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 \(P(X\geq 1)=1-P(X\leq 1)=1-P(0)\)
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 \(P(X\geq 1)=1-.777=0.283\), so \(28\%\) 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?
\(P\left ( 0\right ) =0.5=\frac {\omega 0}{0!}e^{-\omega }\rightarrow 0.5=e^{-\omega }\) hence \(\ln 0.5=-\omega \rightarrow \omega =0.693\), so \(\lambda \frac {x}{60}=0.693\rightarrow x=20.7\) 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 \(F\left ( x\right ) \geq 0\) for all \(x.\) Do this by showing \(F^{\prime }\left ( x\right ) \geq 0\), and show limit \(F\left ( x\right ) \rightarrow 1\) as \(x\rightarrow \infty \) and limit \(F\left ( x\right ) \rightarrow 0\) as \(x\rightarrow -\infty \). And \(P\left ( k_{1}\leq T<k_{2}\right ) =F\left ( k_{2}\right ) -F\left ( k_{1}\right ) \)
properties of pdf:
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 \[ \sum _{k=1}^{n}k=\frac {1}{2}n\left ( n+1\right ) \] Geometric sum \[ {\sum \limits _{k=0}^{n}}r^{k}=\frac {1-r^{n+1}}{1-r}\] if \(-1<r<1\), then \[ {\sum \limits _{k=0}^{\infty }}r^{k}=\frac {1}{1-r}\] if the sum is from 1 then \[ {\sum \limits _{k=1}^{n}}r^{k}=\frac {r\left ( 1-r^{n+1}\right ) }{1-r}\] if \(-1<r<1\), then \begin {align*} {\sum \limits _{k=1}^{\infty }}r^{k} & =\frac {r}{1-r}\\ \Gamma \left ( x\right ) & =\int _{0}^{\infty }u^{x-1}e^{-u}du\\ \Gamma \left ( n\right ) & =\left ( n-1\right ) !\\ \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 {align*}
And
\[\begin {pmatrix} & n & \\ n_{1} & n_{2} & n_{3}\end {pmatrix} =\frac {n!}{n_{1}!\ n_{2}!\ n_{3}!}\] If given joint density \(f_{XY}\left ( x,y\right ) \) and asked to find conditional \(P\left ( X|Y\right ) =\frac {f_{XY}\left ( x,y\right ) }{f_{Y}\left ( y\right ) }\) so need to find marginals. Marginal is found from \(f_{Y}\left ( y\right ) =\int _{x}f_{XY}\left ( x,y\right ) dx\), and \(f_{X}\left ( x\right ) =\int _{y}f_{XY}\left ( x,y\right ) dy\)
To convert from \(x,y\,\ \)to polar, example: given \(f\left ( x,y\right ) =c\sqrt {1-\left ( x^{2}+y^{2}\right ) }\) find \(c\), where \(x^{2}+y^{2}\leq 1\), 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 \(f\left ( Y\right ) =\int _{-\infty }^{\infty }f_{Y|X}\left ( y|x\right ) f_{X}\left ( x\right ) dx\)
\[ Z=\frac {\bar {X}_{n}-\mu }{\sigma /\sqrt {n}}\rightarrow N(0,1) \]
\[ T=\frac {\bar {X}_{n}-\mu }{S_{n}/\sqrt {n}}\rightarrow T(n) \] where \(S_{n}\ \)is \(std\) of the sample.
Note Var(sample) has chi square (n) distribution.
CI for T: \begin {align*} \Pr \left ( -A<\frac {\bar {X}_{n}-\mu }{S_{n}/\sqrt {n}}<A\right ) & =1-\alpha \\ \Pr \left ( \bar {X}_{n}-A\frac {S_{n}}{\sqrt {n}}<\mu <\bar {X}_{n}+A\frac {S_{n}}{\sqrt {n}}\right ) & =1-\alpha \end {align*}