Chebyshev inequality, lower bound on $P(X ge 200)$

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$begingroup$


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







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  • 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












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
















1












1








1


1



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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
















  • 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












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







share|cite|improve this answer









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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







    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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







      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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







        share|cite|improve this answer









        $endgroup$



        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








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        answered Jan 22 at 9:20









        HenryHenry

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