using Central Limit Theorem to approximate a probability with large 'n'
$begingroup$
For $i>1$ , let $X_i$ ~ $G_{1/2}$ be distributed Geometrically with parameter $1/2$.
$$ Y_n= frac{1}{sqrt n} sum_{r=1}^n X_r-2 $$
Approximate $P(-1le Y_nle 2)$ with large n.($Y_n$ is not properly normalized)
My attempt-
normalised $Y_n = frac{1}{sqrt n} sum_{r=1}^n frac{(X_r-2)-E[X_r-2]}{V[X-2]*n}$
then approximating $P(-1le Y_nle 2)$ I got the answer as 0.4332 which is incorrect. Can anybody give me the correct solution of this question.
statistics central-limit-theorem
asked Dec 8 '18 at 10:15
Kriti AroraKriti Arora
396
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add a comment |
$begingroup$
For $i>1$ , let $X_i$ ~ $G_{1/2}$ be distributed Geometrically with parameter $1/2$.
$$ Y_n= frac{1}{sqrt n} sum_{r=1}^n X_r-2 $$
Approximate $P(-1le Y_nle 2)$ with large n.($Y_n$ is not properly normalized)
My attempt-
normalised $Y_n = frac{1}{sqrt n} sum_{r=1}^n frac{(X_r-2)-E[X_r-2]}{V[X-2]*n}$
then approximating $P(-1le Y_nle 2)$ I got the answer as 0.4332 which is incorrect. Can anybody give me the correct solution of this question.
statistics central-limit-theorem
asked Dec 8 '18 at 10:15
Kriti AroraKriti Arora
396
$endgroup$
$begingroup$
@Damandeep Its an online course in probability and stats
$endgroup$
– Kriti Arora
Dec 14 '18 at 11:43
add a comment |
$begingroup$
For $i>1$ , let $X_i$ ~ $G_{1/2}$ be distributed Geometrically with parameter $1/2$.
$$ Y_n= frac{1}{sqrt n} sum_{r=1}^n X_r-2 $$
Approximate $P(-1le Y_nle 2)$ with large n.($Y_n$ is not properly normalized)
My attempt-
normalised $Y_n = frac{1}{sqrt n} sum_{r=1}^n frac{(X_r-2)-E[X_r-2]}{V[X-2]*n}$
then approximating $P(-1le Y_nle 2)$ I got the answer as 0.4332 which is incorrect. Can anybody give me the correct solution of this question.
statistics central-limit-theorem
asked Dec 8 '18 at 10:15
Kriti AroraKriti Arora
396
$endgroup$
For $i>1$ , let $X_i$ ~ $G_{1/2}$ be distributed Geometrically with parameter $1/2$.
$$ Y_n= frac{1}{sqrt n} sum_{r=1}^n X_r-2 $$
Approximate $P(-1le Y_nle 2)$ with large n.($Y_n$ is not properly normalized)
My attempt-
normalised $Y_n = frac{1}{sqrt n} sum_{r=1}^n frac{(X_r-2)-E[X_r-2]}{V[X-2]*n}$
then approximating $P(-1le Y_nle 2)$ I got the answer as 0.4332 which is incorrect. Can anybody give me the correct solution of this question.
statistics central-limit-theorem
statistics central-limit-theorem
asked Dec 8 '18 at 10:15
Kriti AroraKriti Arora
396
asked Dec 8 '18 at 10:15
Kriti AroraKriti Arora
396
asked Dec 8 '18 at 10:15
Kriti AroraKriti Arora
396
asked Dec 8 '18 at 10:15
Kriti AroraKriti Arora
396
asked Dec 8 '18 at 10:15
Kriti AroraKriti Arora
396
396
$begingroup$
@Damandeep Its an online course in probability and stats
$endgroup$
– Kriti Arora
Dec 14 '18 at 11:43
add a comment |
$begingroup$
@Damandeep Its an online course in probability and stats
$endgroup$
– Kriti Arora
Dec 14 '18 at 11:43
$begingroup$
@Damandeep Its an online course in probability and stats
$endgroup$
– Kriti Arora
Dec 14 '18 at 11:43
$begingroup$
@Damandeep Its an online course in probability and stats
$endgroup$
– Kriti Arora
Dec 14 '18 at 11:43
add a comment |
1 Answer
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Like $X_r-2$, $Y_n$ has mean $0$ and variance $2$. The answer is therefore $Phi(sqrt{2})-Phi(frac{-1}{sqrt{2}})approx 0.6816$.
answered Dec 8 '18 at 10:27
J.G.J.G.
26.3k22541
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add a comment |
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1 Answer
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$begingroup$
Like $X_r-2$, $Y_n$ has mean $0$ and variance $2$. The answer is therefore $Phi(sqrt{2})-Phi(frac{-1}{sqrt{2}})approx 0.6816$.
answered Dec 8 '18 at 10:27
J.G.J.G.
26.3k22541
$endgroup$
add a comment |
$begingroup$
Like $X_r-2$, $Y_n$ has mean $0$ and variance $2$. The answer is therefore $Phi(sqrt{2})-Phi(frac{-1}{sqrt{2}})approx 0.6816$.
answered Dec 8 '18 at 10:27
J.G.J.G.
26.3k22541
$endgroup$
add a comment |
$begingroup$
Like $X_r-2$, $Y_n$ has mean $0$ and variance $2$. The answer is therefore $Phi(sqrt{2})-Phi(frac{-1}{sqrt{2}})approx 0.6816$.
answered Dec 8 '18 at 10:27
J.G.J.G.
26.3k22541
$endgroup$
Like $X_r-2$, $Y_n$ has mean $0$ and variance $2$. The answer is therefore $Phi(sqrt{2})-Phi(frac{-1}{sqrt{2}})approx 0.6816$.
answered Dec 8 '18 at 10:27
J.G.J.G.
26.3k22541
answered Dec 8 '18 at 10:27
J.G.J.G.
26.3k22541
answered Dec 8 '18 at 10:27
J.G.J.G.
26.3k22541
answered Dec 8 '18 at 10:27
J.G.J.G.
26.3k22541
26.3k22541
add a comment |
add a comment |
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$begingroup$
@Damandeep Its an online course in probability and stats
$endgroup$
– Kriti Arora
Dec 14 '18 at 11:43