Chebyshev inequality, lower bound on $P(X ge 200)$
$begingroup$
We throw a die $100$ times. Using Chebyshev inequality find the lower bound on the probability that the sum of spots in these $100$ throws is bigger than $200$.
Let $X = $ number of spots after $100$ throws
We are looking for $t$ such that $P(X ge 200) ge t$
This is a Bernoulli distribution with $n=100$ but what will $p$ be in this case?
Also, Chebyshev inequality immediately gives us this estimation: $P(X ge 200) le t$ and consequently, $P(X < 200) ge t$ but not $P(X ge 200) ge t$.
Could you help me with that $p$ and the estimation?
Thank you!
probability probability-distributions
asked Mar 1 '15 at 9:01
AmithAmith
969
$endgroup$
add a comment |
$begingroup$
We throw a die $100$ times. Using Chebyshev inequality find the lower bound on the probability that the sum of spots in these $100$ throws is bigger than $200$.
Let $X = $ number of spots after $100$ throws
We are looking for $t$ such that $P(X ge 200) ge t$
This is a Bernoulli distribution with $n=100$ but what will $p$ be in this case?
Also, Chebyshev inequality immediately gives us this estimation: $P(X ge 200) le t$ and consequently, $P(X < 200) ge t$ but not $P(X ge 200) ge t$.
Could you help me with that $p$ and the estimation?
Thank you!
probability probability-distributions
asked Mar 1 '15 at 9:01
AmithAmith
969
$endgroup$
1
$begingroup$
The result of the individual throws can be $1,2,3,4,5,6$ with equal probability. That is, this is not a Bernoulli distribution.
$endgroup$
– zoli
Mar 1 '15 at 10:25
$begingroup$
Oh, ok. It is a $6$ point distribution! And here expected value of an individual throw is $1/6(1+2+3+4+5+6)=21/6=3,5$. Is that correct? And standard deviation is $sqrt{21/36}=1/2 sqrt{7/3}$
$endgroup$
– Amith
Mar 1 '15 at 10:46
1
$begingroup$
Corrrect. And the variance can also be calculated. However, I am having the same problem with Cheby that you are... fo rthe time being.
$endgroup$
– zoli
Mar 1 '15 at 10:48
$begingroup$
Ok. But thanks for pointing out that Bernoulli distribution so far.
$endgroup$
– Amith
Mar 1 '15 at 10:49
add a comment |
$begingroup$
We throw a die $100$ times. Using Chebyshev inequality find the lower bound on the probability that the sum of spots in these $100$ throws is bigger than $200$.
Let $X = $ number of spots after $100$ throws
We are looking for $t$ such that $P(X ge 200) ge t$
This is a Bernoulli distribution with $n=100$ but what will $p$ be in this case?
Also, Chebyshev inequality immediately gives us this estimation: $P(X ge 200) le t$ and consequently, $P(X < 200) ge t$ but not $P(X ge 200) ge t$.
Could you help me with that $p$ and the estimation?
Thank you!
probability probability-distributions
asked Mar 1 '15 at 9:01
AmithAmith
969
$endgroup$
We throw a die $100$ times. Using Chebyshev inequality find the lower bound on the probability that the sum of spots in these $100$ throws is bigger than $200$.
Let $X = $ number of spots after $100$ throws
We are looking for $t$ such that $P(X ge 200) ge t$
This is a Bernoulli distribution with $n=100$ but what will $p$ be in this case?
Also, Chebyshev inequality immediately gives us this estimation: $P(X ge 200) le t$ and consequently, $P(X < 200) ge t$ but not $P(X ge 200) ge t$.
Could you help me with that $p$ and the estimation?
Thank you!
probability probability-distributions
probability probability-distributions
asked Mar 1 '15 at 9:01
AmithAmith
969
asked Mar 1 '15 at 9:01
AmithAmith
969
asked Mar 1 '15 at 9:01
AmithAmith
969
asked Mar 1 '15 at 9:01
AmithAmith
969
asked Mar 1 '15 at 9:01
AmithAmith
969
969
1
$begingroup$
The result of the individual throws can be $1,2,3,4,5,6$ with equal probability. That is, this is not a Bernoulli distribution.
$endgroup$
– zoli
Mar 1 '15 at 10:25
$begingroup$
Oh, ok. It is a $6$ point distribution! And here expected value of an individual throw is $1/6(1+2+3+4+5+6)=21/6=3,5$. Is that correct? And standard deviation is $sqrt{21/36}=1/2 sqrt{7/3}$
$endgroup$
– Amith
Mar 1 '15 at 10:46
1
$begingroup$
Corrrect. And the variance can also be calculated. However, I am having the same problem with Cheby that you are... fo rthe time being.
$endgroup$
– zoli
Mar 1 '15 at 10:48
$begingroup$
Ok. But thanks for pointing out that Bernoulli distribution so far.
$endgroup$
– Amith
Mar 1 '15 at 10:49
add a comment |
1
$begingroup$
The result of the individual throws can be $1,2,3,4,5,6$ with equal probability. That is, this is not a Bernoulli distribution.
$endgroup$
– zoli
Mar 1 '15 at 10:25
$begingroup$
Oh, ok. It is a $6$ point distribution! And here expected value of an individual throw is $1/6(1+2+3+4+5+6)=21/6=3,5$. Is that correct? And standard deviation is $sqrt{21/36}=1/2 sqrt{7/3}$
$endgroup$
– Amith
Mar 1 '15 at 10:46
1
$begingroup$
Corrrect. And the variance can also be calculated. However, I am having the same problem with Cheby that you are... fo rthe time being.
$endgroup$
– zoli
Mar 1 '15 at 10:48
$begingroup$
Ok. But thanks for pointing out that Bernoulli distribution so far.
$endgroup$
– Amith
Mar 1 '15 at 10:49
1
1
$begingroup$
The result of the individual throws can be $1,2,3,4,5,6$ with equal probability. That is, this is not a Bernoulli distribution.
$endgroup$
– zoli
Mar 1 '15 at 10:25
$begingroup$
The result of the individual throws can be $1,2,3,4,5,6$ with equal probability. That is, this is not a Bernoulli distribution.
$endgroup$
– zoli
Mar 1 '15 at 10:25
$begingroup$
Oh, ok. It is a $6$ point distribution! And here expected value of an individual throw is $1/6(1+2+3+4+5+6)=21/6=3,5$. Is that correct? And standard deviation is $sqrt{21/36}=1/2 sqrt{7/3}$
$endgroup$
– Amith
Mar 1 '15 at 10:46
$begingroup$
Oh, ok. It is a $6$ point distribution! And here expected value of an individual throw is $1/6(1+2+3+4+5+6)=21/6=3,5$. Is that correct? And standard deviation is $sqrt{21/36}=1/2 sqrt{7/3}$
$endgroup$
– Amith
Mar 1 '15 at 10:46
1
1
$begingroup$
Corrrect. And the variance can also be calculated. However, I am having the same problem with Cheby that you are... fo rthe time being.
$endgroup$
– zoli
Mar 1 '15 at 10:48
$begingroup$
Corrrect. And the variance can also be calculated. However, I am having the same problem with Cheby that you are... fo rthe time being.
$endgroup$
– zoli
Mar 1 '15 at 10:48
$begingroup$
Ok. But thanks for pointing out that Bernoulli distribution so far.
$endgroup$
– Amith
Mar 1 '15 at 10:49
$begingroup$
Ok. But thanks for pointing out that Bernoulli distribution so far.
$endgroup$
– Amith
Mar 1 '15 at 10:49
add a comment |
1 Answer
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oldest
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$begingroup$
Various points:
As you say in your comment, the expected value of each throw is $frac{21}{6} = 3.5$
You seem to have miscalculated the variance of each throw, which is $frac{105}{36}$ rather than $frac{21}{36}$
So with $100$ throws the expected value of the sum is $frac{2100}{6} = 350$ while the variance of the sum is $frac{10500}{36} approx 291.667$ and the standard deviation is the square root of this so about $17.08$.
We can say that $200$ is about $frac{350-200}{17.08} approx 8.783$ standard deviations below the mean, but you might find it more useful to say that $199$ is about $frac{350-199}{17.08} approx 8.842$ below the mean to deal with your $P(X ge 200)=P(X gt 199)$ point
One version of Chebyshev's inequality is $Pr(|X-mu|geq ksigma) leq frac{1}{k^2}$ and you can rewrite this as $Pr(|X-mu|lt ksigma) geq 1-frac{1}{k^2}$ so the probability of being strictly within $8.842$ standard deviations of the mean is at least about $1-frac1{8.842^2}approx 0.987$ and in a sense this is an answer to your question looking for a lower bound
You might prefer a one-sided version of Chebyshev's inequality, such as $Pr(X-mu geq ksigma) leq frac{1}{1+k^2}$, though since you are looking at the lower tail you would want $Pr(mu - X geq ksigma) leq frac{1}{1+k^2}$ or $Pr(mu - X lt ksigma) geq 1-frac{1}{1+k^2}$ and here this would give about $1-frac1{1+8.842^2}approx 0.987$ as a lower bound, so close to the two-sided result
Chebyshev's inequality often a poor approximation, and it is here. In fact $P(X lt 200) approx 2.5234times 10^{-20}$ so $P(X ge 200)$ is very much closer to $1$ than Chebyshev's inequality might suggest
answered Jan 22 at 9:20
HenryHenry
100k481168
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$begingroup$
Various points:
As you say in your comment, the expected value of each throw is $frac{21}{6} = 3.5$
You seem to have miscalculated the variance of each throw, which is $frac{105}{36}$ rather than $frac{21}{36}$
So with $100$ throws the expected value of the sum is $frac{2100}{6} = 350$ while the variance of the sum is $frac{10500}{36} approx 291.667$ and the standard deviation is the square root of this so about $17.08$.
We can say that $200$ is about $frac{350-200}{17.08} approx 8.783$ standard deviations below the mean, but you might find it more useful to say that $199$ is about $frac{350-199}{17.08} approx 8.842$ below the mean to deal with your $P(X ge 200)=P(X gt 199)$ point
One version of Chebyshev's inequality is $Pr(|X-mu|geq ksigma) leq frac{1}{k^2}$ and you can rewrite this as $Pr(|X-mu|lt ksigma) geq 1-frac{1}{k^2}$ so the probability of being strictly within $8.842$ standard deviations of the mean is at least about $1-frac1{8.842^2}approx 0.987$ and in a sense this is an answer to your question looking for a lower bound
You might prefer a one-sided version of Chebyshev's inequality, such as $Pr(X-mu geq ksigma) leq frac{1}{1+k^2}$, though since you are looking at the lower tail you would want $Pr(mu - X geq ksigma) leq frac{1}{1+k^2}$ or $Pr(mu - X lt ksigma) geq 1-frac{1}{1+k^2}$ and here this would give about $1-frac1{1+8.842^2}approx 0.987$ as a lower bound, so close to the two-sided result
Chebyshev's inequality often a poor approximation, and it is here. In fact $P(X lt 200) approx 2.5234times 10^{-20}$ so $P(X ge 200)$ is very much closer to $1$ than Chebyshev's inequality might suggest
answered Jan 22 at 9:20
HenryHenry
100k481168
$endgroup$
add a comment |
$begingroup$
Various points:
As you say in your comment, the expected value of each throw is $frac{21}{6} = 3.5$
You seem to have miscalculated the variance of each throw, which is $frac{105}{36}$ rather than $frac{21}{36}$
So with $100$ throws the expected value of the sum is $frac{2100}{6} = 350$ while the variance of the sum is $frac{10500}{36} approx 291.667$ and the standard deviation is the square root of this so about $17.08$.
We can say that $200$ is about $frac{350-200}{17.08} approx 8.783$ standard deviations below the mean, but you might find it more useful to say that $199$ is about $frac{350-199}{17.08} approx 8.842$ below the mean to deal with your $P(X ge 200)=P(X gt 199)$ point
One version of Chebyshev's inequality is $Pr(|X-mu|geq ksigma) leq frac{1}{k^2}$ and you can rewrite this as $Pr(|X-mu|lt ksigma) geq 1-frac{1}{k^2}$ so the probability of being strictly within $8.842$ standard deviations of the mean is at least about $1-frac1{8.842^2}approx 0.987$ and in a sense this is an answer to your question looking for a lower bound
You might prefer a one-sided version of Chebyshev's inequality, such as $Pr(X-mu geq ksigma) leq frac{1}{1+k^2}$, though since you are looking at the lower tail you would want $Pr(mu - X geq ksigma) leq frac{1}{1+k^2}$ or $Pr(mu - X lt ksigma) geq 1-frac{1}{1+k^2}$ and here this would give about $1-frac1{1+8.842^2}approx 0.987$ as a lower bound, so close to the two-sided result
Chebyshev's inequality often a poor approximation, and it is here. In fact $P(X lt 200) approx 2.5234times 10^{-20}$ so $P(X ge 200)$ is very much closer to $1$ than Chebyshev's inequality might suggest
answered Jan 22 at 9:20
HenryHenry
100k481168
$endgroup$
add a comment |
$begingroup$
Various points:
As you say in your comment, the expected value of each throw is $frac{21}{6} = 3.5$
You seem to have miscalculated the variance of each throw, which is $frac{105}{36}$ rather than $frac{21}{36}$
So with $100$ throws the expected value of the sum is $frac{2100}{6} = 350$ while the variance of the sum is $frac{10500}{36} approx 291.667$ and the standard deviation is the square root of this so about $17.08$.
We can say that $200$ is about $frac{350-200}{17.08} approx 8.783$ standard deviations below the mean, but you might find it more useful to say that $199$ is about $frac{350-199}{17.08} approx 8.842$ below the mean to deal with your $P(X ge 200)=P(X gt 199)$ point
One version of Chebyshev's inequality is $Pr(|X-mu|geq ksigma) leq frac{1}{k^2}$ and you can rewrite this as $Pr(|X-mu|lt ksigma) geq 1-frac{1}{k^2}$ so the probability of being strictly within $8.842$ standard deviations of the mean is at least about $1-frac1{8.842^2}approx 0.987$ and in a sense this is an answer to your question looking for a lower bound
You might prefer a one-sided version of Chebyshev's inequality, such as $Pr(X-mu geq ksigma) leq frac{1}{1+k^2}$, though since you are looking at the lower tail you would want $Pr(mu - X geq ksigma) leq frac{1}{1+k^2}$ or $Pr(mu - X lt ksigma) geq 1-frac{1}{1+k^2}$ and here this would give about $1-frac1{1+8.842^2}approx 0.987$ as a lower bound, so close to the two-sided result
Chebyshev's inequality often a poor approximation, and it is here. In fact $P(X lt 200) approx 2.5234times 10^{-20}$ so $P(X ge 200)$ is very much closer to $1$ than Chebyshev's inequality might suggest
answered Jan 22 at 9:20
HenryHenry
100k481168
$endgroup$
Various points:
As you say in your comment, the expected value of each throw is $frac{21}{6} = 3.5$
You seem to have miscalculated the variance of each throw, which is $frac{105}{36}$ rather than $frac{21}{36}$
So with $100$ throws the expected value of the sum is $frac{2100}{6} = 350$ while the variance of the sum is $frac{10500}{36} approx 291.667$ and the standard deviation is the square root of this so about $17.08$.
We can say that $200$ is about $frac{350-200}{17.08} approx 8.783$ standard deviations below the mean, but you might find it more useful to say that $199$ is about $frac{350-199}{17.08} approx 8.842$ below the mean to deal with your $P(X ge 200)=P(X gt 199)$ point
One version of Chebyshev's inequality is $Pr(|X-mu|geq ksigma) leq frac{1}{k^2}$ and you can rewrite this as $Pr(|X-mu|lt ksigma) geq 1-frac{1}{k^2}$ so the probability of being strictly within $8.842$ standard deviations of the mean is at least about $1-frac1{8.842^2}approx 0.987$ and in a sense this is an answer to your question looking for a lower bound
You might prefer a one-sided version of Chebyshev's inequality, such as $Pr(X-mu geq ksigma) leq frac{1}{1+k^2}$, though since you are looking at the lower tail you would want $Pr(mu - X geq ksigma) leq frac{1}{1+k^2}$ or $Pr(mu - X lt ksigma) geq 1-frac{1}{1+k^2}$ and here this would give about $1-frac1{1+8.842^2}approx 0.987$ as a lower bound, so close to the two-sided result
Chebyshev's inequality often a poor approximation, and it is here. In fact $P(X lt 200) approx 2.5234times 10^{-20}$ so $P(X ge 200)$ is very much closer to $1$ than Chebyshev's inequality might suggest
answered Jan 22 at 9:20
HenryHenry
100k481168
answered Jan 22 at 9:20
HenryHenry
100k481168
answered Jan 22 at 9:20
HenryHenry
100k481168
answered Jan 22 at 9:20
HenryHenry
100k481168
100k481168
add a comment |
add a comment |
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$begingroup$
The result of the individual throws can be $1,2,3,4,5,6$ with equal probability. That is, this is not a Bernoulli distribution.
$endgroup$
– zoli
Mar 1 '15 at 10:25
$begingroup$
Oh, ok. It is a $6$ point distribution! And here expected value of an individual throw is $1/6(1+2+3+4+5+6)=21/6=3,5$. Is that correct? And standard deviation is $sqrt{21/36}=1/2 sqrt{7/3}$
$endgroup$
– Amith
Mar 1 '15 at 10:46
1
$begingroup$
Corrrect. And the variance can also be calculated. However, I am having the same problem with Cheby that you are... fo rthe time being.
$endgroup$
– zoli
Mar 1 '15 at 10:48
$begingroup$
Ok. But thanks for pointing out that Bernoulli distribution so far.
$endgroup$
– Amith
Mar 1 '15 at 10:49