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

1038




1038












  • $begingroup$
    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












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1 Answer
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1 Answer
1






active

oldest

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active

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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

15.4k1835




15.4k1835












  • $begingroup$
    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


















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