Asymptotic growth of translation in random walk.












-1












$begingroup$


Let's consider a random walk with translation $$F(N)=sum_{n=1}^{N}X(n)$$
Where $X(n)$ is a random variable distribiuted independently and equal to $0$ or $1$.



What is asymptotic growth of $F(N)$?



i.e.



I am looking for function $f$ which satisfy:$$F(N)=O(f(N))$$
Is this true that $f(N)=N^{0.5}$



I hope that someone will help me.






asymptotics random-walk







share|cite|improve this question









$endgroup$












  • $begingroup$
    To be clear, $X$ has a probability of $1/2$ being $0$ and $1/2$ being $1$, correct? It's not really clear from your wording.
    $endgroup$
    – Noble Mushtak
    Dec 28 '18 at 21:35
















-1












$begingroup$


Let's consider a random walk with translation $$F(N)=sum_{n=1}^{N}X(n)$$
Where $X(n)$ is a random variable distribiuted independently and equal to $0$ or $1$.



What is asymptotic growth of $F(N)$?



i.e.



I am looking for function $f$ which satisfy:$$F(N)=O(f(N))$$
Is this true that $f(N)=N^{0.5}$



I hope that someone will help me.






asymptotics random-walk







share|cite|improve this question









$endgroup$












  • $begingroup$
    To be clear, $X$ has a probability of $1/2$ being $0$ and $1/2$ being $1$, correct? It's not really clear from your wording.
    $endgroup$
    – Noble Mushtak
    Dec 28 '18 at 21:35














-1












-1








-1





$begingroup$


Let's consider a random walk with translation $$F(N)=sum_{n=1}^{N}X(n)$$
Where $X(n)$ is a random variable distribiuted independently and equal to $0$ or $1$.



What is asymptotic growth of $F(N)$?



i.e.



I am looking for function $f$ which satisfy:$$F(N)=O(f(N))$$
Is this true that $f(N)=N^{0.5}$



I hope that someone will help me.






asymptotics random-walk







share|cite|improve this question









$endgroup$




Let's consider a random walk with translation $$F(N)=sum_{n=1}^{N}X(n)$$
Where $X(n)$ is a random variable distribiuted independently and equal to $0$ or $1$.



What is asymptotic growth of $F(N)$?



i.e.



I am looking for function $f$ which satisfy:$$F(N)=O(f(N))$$
Is this true that $f(N)=N^{0.5}$



I hope that someone will help me.






asymptotics random-walk




asymptotics random-walk






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 28 '18 at 21:30









mkultramkultra

708




708












  • $begingroup$
    To be clear, $X$ has a probability of $1/2$ being $0$ and $1/2$ being $1$, correct? It's not really clear from your wording.
    $endgroup$
    – Noble Mushtak
    Dec 28 '18 at 21:35


















  • $begingroup$
    To be clear, $X$ has a probability of $1/2$ being $0$ and $1/2$ being $1$, correct? It's not really clear from your wording.
    $endgroup$
    – Noble Mushtak
    Dec 28 '18 at 21:35
















$begingroup$
To be clear, $X$ has a probability of $1/2$ being $0$ and $1/2$ being $1$, correct? It's not really clear from your wording.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 21:35




$begingroup$
To be clear, $X$ has a probability of $1/2$ being $0$ and $1/2$ being $1$, correct? It's not really clear from your wording.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 21:35










1 Answer
1






active

oldest

votes


















1












$begingroup$

First, notice that $mu_X=frac 1 2$ and $sigma_X=frac 1 2(0-frac 1 2)^2+frac 1 2(1-frac 1 2)^2=frac 1 4$.



Let's consider:



$$G(N)=frac{F(N)}{N}=frac 1 N sum_{n=1}^N X(n)$$



By the Central Limit Theorem, as $Nrightarrow infty $, $G$ becomes normally distributed with $mu_G=mu_X=frac 1 2$ and $sigma_G=frac{sigma_X}{sqrt N}=frac{1}{4sqrt N}$



Now, $F=NG$. Therefore, as $Nrightarrow infty$, $F$ becomes normally distributed with $mu_F=Nmu_G=frac N 2$ and $sigma_F=Nsigma_G=frac{sqrt N}{4}$. Therefore, the expected value of $F$ is $O(N)$ and the standard deviation of $F$ is $O(sqrt N)$.



In short, in the average-case scenario, $F$ is $O(N)$. Also, in the worst-case scenario, $F=N$, so $F$ is $O(N)$ in worst-case scenario as well.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But how the expected value determines "big O" notation?
    $endgroup$
    – mkultra
    Dec 29 '18 at 12:20










  • $begingroup$
    @mkultra Well, really the worst-case scenario determines big O notation, but sometimes, you will see people talk about average-case complexity as well: en.wikipedia.org/wiki/Average-case_complexity
    $endgroup$
    – Noble Mushtak
    Dec 29 '18 at 15:09











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055308%2fasymptotic-growth-of-translation-in-random-walk%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

First, notice that $mu_X=frac 1 2$ and $sigma_X=frac 1 2(0-frac 1 2)^2+frac 1 2(1-frac 1 2)^2=frac 1 4$.



Let's consider:



$$G(N)=frac{F(N)}{N}=frac 1 N sum_{n=1}^N X(n)$$



By the Central Limit Theorem, as $Nrightarrow infty $, $G$ becomes normally distributed with $mu_G=mu_X=frac 1 2$ and $sigma_G=frac{sigma_X}{sqrt N}=frac{1}{4sqrt N}$



Now, $F=NG$. Therefore, as $Nrightarrow infty$, $F$ becomes normally distributed with $mu_F=Nmu_G=frac N 2$ and $sigma_F=Nsigma_G=frac{sqrt N}{4}$. Therefore, the expected value of $F$ is $O(N)$ and the standard deviation of $F$ is $O(sqrt N)$.



In short, in the average-case scenario, $F$ is $O(N)$. Also, in the worst-case scenario, $F=N$, so $F$ is $O(N)$ in worst-case scenario as well.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But how the expected value determines "big O" notation?
    $endgroup$
    – mkultra
    Dec 29 '18 at 12:20










  • $begingroup$
    @mkultra Well, really the worst-case scenario determines big O notation, but sometimes, you will see people talk about average-case complexity as well: en.wikipedia.org/wiki/Average-case_complexity
    $endgroup$
    – Noble Mushtak
    Dec 29 '18 at 15:09
















1












$begingroup$

First, notice that $mu_X=frac 1 2$ and $sigma_X=frac 1 2(0-frac 1 2)^2+frac 1 2(1-frac 1 2)^2=frac 1 4$.



Let's consider:



$$G(N)=frac{F(N)}{N}=frac 1 N sum_{n=1}^N X(n)$$



By the Central Limit Theorem, as $Nrightarrow infty $, $G$ becomes normally distributed with $mu_G=mu_X=frac 1 2$ and $sigma_G=frac{sigma_X}{sqrt N}=frac{1}{4sqrt N}$



Now, $F=NG$. Therefore, as $Nrightarrow infty$, $F$ becomes normally distributed with $mu_F=Nmu_G=frac N 2$ and $sigma_F=Nsigma_G=frac{sqrt N}{4}$. Therefore, the expected value of $F$ is $O(N)$ and the standard deviation of $F$ is $O(sqrt N)$.



In short, in the average-case scenario, $F$ is $O(N)$. Also, in the worst-case scenario, $F=N$, so $F$ is $O(N)$ in worst-case scenario as well.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But how the expected value determines "big O" notation?
    $endgroup$
    – mkultra
    Dec 29 '18 at 12:20










  • $begingroup$
    @mkultra Well, really the worst-case scenario determines big O notation, but sometimes, you will see people talk about average-case complexity as well: en.wikipedia.org/wiki/Average-case_complexity
    $endgroup$
    – Noble Mushtak
    Dec 29 '18 at 15:09














1












1








1





$begingroup$

First, notice that $mu_X=frac 1 2$ and $sigma_X=frac 1 2(0-frac 1 2)^2+frac 1 2(1-frac 1 2)^2=frac 1 4$.



Let's consider:



$$G(N)=frac{F(N)}{N}=frac 1 N sum_{n=1}^N X(n)$$



By the Central Limit Theorem, as $Nrightarrow infty $, $G$ becomes normally distributed with $mu_G=mu_X=frac 1 2$ and $sigma_G=frac{sigma_X}{sqrt N}=frac{1}{4sqrt N}$



Now, $F=NG$. Therefore, as $Nrightarrow infty$, $F$ becomes normally distributed with $mu_F=Nmu_G=frac N 2$ and $sigma_F=Nsigma_G=frac{sqrt N}{4}$. Therefore, the expected value of $F$ is $O(N)$ and the standard deviation of $F$ is $O(sqrt N)$.



In short, in the average-case scenario, $F$ is $O(N)$. Also, in the worst-case scenario, $F=N$, so $F$ is $O(N)$ in worst-case scenario as well.






share|cite|improve this answer









$endgroup$



First, notice that $mu_X=frac 1 2$ and $sigma_X=frac 1 2(0-frac 1 2)^2+frac 1 2(1-frac 1 2)^2=frac 1 4$.



Let's consider:



$$G(N)=frac{F(N)}{N}=frac 1 N sum_{n=1}^N X(n)$$



By the Central Limit Theorem, as $Nrightarrow infty $, $G$ becomes normally distributed with $mu_G=mu_X=frac 1 2$ and $sigma_G=frac{sigma_X}{sqrt N}=frac{1}{4sqrt N}$



Now, $F=NG$. Therefore, as $Nrightarrow infty$, $F$ becomes normally distributed with $mu_F=Nmu_G=frac N 2$ and $sigma_F=Nsigma_G=frac{sqrt N}{4}$. Therefore, the expected value of $F$ is $O(N)$ and the standard deviation of $F$ is $O(sqrt N)$.



In short, in the average-case scenario, $F$ is $O(N)$. Also, in the worst-case scenario, $F=N$, so $F$ is $O(N)$ in worst-case scenario as well.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 28 '18 at 21:43









Noble MushtakNoble Mushtak

15.3k1835




15.3k1835












  • $begingroup$
    But how the expected value determines "big O" notation?
    $endgroup$
    – mkultra
    Dec 29 '18 at 12:20










  • $begingroup$
    @mkultra Well, really the worst-case scenario determines big O notation, but sometimes, you will see people talk about average-case complexity as well: en.wikipedia.org/wiki/Average-case_complexity
    $endgroup$
    – Noble Mushtak
    Dec 29 '18 at 15:09


















  • $begingroup$
    But how the expected value determines "big O" notation?
    $endgroup$
    – mkultra
    Dec 29 '18 at 12:20










  • $begingroup$
    @mkultra Well, really the worst-case scenario determines big O notation, but sometimes, you will see people talk about average-case complexity as well: en.wikipedia.org/wiki/Average-case_complexity
    $endgroup$
    – Noble Mushtak
    Dec 29 '18 at 15:09
















$begingroup$
But how the expected value determines "big O" notation?
$endgroup$
– mkultra
Dec 29 '18 at 12:20




$begingroup$
But how the expected value determines "big O" notation?
$endgroup$
– mkultra
Dec 29 '18 at 12:20












$begingroup$
@mkultra Well, really the worst-case scenario determines big O notation, but sometimes, you will see people talk about average-case complexity as well: en.wikipedia.org/wiki/Average-case_complexity
$endgroup$
– Noble Mushtak
Dec 29 '18 at 15:09




$begingroup$
@mkultra Well, really the worst-case scenario determines big O notation, but sometimes, you will see people talk about average-case complexity as well: en.wikipedia.org/wiki/Average-case_complexity
$endgroup$
– Noble Mushtak
Dec 29 '18 at 15:09


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055308%2fasymptotic-growth-of-translation-in-random-walk%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more