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1 my first cheat sheet

2 second cheat sheet

problem: phone calls received at rate $\lambda =2$ per hr. If person wants to take 10 min shower, what is probability a phone will ring during that time?

answer: first change to $\omega =\lambda \frac{10}{60}=2\frac{10}{60}=.3333$, now we want $P(X\ge 1)=1-P(X\le 1)=1-P(0)$

but $P\left(k\right)=\frac{{\lambda }^k}{k!}{e}^{-\lambda }$, but remember, we are using $\omega$, so $P\left(k\right)=\frac{{\omega }^k}{k!}{e}^{-\omega }$ so $P\left(0\right)=\frac{{.3333}^0}{0!}{e}^{-.3333}=0.777$

so $P(X\ge 1)=1-.777=0.283$, so $28\mathrm{\%}$ change the phone will ring.

How long can shower be if they wish probability of receiving no phone calls to be at most 0.5?

$P\left(0\right)=0.5=\frac{\omega 0}{0!}{e}^{-\omega }\to 0.5=e^{-\omega }$ hence $\ln 0.5=-\omega \to \omega =0.693$, so $\lambda \frac{x}{60}=0.693\to x=20.7$ minutes

To find quantile, say $\frac{1}{4}$, first find an expression for $F\left(x\right)$ as function of $x$, then solve for $x$ in $F\left(x\right)=.25$

For median, solve for $x$ in $F\left(x\right)=.5$

properties of CDF: 1. Show $F\left(x\right)\ge 0$ for all $x.$ Do this by showing $F^{\prime }\left(x\right)\ge 0$, and show limit $F\left(x\right)\to 1$ as $x\to \mathrm{\infty }$ and limit $F\left(x\right)\to 0$ as $x\to -\mathrm{\infty }$. And $P(k_1\le T<k_2)=F\left(k_2\right)-F\left(k_1\right)$

properties of pdf:

1. piecewise continuous
2. pdf$\left(x\right)\ge 0$
3. ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}pdf\left(x\right)=1$

remember $\frac{d}{dx}{\tan }^{-1}x=\frac{1}{1+x^2}$

The geometric distribution is the only discrete memoryless random distribution. It is a discrete analog of the exponential distribution. continuous

Some relations  $$\sum _{k=1}^{n}k=\frac{1}{2}n(n+1)$$ Geometric sum $$\sum _{k=0}^n{r}^k=\frac{1-r^{n+1}}{1-r}$$ if $-1<r<1$, then $$\sum _{k=0}^{\mathrm{\infty }}{r}^k=\frac{1}{1-r}$$ if the sum is from 1 then $$\sum _{k=1}^n{r}^k=\frac{r(1-r^{n+1})}{1-r}$$ if $-1<r<1$, then $$\begin{array}{rl}\sum _{k=1}^{\mathrm{\infty }}{r}^k& =\frac{r}{1-r}\\ \mathrm{\Gamma }\left(x\right)& ={\int }_0^{\mathrm{\infty }}{u}^{x-1}{e}^{-u}du\\ \mathrm{\Gamma }\left(n\right)& =(n-1)!\\ \frac{d}{dx}\ln \left(x\right)& =\frac{1}{x}\\ \int \ln \left(y\right)dy& =-y+y\ln \left(y\right)\\ \int \frac{1}{y}dy& =\ln \left(y\right)\end{array}$$

And

$$\left(\begin{array}{ccc}& n& \\ n_1& n_2& n_3\end{array}\right)=\frac{n!}{n_1!n_2!n_3!}$$ If given joint density $f_{XY}(x,y)$ and asked to find conditional $P\left(X|Y\right)=\frac{f_{XY}(x,y)}{f_Y\left(y\right)}$ so need to find marginals. Marginal is found from $f_Y\left(y\right)={\int }_x{f}_{XY}(x,y)dx$, and $f_X\left(x\right)={\int }_y{f}_{XY}(x,y)dy$

To convert from $x,y$to polar, example: given $f(x,y)=c\sqrt{1-(x^2+y^2)}$ find $c$, where $x^2+y^2\le 1$, then write

$$c{\int }_{\theta =-\pi }^{\theta =\pi }{\int }_{r=0}^{r=1}\sqrt{1-r^2}rdrd\theta$$ Use identity above.

law of total probablity: if we know $Y|X$ and $X$ and want to know distribution of $Y$, then $f\left(Y\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{f}_{Y|X}\left(y|x\right)f_X\left(x\right)dx$

$$Z=\frac{{\overline{X}}_n-\mu }{\sigma /\sqrt{n}}\to N(0,1)$$

$$T=\frac{{\overline{X}}_n-\mu }{S_n/\sqrt{n}}\to T(n)$$ where $S_n$is $std$ of the sample.

Note Var(sample) has chi square (n) distribution.

CI for T: $$\begin{array}{rl}Pr(-A<\frac{{\overline{X}}_n-\mu }{S_n/\sqrt{n}}<A)& =1-\alpha \\ Pr({\overline{X}}_n-A\frac{S_n}{\sqrt{n}}<\mu <{\overline{X}}_n+A\frac{S_n}{\sqrt{n}})& =1-\alpha \end{array}$$