|
Autor |
Zufallsvariable exponentialverteilt |
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Carly2004
Wenig Aktiv  Dabei seit: 26.06.2022 Mitteilungen: 37
 | Themenstart: 2023-06-05
|
Seien ((X_n))_(n\el\ \IN) unabhängige Zufallsvariablen, die jeweils exponentialverteilt mit Parameter \lambda > 0
sind, und S_n = sum(X_k,k=1,n)
Sei außerdem \delta > 0. Finden Sie ein möglichst großes c > 0, sodass für
alle n\el\ \IN gilt:
P(1/n S_n >=(1+\delta)E(X_1))<=e^(-cn)
Ich habe leider keinen Ansatz bzw Idee wie ich vorgehen könnte, kann mir einer ein Tipp geben?
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Profil
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luis52
Senior  Dabei seit: 24.12.2018 Mitteilungen: 1111
 | Beitrag No.1, eingetragen 2023-06-05
|
\(\begingroup\)\(\newcommand{\diag}{\operatorname{diag}} % diag
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\newcommand{\ARMA}{\operatorname{ARMA}} % ARMA-Prozess
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%****************************************************************
%************************** Abkuerzungen ************************
%****************************************************************
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%****************************************************************
%************************** Miszellanien ************************
%****************************************************************
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\newcommand{\spn}{\operatorname{span}} % span
%****************************************************************
%************************** Abkuerzungen ************************
%****************************************************************
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%****************************************************************
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%****************************************************************
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%------------- Fette grosse Buchstaben im Mathemodus -----------
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%------------- Fette kleine Buchstaben im Mathemodus -----------
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%----- Fette Null und fette Eins --------------------------
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%----- Fettes Nabla --------------------------
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%----- Fette grosse griechische Buchstaben im Mathemodus -----
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%----- Fette kleine griechische Buchstaben im Mathemodus -----
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%----------- Kalligraphische Buchstaben im Mathemodus
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\newcommand{\mcO}{\mathcal{O}}
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\newcommand{\mcS}{\mathcal{S}}
\newcommand{\mcT}{\mathcal{T}}
\newcommand{\mcU}{\mathcal{U}}
\newcommand{\mcV}{\mathcal{V}}
\newcommand{\mcW}{\mathcal{W}}
\newcommand{\mcX}{\mathcal{X}}
\newcommand{\mcY}{\mathcal{Y}}
\newcommand{\mcZ}{\mathcal{Z}}
%----------- Kalligraphische Buchstaben im Mathemodus
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\newcommand{\mcy}{\mathcal{Y}}
\newcommand{\mcz}{\mathcal{Z}}
%---------- Doppelsumme ------------------------------------------
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%---------- \argmin und \argmax ------------------------------
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%---------- Folgen ------------------------------------------
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%---------- Geordnete Werte ---------------------------------------
\newcommand{\ord}[2]{\ensuremath{#1_{(1)},\ldots,#1_{(#2)}}}
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\newcommand{\ordne}[3]{\ensuremath{#1_{(1)}#3 #1_{(2)}#3\dots#3#1_{(#2)}}}
\)
\quoteon(2023-06-05 03:50 - Carly2004 im Themenstart)
kann mir einer ein Tipp geben?
\quoteoff
$S_n$ ist Erlang-verteilt.\(\endgroup\)
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Profil
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Carly2004
Wenig Aktiv  Dabei seit: 26.06.2022 Mitteilungen: 37
 | Beitrag No.2, vom Themenstarter, eingetragen 2023-06-05
|
Ich kann damit leider immernoch nicht viel Anfangen. :(
|
Profil
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luis52
Senior  Dabei seit: 24.12.2018 Mitteilungen: 1111
 | Beitrag No.3, eingetragen 2023-06-05
|
\(\begingroup\)\(\newcommand{\diag}{\operatorname{diag}} % diag
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\newcommand{\rg}{\operatorname{rg}} % Rang
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\newcommand{\sign}{\operatorname{sign}} % logit
\newcommand{\AR}{\operatorname{AR}} % AR-Prozess
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\newcommand{\ARMA}{\operatorname{ARMA}} % ARMA-Prozess
\newcommand{\VAR}{\operatorname{VAR}} % VAR-Prozess
\newcommand{\spn}{\operatorname{span}} % span
%****************************************************************
%************************** Abkuerzungen ************************
%****************************************************************
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%****************************************************************
%************************** Miszellanien ************************
%****************************************************************
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\newcommand{\spn}{\operatorname{span}} % span
%****************************************************************
%************************** Abkuerzungen ************************
%****************************************************************
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\newcommand{\veps}{\varepsilon}
%****************************************************************
%************************** Miszellanien ************************
%****************************************************************
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%************** Wahrscheinlichkeitsrechnung *********************
%****************************************************************
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%------------- Fette grosse Buchstaben im Mathemodus -----------
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%------------- Fette kleine Buchstaben im Mathemodus -----------
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%----- Fette Null und fette Eins --------------------------
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%----- Fettes Nabla --------------------------
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%----- Fette grosse griechische Buchstaben im Mathemodus -----
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%----- Fette kleine griechische Buchstaben im Mathemodus -----
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%----------- Kalligraphische Buchstaben im Mathemodus
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\newcommand{\mcV}{\mathcal{V}}
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\newcommand{\mcX}{\mathcal{X}}
\newcommand{\mcY}{\mathcal{Y}}
\newcommand{\mcZ}{\mathcal{Z}}
%----------- Kalligraphische Buchstaben im Mathemodus
\newcommand{\mca}{\mathcal{A}}
\newcommand{\mcb}{\mathcal{B}}
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\newcommand{\mcv}{\mathcal{V}}
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\newcommand{\mcx}{\mathcal{X}}
\newcommand{\mcy}{\mathcal{Y}}
\newcommand{\mcz}{\mathcal{Z}}
%---------- Doppelsumme ------------------------------------------
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%---------- \argmin und \argmax ------------------------------
\newcommand{\argmin}{\ensuremath{\operatorname*{arg{}min}}}
\newcommand{\argmax}{\ensuremath{\operatorname*{arg{}max}}}
%---------- Folgen ------------------------------------------
\newcommand{\avec}[2]{#1_1,\ldots,#1_{#2}}
\newcommand{\avecc}[2]{\ensuremath{#1_1,#1_2,\ldots,#1_{#2}}}
\newcommand{\aveccc}[3]{\ensuremath{(#1_{#2},\ldots,#1_{#3})}}
%---------- Geordnete Werte ---------------------------------------
\newcommand{\ord}[2]{\ensuremath{#1_{(1)},\ldots,#1_{(#2)}}}
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\newcommand{\ordne}[3]{\ensuremath{#1_{(1)}#3 #1_{(2)}#3\dots#3#1_{(#2)}}}
\)
\quoteon(2023-06-05 15:09 - Carly2004 in Beitrag No. 2)
Ich kann damit leider immernoch nicht viel Anfangen. :(
\quoteoff
Die Verteilungsfunktion ist damit
\[P(S_n\le x)=1-\sum_{j=0}^{n-1}\frac{(\lambda x)^j}{j!}e^{-\lambda x}\,,\quad x>0\,.\]
\(\endgroup\)
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Profil
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luis52
Senior  Dabei seit: 24.12.2018 Mitteilungen: 1111
 | Beitrag No.4, eingetragen 2023-06-06
|
Vielleicht kommt man auch alternativ mit der Markowschen Ungleichung weiter.
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Profil
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Carly2004
Wenig Aktiv  Dabei seit: 26.06.2022 Mitteilungen: 37
 | Beitrag No.5, vom Themenstarter, eingetragen 2023-06-08
|
Also ich ahbe es durchgerechnet, ich habe als Ergebnis:
c=-log(1+\delta)-\lambda+1+\delta
Kann das so stimmen
|
Profil
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luis52
Senior  Dabei seit: 24.12.2018 Mitteilungen: 1111
 | Beitrag No.6, eingetragen 2023-06-08
|
\(\begingroup\)\(\newcommand{\diag}{\operatorname{diag}} % diag
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%****************************************************************
%************************** Abkuerzungen ************************
%****************************************************************
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%****************************************************************
%************************** Miszellanien ************************
%****************************************************************
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\newcommand{\VAR}{\operatorname{VAR}} % VAR-Prozess
\newcommand{\spn}{\operatorname{span}} % span
%****************************************************************
%************************** Abkuerzungen ************************
%****************************************************************
\newcommand{\eps}{\epsilon}
\newcommand{\veps}{\varepsilon}
%****************************************************************
%************************** Miszellanien ************************
%****************************************************************
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%****************************************************************
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%****************************************************************
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%------------- Fette grosse Buchstaben im Mathemodus -----------
\newcommand{\mbA}{\mathbf{A}}
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\newcommand{\mbZ}{\mathbf{Z}}
%------------- Fette kleine Buchstaben im Mathemodus -----------
\newcommand{\mba}{\mathbf{a}}
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\newcommand{\mbz}{\mathbf{z}}
%----- Fette Null und fette Eins --------------------------
\newcommand{\fatnull}{\mathbf{0}}
\newcommand{\fateins}{\mathbf{1}}
%----- Fettes Nabla --------------------------
\newcommand{\bnabla}{\boldsymbol{\nabla}}
%----- Fette grosse griechische Buchstaben im Mathemodus -----
\newcommand{\bGamma}{\boldsymbol{\Gamma}}
\newcommand{\bDelta}{\boldsymbol{\Delta}}
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\newcommand{\bLambda}{\boldsymbol{\Lambda}}
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\newcommand{\bSigma}{\boldsymbol{\Sigma}}
\newcommand{\bUpsilon}{\boldsymbol{\Upsilon}}
\newcommand{\bPhi}{\boldsymbol{\Phi}}
\newcommand{\bPsi}{\boldsymbol{\Psi}}
\newcommand{\bOmega}{\boldsymbol{\Omega}}
%----- Fette kleine griechische Buchstaben im Mathemodus -----
\newcommand{\balpha}{\boldsymbol{\alpha}}
\newcommand{\bbeta}{\boldsymbol{\beta}}
\newcommand{\bgamma}{\boldsymbol{\gamma}}
\newcommand{\bdelta}{\boldsymbol{\delta}}
\newcommand{\beps}{\boldsymbol{\varepsilon}}
\newcommand{\bzeta}{\boldsymbol{\zeta}}
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\newcommand{\bvartheta}{\boldsymbol{\vartheta}}
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\newcommand{\bvphi}{\boldsymbol{\varphi}}
\newcommand{\bchi}{\boldsymbol{\chi}}
\newcommand{\bpsi}{\boldsymbol{\psi}}
\newcommand{\bomega}{\boldsymbol{\omega}}
%----------- Kalligraphische Buchstaben im Mathemodus
\newcommand{\mcA}{\mathcal{A}}
\newcommand{\mcB}{\mathcal{B}}
\newcommand{\mcC}{\mathcal{C}}
\newcommand{\mcD}{\mathcal{D}}
\newcommand{\mcE}{\mathcal{E}}
\newcommand{\mcF}{\mathcal{F}}
\newcommand{\mcG}{\mathcal{G}}
\newcommand{\mcH}{\mathcal{H}}
\newcommand{\mcI}{\mathcal{I}}
\newcommand{\mcJ}{\mathcal{J}}
\newcommand{\mcK}{\mathcal{K}}
\newcommand{\mcL}{\mathcal{L}}
\newcommand{\mcM}{\mathcal{M}}
\newcommand{\mcN}{\mathcal{N}}
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\newcommand{\mcP}{\mathcal{P}}
\newcommand{\mcQ}{\mathcal{Q}}
\newcommand{\mcR}{\mathcal{R}}
\newcommand{\mcS}{\mathcal{S}}
\newcommand{\mcT}{\mathcal{T}}
\newcommand{\mcU}{\mathcal{U}}
\newcommand{\mcV}{\mathcal{V}}
\newcommand{\mcW}{\mathcal{W}}
\newcommand{\mcX}{\mathcal{X}}
\newcommand{\mcY}{\mathcal{Y}}
\newcommand{\mcZ}{\mathcal{Z}}
%----------- Kalligraphische Buchstaben im Mathemodus
\newcommand{\mca}{\mathcal{A}}
\newcommand{\mcb}{\mathcal{B}}
\newcommand{\mcc}{\mathcal{C}}
\newcommand{\mcd}{\mathcal{D}}
\newcommand{\mce}{\mathcal{E}}
\newcommand{\mcf}{\mathcal{F}}
\newcommand{\mcg}{\mathcal{G}}
\newcommand{\mch}{\mathcal{H}}
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\newcommand{\mcj}{\mathcal{J}}
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\newcommand{\mcl}{\mathcal{L}}
\newcommand{\mcm}{\mathcal{M}}
\newcommand{\mcn}{\mathcal{N}}
\newcommand{\mco}{\mathcal{O}}
\newcommand{\mcp}{\mathcal{P}}
\newcommand{\mcq}{\mathcal{Q}}
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\newcommand{\mcs}{\mathcal{S}}
\newcommand{\mct}{\mathcal{T}}
\newcommand{\mcu}{\mathcal{U}}
\newcommand{\mcv}{\mathcal{V}}
\newcommand{\mcw}{\mathcal{W}}
\newcommand{\mcx}{\mathcal{X}}
\newcommand{\mcy}{\mathcal{Y}}
\newcommand{\mcz}{\mathcal{Z}}
%---------- Doppelsumme ------------------------------------------
\newcommand{\dblsum}[1]{\ensuremath{\mathop{\sum\sum}\limits_{#1}}}
%---------- \argmin und \argmax ------------------------------
\newcommand{\argmin}{\ensuremath{\operatorname*{arg{}min}}}
\newcommand{\argmax}{\ensuremath{\operatorname*{arg{}max}}}
%---------- Folgen ------------------------------------------
\newcommand{\avec}[2]{#1_1,\ldots,#1_{#2}}
\newcommand{\avecc}[2]{\ensuremath{#1_1,#1_2,\ldots,#1_{#2}}}
\newcommand{\aveccc}[3]{\ensuremath{(#1_{#2},\ldots,#1_{#3})}}
%---------- Geordnete Werte ---------------------------------------
\newcommand{\ord}[2]{\ensuremath{#1_{(1)},\ldots,#1_{(#2)}}}
\newcommand{\ordd}[2]{\ensuremath{#1_{(1)}\le#1_{(2)}\le\cdots\le#1_{(#2)}}}
\newcommand{\ordstat}[2]{\ensuremath{#1_{(#2)}}}
\newcommand{\ordne}[3]{\ensuremath{#1_{(1)}#3 #1_{(2)}#3\dots#3#1_{(#2)}}}
\)
\quoteon(2023-06-08 17:03 - Carly2004 in Beitrag No. 5)
Also ich ahbe es durchgerechnet, ich habe als Ergebnis:
c=-log(1+\delta)-\lambda+1+\delta
Kann das so stimmen
\quoteoff
Hm, "deine" obere Schranke lautet also
\[\exp(-cn)=\exp(-(-\log(1+\delta)-\lambda+1+\delta)n))\,,\]
korrekt?
Fuer $n=2$, $\lambda=2$ und $\delta=1$ erhaelt man so:
\sourceon R
R> n <-2
R> lambda <- 2
R> delta <-1
R> exp(-(-log(1+delta)-lambda+1+delta)*n)
[1] 4
\sourceoff
Arg konservativ, findest du nicht?
Wie waere es, wenn du deine Herleitung mal vorstellst?
\(\endgroup\)
|
Profil
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Carly2004
Wenig Aktiv  Dabei seit: 26.06.2022 Mitteilungen: 37
 | Beitrag No.7, vom Themenstarter, eingetragen 2023-06-09
|
Ok.
P(1/n*S_n>=(1+\delta)E(X_1))
=P(S_n>=n*(1+\delta)E(X_1))
<=exp[-tn(1+\delta)E(X_1)]E(exp[t*S_n])
=exp[-tn(1+\delta)E(X_1)]M_(X1)^n
=exp[n(logM_(X1)(t))-t(1+\delta)E(X_1))]
Minimiere g(t)=logM_(X1)(t))-t(1+\delta)E(X_1))=log(\lambda/(\lambda-t))-t(1+\delta)1/\lambda
Da habe ich heraus t=\lambda-\lambda^2/(1+\delta)<\lambda (muss wegen M_(X1) gelten)
Dann habe ich das t eingesetzt:
=exp[n(log(\lambda/(\lambda+\lambda^2/(1+\delta)-\lambda)+(\lambda^2/(1+\delta)-\lambda)(1+\delta)1/\lambda)]
=exp[n*log(1+\delta)+\lambda-(1+\delta))]
=exp[-(-log(1+\delta)-\lambda+(1+\delta))n]
=>c=-log(1+\delta)-\lambda+(1+\delta)
|
Profil
|
luis52
Senior  Dabei seit: 24.12.2018 Mitteilungen: 1111
 | Beitrag No.8, eingetragen 2023-06-09
|
\(\begingroup\)\(\newcommand{\diag}{\operatorname{diag}} % diag
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%****************************************************************
%************************** Abkuerzungen ************************
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%************************** Miszellanien ************************
%****************************************************************
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\newcommand{\spn}{\operatorname{span}} % span
%****************************************************************
%************************** Abkuerzungen ************************
%****************************************************************
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%****************************************************************
%************************** Miszellanien ************************
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%------------- Fette grosse Buchstaben im Mathemodus -----------
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%------------- Fette kleine Buchstaben im Mathemodus -----------
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%----- Fette grosse griechische Buchstaben im Mathemodus -----
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%----- Fette kleine griechische Buchstaben im Mathemodus -----
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%----------- Kalligraphische Buchstaben im Mathemodus
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\newcommand{\mcX}{\mathcal{X}}
\newcommand{\mcY}{\mathcal{Y}}
\newcommand{\mcZ}{\mathcal{Z}}
%----------- Kalligraphische Buchstaben im Mathemodus
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\newcommand{\mcy}{\mathcal{Y}}
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%---------- Doppelsumme ------------------------------------------
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%---------- \argmin und \argmax ------------------------------
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%---------- Folgen ------------------------------------------
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%---------- Geordnete Werte ---------------------------------------
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\)
Die erste Ungleichung
\[P(S_n\ge n(1+\delta)\E(X_1))\le
\exp[-tn(1+\delta)\E(X_1)]\E(\exp[tS_n])\]
verstehe ich ueberhaupt nicht. Wo kommt sie her? Was ist $t$? Uebrigens $\E(X_1)=1/\lambda$ ... \(\endgroup\)
|
Profil
|
Carly2004
Wenig Aktiv  Dabei seit: 26.06.2022 Mitteilungen: 37
 | Beitrag No.9, vom Themenstarter, eingetragen 2023-06-09
|
Aus der Markov-Ungleichung mit
f(x)=e^tx mit t>0
|
Profil
|
luis52
Senior  Dabei seit: 24.12.2018 Mitteilungen: 1111
 | Beitrag No.10, eingetragen 2023-06-09
|
\(\begingroup\)\(\newcommand{\diag}{\operatorname{diag}} % diag
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\newcommand{\spn}{\operatorname{span}} % span
%****************************************************************
%************************** Abkuerzungen ************************
%****************************************************************
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%****************************************************************
%************************** Miszellanien ************************
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%****************************************************************
%************************** Abkuerzungen ************************
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%****************************************************************
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%------------- Fette grosse Buchstaben im Mathemodus -----------
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%----- Fette Null und fette Eins --------------------------
\newcommand{\fatnull}{\mathbf{0}}
\newcommand{\fateins}{\mathbf{1}}
%----- Fettes Nabla --------------------------
\newcommand{\bnabla}{\boldsymbol{\nabla}}
%----- Fette grosse griechische Buchstaben im Mathemodus -----
\newcommand{\bGamma}{\boldsymbol{\Gamma}}
\newcommand{\bDelta}{\boldsymbol{\Delta}}
\newcommand{\bTheta}{\boldsymbol{\Theta}}
\newcommand{\bLambda}{\boldsymbol{\Lambda}}
\newcommand{\bXi}{\boldsymbol{\Xi}}
\newcommand{\bPi}{\boldsymbol{\Pi}}
\newcommand{\bSigma}{\boldsymbol{\Sigma}}
\newcommand{\bUpsilon}{\boldsymbol{\Upsilon}}
\newcommand{\bPhi}{\boldsymbol{\Phi}}
\newcommand{\bPsi}{\boldsymbol{\Psi}}
\newcommand{\bOmega}{\boldsymbol{\Omega}}
%----- Fette kleine griechische Buchstaben im Mathemodus -----
\newcommand{\balpha}{\boldsymbol{\alpha}}
\newcommand{\bbeta}{\boldsymbol{\beta}}
\newcommand{\bgamma}{\boldsymbol{\gamma}}
\newcommand{\bdelta}{\boldsymbol{\delta}}
\newcommand{\beps}{\boldsymbol{\varepsilon}}
\newcommand{\bzeta}{\boldsymbol{\zeta}}
\newcommand{\boldeta}{\boldsymbol{\eta}}
\newcommand{\btheta}{\boldsymbol{\theta}}
\newcommand{\bvartheta}{\boldsymbol{\vartheta}}
\newcommand{\biota}{\boldsymbol{\iota}}
\newcommand{\bkappa}{\boldsymbol{\kappa}}
\newcommand{\blambda}{\boldsymbol{\lambda}}
\newcommand{\bmu}{\boldsymbol{\mu}}
\newcommand{\bnu}{\boldsymbol{\nu}}
\newcommand{\bxi}{\boldsymbol{\xi}}
\newcommand{\bpi}{\boldsymbol{\pi}}
\newcommand{\brho}{\boldsymbol{\rho}}
\newcommand{\bvrho}{\boldsymbol{\varrho}}
\newcommand{\bsigma}{\boldsymbol{\sigma}}
\newcommand{\btau}{\boldsymbol{\tau}}
\newcommand{\bupsilon}{\boldsymbol{\upsilon}}
\newcommand{\bphi}{\boldsymbol{\phi}}
\newcommand{\bvphi}{\boldsymbol{\varphi}}
\newcommand{\bchi}{\boldsymbol{\chi}}
\newcommand{\bpsi}{\boldsymbol{\psi}}
\newcommand{\bomega}{\boldsymbol{\omega}}
%----------- Kalligraphische Buchstaben im Mathemodus
\newcommand{\mcA}{\mathcal{A}}
\newcommand{\mcB}{\mathcal{B}}
\newcommand{\mcC}{\mathcal{C}}
\newcommand{\mcD}{\mathcal{D}}
\newcommand{\mcE}{\mathcal{E}}
\newcommand{\mcF}{\mathcal{F}}
\newcommand{\mcG}{\mathcal{G}}
\newcommand{\mcH}{\mathcal{H}}
\newcommand{\mcI}{\mathcal{I}}
\newcommand{\mcJ}{\mathcal{J}}
\newcommand{\mcK}{\mathcal{K}}
\newcommand{\mcL}{\mathcal{L}}
\newcommand{\mcM}{\mathcal{M}}
\newcommand{\mcN}{\mathcal{N}}
\newcommand{\mcO}{\mathcal{O}}
\newcommand{\mcP}{\mathcal{P}}
\newcommand{\mcQ}{\mathcal{Q}}
\newcommand{\mcR}{\mathcal{R}}
\newcommand{\mcS}{\mathcal{S}}
\newcommand{\mcT}{\mathcal{T}}
\newcommand{\mcU}{\mathcal{U}}
\newcommand{\mcV}{\mathcal{V}}
\newcommand{\mcW}{\mathcal{W}}
\newcommand{\mcX}{\mathcal{X}}
\newcommand{\mcY}{\mathcal{Y}}
\newcommand{\mcZ}{\mathcal{Z}}
%----------- Kalligraphische Buchstaben im Mathemodus
\newcommand{\mca}{\mathcal{A}}
\newcommand{\mcb}{\mathcal{B}}
\newcommand{\mcc}{\mathcal{C}}
\newcommand{\mcd}{\mathcal{D}}
\newcommand{\mce}{\mathcal{E}}
\newcommand{\mcf}{\mathcal{F}}
\newcommand{\mcg}{\mathcal{G}}
\newcommand{\mch}{\mathcal{H}}
\newcommand{\mci}{\mathcal{I}}
\newcommand{\mcj}{\mathcal{J}}
\newcommand{\mck}{\mathcal{K}}
\newcommand{\mcl}{\mathcal{L}}
\newcommand{\mcm}{\mathcal{M}}
\newcommand{\mcn}{\mathcal{N}}
\newcommand{\mco}{\mathcal{O}}
\newcommand{\mcp}{\mathcal{P}}
\newcommand{\mcq}{\mathcal{Q}}
\newcommand{\mcr}{\mathcal{R}}
\newcommand{\mcs}{\mathcal{S}}
\newcommand{\mct}{\mathcal{T}}
\newcommand{\mcu}{\mathcal{U}}
\newcommand{\mcv}{\mathcal{V}}
\newcommand{\mcw}{\mathcal{W}}
\newcommand{\mcx}{\mathcal{X}}
\newcommand{\mcy}{\mathcal{Y}}
\newcommand{\mcz}{\mathcal{Z}}
%---------- Doppelsumme ------------------------------------------
\newcommand{\dblsum}[1]{\ensuremath{\mathop{\sum\sum}\limits_{#1}}}
%---------- \argmin und \argmax ------------------------------
\newcommand{\argmin}{\ensuremath{\operatorname*{arg{}min}}}
\newcommand{\argmax}{\ensuremath{\operatorname*{arg{}max}}}
%---------- Folgen ------------------------------------------
\newcommand{\avec}[2]{#1_1,\ldots,#1_{#2}}
\newcommand{\avecc}[2]{\ensuremath{#1_1,#1_2,\ldots,#1_{#2}}}
\newcommand{\aveccc}[3]{\ensuremath{(#1_{#2},\ldots,#1_{#3})}}
%---------- Geordnete Werte ---------------------------------------
\newcommand{\ord}[2]{\ensuremath{#1_{(1)},\ldots,#1_{(#2)}}}
\newcommand{\ordd}[2]{\ensuremath{#1_{(1)}\le#1_{(2)}\le\cdots\le#1_{(#2)}}}
\newcommand{\ordstat}[2]{\ensuremath{#1_{(#2)}}}
\newcommand{\ordne}[3]{\ensuremath{#1_{(1)}#3 #1_{(2)}#3\dots#3#1_{(#2)}}}
\)
\quoteon(2023-06-09 17:10 - Carly2004 in Beitrag No. 9)
Aus der Markov-Ungleichung mit
f(x)=e^tx mit t>0
\quoteoff
Das ergibt fuer mich keinen Sinn.
Hier finde ich fuer eine Zufallsvariable $X$:
\[P(|X|\ge a)\le\frac{\E[|X|]}{a}\]
mit $a>0$. Angewendet auf $S_n=|S_n|$ ergibt sich
\[P\left(S_n\ge\dfrac{n(1+\delta)}{\lambda}\right)\le\frac{\E[S_n]\lambda}{n(1+\delta)}\,.\]
\(\endgroup\)
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