Convergence in distribution of a certain kind of sequence of random variables
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Let ${X_n}$ be a sequence of independent random variables such that $P(X_n=1)=P(X_n=-1)=dfrac 12 -dfrac 1{2n^2}$ and $P(X_n=n)=P(X_n=-n)=dfrac 1{2n^2}$.
Then does the sequence $Y_n=dfrac 1{sqrt n}sum_{k=1}^n X_k$ converge in distribution ? If it does, what is the limiting distribution ?
My work: $E(X_n)=0$ and $Var(X_n)=E(X_n^2)=2-dfrac 1{n^2}$. $E(Y_n^2)=2-dfrac {sum_{k=1}^ndfrac 1{k^2}}{n}$ . So at least there is no constant $c$ such that $Y_n to c$ in $L^2$. I don't know what to do next. Please help
probability-theory probability-distributions random-variables
asked Jan 12 at 11:27
user521337user521337
1,0501415
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add a comment |
$begingroup$
Let ${X_n}$ be a sequence of independent random variables such that $P(X_n=1)=P(X_n=-1)=dfrac 12 -dfrac 1{2n^2}$ and $P(X_n=n)=P(X_n=-n)=dfrac 1{2n^2}$.
Then does the sequence $Y_n=dfrac 1{sqrt n}sum_{k=1}^n X_k$ converge in distribution ? If it does, what is the limiting distribution ?
My work: $E(X_n)=0$ and $Var(X_n)=E(X_n^2)=2-dfrac 1{n^2}$. $E(Y_n^2)=2-dfrac {sum_{k=1}^ndfrac 1{k^2}}{n}$ . So at least there is no constant $c$ such that $Y_n to c$ in $L^2$. I don't know what to do next. Please help
probability-theory probability-distributions random-variables
asked Jan 12 at 11:27
user521337user521337
1,0501415
$endgroup$
add a comment |
$begingroup$
Let ${X_n}$ be a sequence of independent random variables such that $P(X_n=1)=P(X_n=-1)=dfrac 12 -dfrac 1{2n^2}$ and $P(X_n=n)=P(X_n=-n)=dfrac 1{2n^2}$.
Then does the sequence $Y_n=dfrac 1{sqrt n}sum_{k=1}^n X_k$ converge in distribution ? If it does, what is the limiting distribution ?
My work: $E(X_n)=0$ and $Var(X_n)=E(X_n^2)=2-dfrac 1{n^2}$. $E(Y_n^2)=2-dfrac {sum_{k=1}^ndfrac 1{k^2}}{n}$ . So at least there is no constant $c$ such that $Y_n to c$ in $L^2$. I don't know what to do next. Please help
probability-theory probability-distributions random-variables
asked Jan 12 at 11:27
user521337user521337
1,0501415
$endgroup$
Let ${X_n}$ be a sequence of independent random variables such that $P(X_n=1)=P(X_n=-1)=dfrac 12 -dfrac 1{2n^2}$ and $P(X_n=n)=P(X_n=-n)=dfrac 1{2n^2}$.
Then does the sequence $Y_n=dfrac 1{sqrt n}sum_{k=1}^n X_k$ converge in distribution ? If it does, what is the limiting distribution ?
My work: $E(X_n)=0$ and $Var(X_n)=E(X_n^2)=2-dfrac 1{n^2}$. $E(Y_n^2)=2-dfrac {sum_{k=1}^ndfrac 1{k^2}}{n}$ . So at least there is no constant $c$ such that $Y_n to c$ in $L^2$. I don't know what to do next. Please help
probability-theory probability-distributions random-variables
probability-theory probability-distributions random-variables
asked Jan 12 at 11:27
user521337user521337
1,0501415
asked Jan 12 at 11:27
user521337user521337
1,0501415
asked Jan 12 at 11:27
user521337user521337
1,0501415
asked Jan 12 at 11:27
user521337user521337
1,0501415
asked Jan 12 at 11:27
user521337user521337
1,0501415
1,0501415
add a comment |
add a comment |
1 Answer
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$begingroup$
${Y_n}$ converges to $N(0,1)$ in distribution. To prove this let $Z_n=1$ if $X_nin {1,n}$ and $Z_n=-1$ if $X_nin {-1,-n}$. It is easy to check that $Z_n$ takes the values $pm 1$ with probability $frac 1 2 $ each. Of course, $Z_n$'s are independent. Hence, by CLT, $frac 1 {sqrt n} sumlimits_{k=1}^{n} Z_k to N(0,1)$ in distribution. Now consider $frac 1 {sqrt n} sumlimits_{k=1}^{n} (X_k-Z_k)$. Observe that $sumlimits_{k=1}^{infty} P{X_k neq Z_n} <infty$. By Borel Cantelli Lemma it follows that $X_k=Z_k$ eventually with probability $1$, so $frac 1 {sqrt n} sumlimits_{k=1}^{n} (X_k-Z_k) to 0$ with probability $1$. This completes the proof.
answered Jan 12 at 11:50
Kavi Rama MurthyKavi Rama Murthy
56k42158
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1 Answer
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1 Answer
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$begingroup$
${Y_n}$ converges to $N(0,1)$ in distribution. To prove this let $Z_n=1$ if $X_nin {1,n}$ and $Z_n=-1$ if $X_nin {-1,-n}$. It is easy to check that $Z_n$ takes the values $pm 1$ with probability $frac 1 2 $ each. Of course, $Z_n$'s are independent. Hence, by CLT, $frac 1 {sqrt n} sumlimits_{k=1}^{n} Z_k to N(0,1)$ in distribution. Now consider $frac 1 {sqrt n} sumlimits_{k=1}^{n} (X_k-Z_k)$. Observe that $sumlimits_{k=1}^{infty} P{X_k neq Z_n} <infty$. By Borel Cantelli Lemma it follows that $X_k=Z_k$ eventually with probability $1$, so $frac 1 {sqrt n} sumlimits_{k=1}^{n} (X_k-Z_k) to 0$ with probability $1$. This completes the proof.
answered Jan 12 at 11:50
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
add a comment |
$begingroup$
${Y_n}$ converges to $N(0,1)$ in distribution. To prove this let $Z_n=1$ if $X_nin {1,n}$ and $Z_n=-1$ if $X_nin {-1,-n}$. It is easy to check that $Z_n$ takes the values $pm 1$ with probability $frac 1 2 $ each. Of course, $Z_n$'s are independent. Hence, by CLT, $frac 1 {sqrt n} sumlimits_{k=1}^{n} Z_k to N(0,1)$ in distribution. Now consider $frac 1 {sqrt n} sumlimits_{k=1}^{n} (X_k-Z_k)$. Observe that $sumlimits_{k=1}^{infty} P{X_k neq Z_n} <infty$. By Borel Cantelli Lemma it follows that $X_k=Z_k$ eventually with probability $1$, so $frac 1 {sqrt n} sumlimits_{k=1}^{n} (X_k-Z_k) to 0$ with probability $1$. This completes the proof.
answered Jan 12 at 11:50
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
add a comment |
$begingroup$
${Y_n}$ converges to $N(0,1)$ in distribution. To prove this let $Z_n=1$ if $X_nin {1,n}$ and $Z_n=-1$ if $X_nin {-1,-n}$. It is easy to check that $Z_n$ takes the values $pm 1$ with probability $frac 1 2 $ each. Of course, $Z_n$'s are independent. Hence, by CLT, $frac 1 {sqrt n} sumlimits_{k=1}^{n} Z_k to N(0,1)$ in distribution. Now consider $frac 1 {sqrt n} sumlimits_{k=1}^{n} (X_k-Z_k)$. Observe that $sumlimits_{k=1}^{infty} P{X_k neq Z_n} <infty$. By Borel Cantelli Lemma it follows that $X_k=Z_k$ eventually with probability $1$, so $frac 1 {sqrt n} sumlimits_{k=1}^{n} (X_k-Z_k) to 0$ with probability $1$. This completes the proof.
answered Jan 12 at 11:50
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
${Y_n}$ converges to $N(0,1)$ in distribution. To prove this let $Z_n=1$ if $X_nin {1,n}$ and $Z_n=-1$ if $X_nin {-1,-n}$. It is easy to check that $Z_n$ takes the values $pm 1$ with probability $frac 1 2 $ each. Of course, $Z_n$'s are independent. Hence, by CLT, $frac 1 {sqrt n} sumlimits_{k=1}^{n} Z_k to N(0,1)$ in distribution. Now consider $frac 1 {sqrt n} sumlimits_{k=1}^{n} (X_k-Z_k)$. Observe that $sumlimits_{k=1}^{infty} P{X_k neq Z_n} <infty$. By Borel Cantelli Lemma it follows that $X_k=Z_k$ eventually with probability $1$, so $frac 1 {sqrt n} sumlimits_{k=1}^{n} (X_k-Z_k) to 0$ with probability $1$. This completes the proof.
answered Jan 12 at 11:50
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Jan 12 at 11:50
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Jan 12 at 11:50
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Jan 12 at 11:50
Kavi Rama MurthyKavi Rama Murthy
56k42158
56k42158
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