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Description of Statistical Computation

Author's title

Author*The author of this computation has been verified*
R Software Modulerwasp_rwalk.wasp
Title produced by softwareLaw of Averages
Date of computationWed, 26 Nov 2008 09:50:47 -0700
Cite this page as followsStatistical Computations at FreeStatistics.org, Office for Research Development and Education, URL https://freestatistics.org/blog/index.php?v=date/2008/Nov/26/t1227719156gtizlmcyra4oqkr.htm/, Retrieved Sun, 19 May 2024 07:19:58 +0000
Statistical Computations at FreeStatistics.org, Office for Research Development and Education, URL https://freestatistics.org/blog/index.php?pk=25663, Retrieved Sun, 19 May 2024 07:19:58 +0000
QR Codes:

Original text written by user:
IsPrivate?No (this computation is public)
User-defined keywords
Estimated Impact181

Tree of Dependent Computations

Family? (F = Feedback message, R = changed R code, M = changed R Module, P = changed Parameters, D = changed Data)
F     [Law of Averages] [Random Walk Simul...] [2008-11-25 17:50:19] [b98453cac15ba1066b407e146608df68]
F         [Law of Averages] [non stationary ti...] [2008-11-26 16:50:47] [4f54996111e63ee83b19b6a8540c6bad] [Current]
-           [Law of Averages] [non stationary ti...] [2008-11-26 17:06:47] [c65b85921bf03b2616bf1bee11088685]
Feedback Forum
2008-12-05 15:25:19 [Nick Wuyts] [reply] 
De simulatie toont het resultaat dat we bekomen bij het opwerpen van een muntstuk, met een 50% kans dat er kop of let wordt bekomen. De 500 opwerpingen zijn random (hence de naam random walk), dwz dat er bij elke reproduce een andere oplossing wordt bekomen (the 'patterns' in random-walk time series are purely due to chance, hence they cannot be explained or predicted). Als we oneindig veel simulaties doen, dan zou het aantal worpen kop en munt gelijk zijn. Bij een beperkt aantal simulaties is dit dus niet.

We gebruiken de formule Y(t) = Y(t-1) + e(t) voor deze vraagstelling, waarbij Y(t-1) staat voor 1 eenheid in de tijd terug, en e(t) het toeval is (gemiddeld is dit 0, gezien de kans op kop en munt hetzelfde is). We leiden af dat de Forecast = Y(t) – e(t) = Y(t-1).

Hieruit concluderen we dat de beste voorspelling een rechte lijn is (2e grafiek telkens). Er is geen seizoenaliteit (vanwege het willekeurige karakter van e(t), maar wel een trendmatig verloop op lange termijn aanwezig.

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Dataset

Tables (Output of Computation)





Summary of computational transaction
Raw Inputview raw input (R code)
Raw Outputview raw output of R engine
Computing time1 seconds
R Server'George Udny Yule' @ 72.249.76.132

\begin{tabular}{lllllllll}
\hline
Summary of computational transaction \tabularnewline
Raw Input & view raw input (R code)  \tabularnewline
Raw Output & view raw output of R engine  \tabularnewline
Computing time & 1 seconds \tabularnewline
R Server & 'George Udny Yule' @ 72.249.76.132 \tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=25663&T=0

[TABLE]
[ROW][C]Summary of computational transaction[/C][/ROW]
[ROW][C]Raw Input[/C][C]view raw input (R code) [/C][/ROW]
[ROW][C]Raw Output[/C][C]view raw output of R engine [/C][/ROW]
[ROW][C]Computing time[/C][C]1 seconds[/C][/ROW]
[ROW][C]R Server[/C][C]'George Udny Yule' @ 72.249.76.132[/C][/ROW]
[/TABLE]
Source: https://freestatistics.org/blog/index.php?pk=25663&T=0

Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=25663&T=0

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The GUIDs for individual cells are displayed in the table below:

Summary of computational transaction
Raw Inputview raw input (R code)
Raw Outputview raw output of R engine
Computing time1 seconds
R Server'George Udny Yule' @ 72.249.76.132


Figures (Output of Computation)

Input Parameters & R Code

Parameters (Session):
par1 = 500 ; par2 = 0.5 ;
Parameters (R input):
par1 = 500 ; par2 = 0.5 ; par3 = ; par4 = ; par5 = ; par6 = ; par7 = ; par8 = ; par9 = ; par10 = ; par11 = ; par12 = ; par13 = ; par14 = ; par15 = ; par16 = ; par17 = ; par18 = ; par19 = ; par20 = ;
R code (references can be found in the software module):
n <- as.numeric(par1)
p <- as.numeric(par2)
heads=rbinom(n-1,1,p)
a=2*(heads)-1
b=diffinv(a,xi=0)
c=1:n
pheads=(diffinv(heads,xi=.5))/c
bitmap(file='test1.png')
op=par(mfrow=c(2,1))
plot(c,b,type='n',main='Law of Averages',xlab='Toss Number',ylab='Excess of Heads',lwd=2,cex.lab=1.5,cex.main=2)
lines(c,b,col='red')
lines(c,rep(0,n),col='black')
plot(c,pheads,type='n',xlab='Toss Number',ylab='Proportion of Heads',lwd=2,cex.lab=1.5)
lines(c,pheads,col='blue')
lines(c,rep(.5,n),col='black')
par(op)
dev.off()