Find Chebyshev's upper bound for this probability












0












$begingroup$


We are given that $X$ is $Uniform[4,10]$ with mean $7$ and variance $3$.



We are first asked to figure out $P(X leq 5 text { or } X geq 9$)




Well this is a uniform graph with height = $1/(10-4)=1/6$, and we can do the following:



$P(X leq 5 text { or } X geq 9) = P(X leq 5) + P(Xgeq 9)=(5-4)/6+(10-9)/6=2/6=1/3$




Now I am asking the following:



Let $X_1, X_2, X_3, dots, X_{16}$ be a random sample from this uniform distribution. We are asked to find Chebyshev's upper bound for $P(overline{X} leq 5 text { or } overline{X} geq 9)$ where $overline{X}=dfrac{sum_{i=1}^{16}X_i}{16}$



Again, we can break up the probability to be as follows:



$P(overline{X} leq 5) + P(overline{X} geq 9)leq text{upper bound}$



Chebyshev Inequality is the following: $P(|X-mu_x| geq a)leq dfrac{var(x)}{a^2}$



So, for $P(overline{X}leq 5)$, we have $P(|overline{X}-7|leq5)=1-dfrac{3}{25}$



For $P(overline{X}geq 9)$, we have $P(|overline{X}-7|geq 9)=dfrac{3}{9^2}$



So in total, the upper bound must be $1-dfrac{3}{25}+dfrac{3}{9^2}=0.917$




Is this correct? I don't see where we used $overline{X}$ in this. There would be nothing different if I considered $X$, instead of $overline{X}$, so I think I am missing something.



Maybe the $var(x)$ should actually be $sigma^2/n=3/16$. But in my notes, I have that this is only the case if $X_1, X_2, dots, X_n$ are $N(mu, sigma^2)$. Here we have that they are uniform, so I am not sure.






probability statistics inequality







share|cite|improve this question









$endgroup$












  • $begingroup$
    In general, $text{var}(bar{X}) = frac{1}{n^2} text{var}(sum_{i=1}^n X_i) = frac{1}{n} text{var}(X_1)$ for i.i.d. $X_i$. You should use this variance when applying Chebychev to $bar{X}$.
    $endgroup$
    – angryavian
    Dec 5 '18 at 21:27
















0












$begingroup$


We are given that $X$ is $Uniform[4,10]$ with mean $7$ and variance $3$.



We are first asked to figure out $P(X leq 5 text { or } X geq 9$)




Well this is a uniform graph with height = $1/(10-4)=1/6$, and we can do the following:



$P(X leq 5 text { or } X geq 9) = P(X leq 5) + P(Xgeq 9)=(5-4)/6+(10-9)/6=2/6=1/3$




Now I am asking the following:



Let $X_1, X_2, X_3, dots, X_{16}$ be a random sample from this uniform distribution. We are asked to find Chebyshev's upper bound for $P(overline{X} leq 5 text { or } overline{X} geq 9)$ where $overline{X}=dfrac{sum_{i=1}^{16}X_i}{16}$



Again, we can break up the probability to be as follows:



$P(overline{X} leq 5) + P(overline{X} geq 9)leq text{upper bound}$



Chebyshev Inequality is the following: $P(|X-mu_x| geq a)leq dfrac{var(x)}{a^2}$



So, for $P(overline{X}leq 5)$, we have $P(|overline{X}-7|leq5)=1-dfrac{3}{25}$



For $P(overline{X}geq 9)$, we have $P(|overline{X}-7|geq 9)=dfrac{3}{9^2}$



So in total, the upper bound must be $1-dfrac{3}{25}+dfrac{3}{9^2}=0.917$




Is this correct? I don't see where we used $overline{X}$ in this. There would be nothing different if I considered $X$, instead of $overline{X}$, so I think I am missing something.



Maybe the $var(x)$ should actually be $sigma^2/n=3/16$. But in my notes, I have that this is only the case if $X_1, X_2, dots, X_n$ are $N(mu, sigma^2)$. Here we have that they are uniform, so I am not sure.






probability statistics inequality







share|cite|improve this question









$endgroup$












  • $begingroup$
    In general, $text{var}(bar{X}) = frac{1}{n^2} text{var}(sum_{i=1}^n X_i) = frac{1}{n} text{var}(X_1)$ for i.i.d. $X_i$. You should use this variance when applying Chebychev to $bar{X}$.
    $endgroup$
    – angryavian
    Dec 5 '18 at 21:27














0












0








0





$begingroup$


We are given that $X$ is $Uniform[4,10]$ with mean $7$ and variance $3$.



We are first asked to figure out $P(X leq 5 text { or } X geq 9$)




Well this is a uniform graph with height = $1/(10-4)=1/6$, and we can do the following:



$P(X leq 5 text { or } X geq 9) = P(X leq 5) + P(Xgeq 9)=(5-4)/6+(10-9)/6=2/6=1/3$




Now I am asking the following:



Let $X_1, X_2, X_3, dots, X_{16}$ be a random sample from this uniform distribution. We are asked to find Chebyshev's upper bound for $P(overline{X} leq 5 text { or } overline{X} geq 9)$ where $overline{X}=dfrac{sum_{i=1}^{16}X_i}{16}$



Again, we can break up the probability to be as follows:



$P(overline{X} leq 5) + P(overline{X} geq 9)leq text{upper bound}$



Chebyshev Inequality is the following: $P(|X-mu_x| geq a)leq dfrac{var(x)}{a^2}$



So, for $P(overline{X}leq 5)$, we have $P(|overline{X}-7|leq5)=1-dfrac{3}{25}$



For $P(overline{X}geq 9)$, we have $P(|overline{X}-7|geq 9)=dfrac{3}{9^2}$



So in total, the upper bound must be $1-dfrac{3}{25}+dfrac{3}{9^2}=0.917$




Is this correct? I don't see where we used $overline{X}$ in this. There would be nothing different if I considered $X$, instead of $overline{X}$, so I think I am missing something.



Maybe the $var(x)$ should actually be $sigma^2/n=3/16$. But in my notes, I have that this is only the case if $X_1, X_2, dots, X_n$ are $N(mu, sigma^2)$. Here we have that they are uniform, so I am not sure.






probability statistics inequality







share|cite|improve this question









$endgroup$




We are given that $X$ is $Uniform[4,10]$ with mean $7$ and variance $3$.



We are first asked to figure out $P(X leq 5 text { or } X geq 9$)




Well this is a uniform graph with height = $1/(10-4)=1/6$, and we can do the following:



$P(X leq 5 text { or } X geq 9) = P(X leq 5) + P(Xgeq 9)=(5-4)/6+(10-9)/6=2/6=1/3$




Now I am asking the following:



Let $X_1, X_2, X_3, dots, X_{16}$ be a random sample from this uniform distribution. We are asked to find Chebyshev's upper bound for $P(overline{X} leq 5 text { or } overline{X} geq 9)$ where $overline{X}=dfrac{sum_{i=1}^{16}X_i}{16}$



Again, we can break up the probability to be as follows:



$P(overline{X} leq 5) + P(overline{X} geq 9)leq text{upper bound}$



Chebyshev Inequality is the following: $P(|X-mu_x| geq a)leq dfrac{var(x)}{a^2}$



So, for $P(overline{X}leq 5)$, we have $P(|overline{X}-7|leq5)=1-dfrac{3}{25}$



For $P(overline{X}geq 9)$, we have $P(|overline{X}-7|geq 9)=dfrac{3}{9^2}$



So in total, the upper bound must be $1-dfrac{3}{25}+dfrac{3}{9^2}=0.917$




Is this correct? I don't see where we used $overline{X}$ in this. There would be nothing different if I considered $X$, instead of $overline{X}$, so I think I am missing something.



Maybe the $var(x)$ should actually be $sigma^2/n=3/16$. But in my notes, I have that this is only the case if $X_1, X_2, dots, X_n$ are $N(mu, sigma^2)$. Here we have that they are uniform, so I am not sure.






probability statistics inequality




probability statistics inequality






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 5 '18 at 21:17









K Split XK Split X

4,19611131




4,19611131












  • $begingroup$
    In general, $text{var}(bar{X}) = frac{1}{n^2} text{var}(sum_{i=1}^n X_i) = frac{1}{n} text{var}(X_1)$ for i.i.d. $X_i$. You should use this variance when applying Chebychev to $bar{X}$.
    $endgroup$
    – angryavian
    Dec 5 '18 at 21:27


















  • $begingroup$
    In general, $text{var}(bar{X}) = frac{1}{n^2} text{var}(sum_{i=1}^n X_i) = frac{1}{n} text{var}(X_1)$ for i.i.d. $X_i$. You should use this variance when applying Chebychev to $bar{X}$.
    $endgroup$
    – angryavian
    Dec 5 '18 at 21:27
















$begingroup$
In general, $text{var}(bar{X}) = frac{1}{n^2} text{var}(sum_{i=1}^n X_i) = frac{1}{n} text{var}(X_1)$ for i.i.d. $X_i$. You should use this variance when applying Chebychev to $bar{X}$.
$endgroup$
– angryavian
Dec 5 '18 at 21:27




$begingroup$
In general, $text{var}(bar{X}) = frac{1}{n^2} text{var}(sum_{i=1}^n X_i) = frac{1}{n} text{var}(X_1)$ for i.i.d. $X_i$. You should use this variance when applying Chebychev to $bar{X}$.
$endgroup$
– angryavian
Dec 5 '18 at 21:27










1 Answer
1






active

oldest

votes


















2












$begingroup$

Hint:



$P(X leq 5 text { or } X geq 9) = P(|X-7| geq 2) $



two away from mean on both sides..






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This would give a different upper bound then what I got
    $endgroup$
    – K Split X
    Dec 5 '18 at 22:40










  • $begingroup$
    Yes, because your calculations steps are not correct. $P(overline{X }leq 5)$, doesn't mean $P(|overline{X}-7| leq 5)$ Just put $X=1$ to check. Your answer is also counterintuive.
    $endgroup$
    – karakfa
    Dec 5 '18 at 22:45












  • $begingroup$
    What is the correct way to solve the $P(overline{X} lt 5)$ ?
    $endgroup$
    – K Split X
    Dec 5 '18 at 22:50












  • $begingroup$
    Follow my hint, you can solve both sides at the same time. This is a direct application of Chebyshev Inequality, nothing else needed.
    $endgroup$
    – karakfa
    Dec 5 '18 at 22:53












  • $begingroup$
    If instead, the mean was something like $8.5$, how would we solve this? In this question, the mean is in the center of $5$ and $9$, but what if its not?
    $endgroup$
    – K Split X
    Dec 6 '18 at 0:27











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027665%2ffind-chebyshevs-upper-bound-for-this-probability%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Hint:



$P(X leq 5 text { or } X geq 9) = P(|X-7| geq 2) $



two away from mean on both sides..






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This would give a different upper bound then what I got
    $endgroup$
    – K Split X
    Dec 5 '18 at 22:40










  • $begingroup$
    Yes, because your calculations steps are not correct. $P(overline{X }leq 5)$, doesn't mean $P(|overline{X}-7| leq 5)$ Just put $X=1$ to check. Your answer is also counterintuive.
    $endgroup$
    – karakfa
    Dec 5 '18 at 22:45












  • $begingroup$
    What is the correct way to solve the $P(overline{X} lt 5)$ ?
    $endgroup$
    – K Split X
    Dec 5 '18 at 22:50












  • $begingroup$
    Follow my hint, you can solve both sides at the same time. This is a direct application of Chebyshev Inequality, nothing else needed.
    $endgroup$
    – karakfa
    Dec 5 '18 at 22:53












  • $begingroup$
    If instead, the mean was something like $8.5$, how would we solve this? In this question, the mean is in the center of $5$ and $9$, but what if its not?
    $endgroup$
    – K Split X
    Dec 6 '18 at 0:27
















2












$begingroup$

Hint:



$P(X leq 5 text { or } X geq 9) = P(|X-7| geq 2) $



two away from mean on both sides..






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This would give a different upper bound then what I got
    $endgroup$
    – K Split X
    Dec 5 '18 at 22:40










  • $begingroup$
    Yes, because your calculations steps are not correct. $P(overline{X }leq 5)$, doesn't mean $P(|overline{X}-7| leq 5)$ Just put $X=1$ to check. Your answer is also counterintuive.
    $endgroup$
    – karakfa
    Dec 5 '18 at 22:45












  • $begingroup$
    What is the correct way to solve the $P(overline{X} lt 5)$ ?
    $endgroup$
    – K Split X
    Dec 5 '18 at 22:50












  • $begingroup$
    Follow my hint, you can solve both sides at the same time. This is a direct application of Chebyshev Inequality, nothing else needed.
    $endgroup$
    – karakfa
    Dec 5 '18 at 22:53












  • $begingroup$
    If instead, the mean was something like $8.5$, how would we solve this? In this question, the mean is in the center of $5$ and $9$, but what if its not?
    $endgroup$
    – K Split X
    Dec 6 '18 at 0:27














2












2








2





$begingroup$

Hint:



$P(X leq 5 text { or } X geq 9) = P(|X-7| geq 2) $



two away from mean on both sides..






share|cite|improve this answer









$endgroup$



Hint:



$P(X leq 5 text { or } X geq 9) = P(|X-7| geq 2) $



two away from mean on both sides..







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 21:57









karakfakarakfa

1,973811




1,973811












  • $begingroup$
    This would give a different upper bound then what I got
    $endgroup$
    – K Split X
    Dec 5 '18 at 22:40










  • $begingroup$
    Yes, because your calculations steps are not correct. $P(overline{X }leq 5)$, doesn't mean $P(|overline{X}-7| leq 5)$ Just put $X=1$ to check. Your answer is also counterintuive.
    $endgroup$
    – karakfa
    Dec 5 '18 at 22:45












  • $begingroup$
    What is the correct way to solve the $P(overline{X} lt 5)$ ?
    $endgroup$
    – K Split X
    Dec 5 '18 at 22:50












  • $begingroup$
    Follow my hint, you can solve both sides at the same time. This is a direct application of Chebyshev Inequality, nothing else needed.
    $endgroup$
    – karakfa
    Dec 5 '18 at 22:53












  • $begingroup$
    If instead, the mean was something like $8.5$, how would we solve this? In this question, the mean is in the center of $5$ and $9$, but what if its not?
    $endgroup$
    – K Split X
    Dec 6 '18 at 0:27


















  • $begingroup$
    This would give a different upper bound then what I got
    $endgroup$
    – K Split X
    Dec 5 '18 at 22:40










  • $begingroup$
    Yes, because your calculations steps are not correct. $P(overline{X }leq 5)$, doesn't mean $P(|overline{X}-7| leq 5)$ Just put $X=1$ to check. Your answer is also counterintuive.
    $endgroup$
    – karakfa
    Dec 5 '18 at 22:45












  • $begingroup$
    What is the correct way to solve the $P(overline{X} lt 5)$ ?
    $endgroup$
    – K Split X
    Dec 5 '18 at 22:50












  • $begingroup$
    Follow my hint, you can solve both sides at the same time. This is a direct application of Chebyshev Inequality, nothing else needed.
    $endgroup$
    – karakfa
    Dec 5 '18 at 22:53












  • $begingroup$
    If instead, the mean was something like $8.5$, how would we solve this? In this question, the mean is in the center of $5$ and $9$, but what if its not?
    $endgroup$
    – K Split X
    Dec 6 '18 at 0:27
















$begingroup$
This would give a different upper bound then what I got
$endgroup$
– K Split X
Dec 5 '18 at 22:40




$begingroup$
This would give a different upper bound then what I got
$endgroup$
– K Split X
Dec 5 '18 at 22:40












$begingroup$
Yes, because your calculations steps are not correct. $P(overline{X }leq 5)$, doesn't mean $P(|overline{X}-7| leq 5)$ Just put $X=1$ to check. Your answer is also counterintuive.
$endgroup$
– karakfa
Dec 5 '18 at 22:45






$begingroup$
Yes, because your calculations steps are not correct. $P(overline{X }leq 5)$, doesn't mean $P(|overline{X}-7| leq 5)$ Just put $X=1$ to check. Your answer is also counterintuive.
$endgroup$
– karakfa
Dec 5 '18 at 22:45














$begingroup$
What is the correct way to solve the $P(overline{X} lt 5)$ ?
$endgroup$
– K Split X
Dec 5 '18 at 22:50






$begingroup$
What is the correct way to solve the $P(overline{X} lt 5)$ ?
$endgroup$
– K Split X
Dec 5 '18 at 22:50














$begingroup$
Follow my hint, you can solve both sides at the same time. This is a direct application of Chebyshev Inequality, nothing else needed.
$endgroup$
– karakfa
Dec 5 '18 at 22:53






$begingroup$
Follow my hint, you can solve both sides at the same time. This is a direct application of Chebyshev Inequality, nothing else needed.
$endgroup$
– karakfa
Dec 5 '18 at 22:53














$begingroup$
If instead, the mean was something like $8.5$, how would we solve this? In this question, the mean is in the center of $5$ and $9$, but what if its not?
$endgroup$
– K Split X
Dec 6 '18 at 0:27




$begingroup$
If instead, the mean was something like $8.5$, how would we solve this? In this question, the mean is in the center of $5$ and $9$, but what if its not?
$endgroup$
– K Split X
Dec 6 '18 at 0:27


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027665%2ffind-chebyshevs-upper-bound-for-this-probability%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Probability when a professor distributes a quiz and homework assignment to a class of n students.

Image
0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
Read more

Aardman Animations

Image
Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
Read more

Are they similar matrix

Image
0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
Read more