
Formula$X~\mathrm{Exp}\left(m\right)$
$f\left(x\right)=m{e}^{-\mathrm{mx}}$ , $m>0,x\ge 0$
Formula$X~N(\mu ,{\sigma}^{2})$
$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi}}{e}^{\frac{{-(x-\mu )}^{2}}{{2\sigma}^{2}}}$ , $\phantom{\rule{12pt}{0ex}}-\infty <x<\infty $
Formula$\Gamma \left(z\right)={\int}_{0}^{\infty}{x}^{z-1}{e}^{-x}\mathrm{dx}$ $z>0$
$\Gamma \left(\frac{1}{2}\right)=\sqrt{\pi}$
$\Gamma (m+1)=m!$ for $m$ , a nonnegative integer
otherwise: $\Gamma (a+1)=\mathrm{a\Gamma}\left(a\right)$
Formula$X~{t}_{\mathrm{df}}$
$f\left(x\right)=\frac{{\left(1+\frac{{x}^{2}}{n}\right)}^{\frac{-(n+1)}{2}}\Gamma \left(\frac{n+1}{2}\right)}{\sqrt{\mathrm{n\pi}}\Gamma \left(\frac{n}{2}\right)}$
$X=\frac{Z}{\sqrt{\frac{Y}{n}}}$
$Z~N\left(\mathrm{0,1}\right)$ , $Y~{{\rm X}}_{\mathrm{df}}^{2}$ , $n$ = degrees of freedom
Formula$X~{{\rm X}}_{\mathrm{df}}^{2}$
$f\left(x\right)=\frac{{x}^{\frac{n-2}{2}}{e}^{\frac{-x}{2}}}{{2}^{\frac{n}{2}}\Gamma \left(\frac{n}{2}\right)}$ , $x>0$ , $n$ = positive integer and degrees of freedom
Formula$X~{F}_{\mathrm{df}\left(n\right),\mathrm{df}\left(d\right)}$
$\mathrm{df}\left(n\right)=$ degrees of freedom for the numerator
$\mathrm{df}\left(d\right)=$ degrees of freedom for the denominator
$f\left(x\right)=\frac{\Gamma \left(\frac{u+v}{2}\right)}{\Gamma \left(\frac{u}{2}\right)\Gamma \left(\frac{v}{2}\right)}{\left(\frac{u}{v}\right)}^{\frac{u}{2}}{x}^{(\frac{u}{2}-1)}[1+\left(\frac{u}{v}\right){x}^{-.5\left(u+v\right)}]$
$X=\frac{{Y}_{u}}{{W}_{v}}$ , $Y$ , $W$ are chi-square
Notification Switch
Would you like to follow the 'Collaborative statistics: custom version modified by r. bloom' conversation and receive update notifications?