Deriving the probability on the Probability of an Event - Chebyshev's Theorem
Question: For the random variable X, where E(X) = 0 and Var(X) = 1, derive the upper bound on the probability of the event ${|X-.6| > .2}$.
My attempt: Using Chebyshev's Inequality- Prob($|X-E(X)| > epsilon) leq frac{Var(X)}{epsilon^2} = frac{1}{(.2)^2} = 25.$
A) With no distributional assumption is this correct? I am thinking that because the probability is greater than 1 then this is a trivial inequality. I have seen alternate definitions of Chebyshev's inequality where $epsilon$ is $k sigma$ and if k is greater than or equal to 1 then this happens. However, 25 seems like an unusual number to get. Am I missing something?
B) As a follow up and for further understanding, how would the probability change if X was normally distributed with mean = 0 and variance = 1? Using the probability normal approximation integral I get the probability to be 3.96953... Although 25 and 3.96 are very different, is accuracy even a question since both probabilities are greater than 1?
probability statistics
add a comment |
Question: For the random variable X, where E(X) = 0 and Var(X) = 1, derive the upper bound on the probability of the event ${|X-.6| > .2}$.
My attempt: Using Chebyshev's Inequality- Prob($|X-E(X)| > epsilon) leq frac{Var(X)}{epsilon^2} = frac{1}{(.2)^2} = 25.$
A) With no distributional assumption is this correct? I am thinking that because the probability is greater than 1 then this is a trivial inequality. I have seen alternate definitions of Chebyshev's inequality where $epsilon$ is $k sigma$ and if k is greater than or equal to 1 then this happens. However, 25 seems like an unusual number to get. Am I missing something?
B) As a follow up and for further understanding, how would the probability change if X was normally distributed with mean = 0 and variance = 1? Using the probability normal approximation integral I get the probability to be 3.96953... Although 25 and 3.96 are very different, is accuracy even a question since both probabilities are greater than 1?
probability statistics
$E[X]=0$ not $0.6$. So the probablity you're calculating is $P(|X|ge.2)$
– Anvit
Nov 25 at 7:07
Also, assuming normal distribution, $P(|X-.6|>.2)=P(X>0.8 or X<0.4)$ which is equal to approx $0.867$
– Anvit
Nov 25 at 7:11
add a comment |
Question: For the random variable X, where E(X) = 0 and Var(X) = 1, derive the upper bound on the probability of the event ${|X-.6| > .2}$.
My attempt: Using Chebyshev's Inequality- Prob($|X-E(X)| > epsilon) leq frac{Var(X)}{epsilon^2} = frac{1}{(.2)^2} = 25.$
A) With no distributional assumption is this correct? I am thinking that because the probability is greater than 1 then this is a trivial inequality. I have seen alternate definitions of Chebyshev's inequality where $epsilon$ is $k sigma$ and if k is greater than or equal to 1 then this happens. However, 25 seems like an unusual number to get. Am I missing something?
B) As a follow up and for further understanding, how would the probability change if X was normally distributed with mean = 0 and variance = 1? Using the probability normal approximation integral I get the probability to be 3.96953... Although 25 and 3.96 are very different, is accuracy even a question since both probabilities are greater than 1?
probability statistics
Question: For the random variable X, where E(X) = 0 and Var(X) = 1, derive the upper bound on the probability of the event ${|X-.6| > .2}$.
My attempt: Using Chebyshev's Inequality- Prob($|X-E(X)| > epsilon) leq frac{Var(X)}{epsilon^2} = frac{1}{(.2)^2} = 25.$
A) With no distributional assumption is this correct? I am thinking that because the probability is greater than 1 then this is a trivial inequality. I have seen alternate definitions of Chebyshev's inequality where $epsilon$ is $k sigma$ and if k is greater than or equal to 1 then this happens. However, 25 seems like an unusual number to get. Am I missing something?
B) As a follow up and for further understanding, how would the probability change if X was normally distributed with mean = 0 and variance = 1? Using the probability normal approximation integral I get the probability to be 3.96953... Although 25 and 3.96 are very different, is accuracy even a question since both probabilities are greater than 1?
probability statistics
probability statistics
asked Nov 25 at 6:34
Joe
182
182
$E[X]=0$ not $0.6$. So the probablity you're calculating is $P(|X|ge.2)$
– Anvit
Nov 25 at 7:07
Also, assuming normal distribution, $P(|X-.6|>.2)=P(X>0.8 or X<0.4)$ which is equal to approx $0.867$
– Anvit
Nov 25 at 7:11
add a comment |
$E[X]=0$ not $0.6$. So the probablity you're calculating is $P(|X|ge.2)$
– Anvit
Nov 25 at 7:07
Also, assuming normal distribution, $P(|X-.6|>.2)=P(X>0.8 or X<0.4)$ which is equal to approx $0.867$
– Anvit
Nov 25 at 7:11
$E[X]=0$ not $0.6$. So the probablity you're calculating is $P(|X|ge.2)$
– Anvit
Nov 25 at 7:07
$E[X]=0$ not $0.6$. So the probablity you're calculating is $P(|X|ge.2)$
– Anvit
Nov 25 at 7:07
Also, assuming normal distribution, $P(|X-.6|>.2)=P(X>0.8 or X<0.4)$ which is equal to approx $0.867$
– Anvit
Nov 25 at 7:11
Also, assuming normal distribution, $P(|X-.6|>.2)=P(X>0.8 or X<0.4)$ which is equal to approx $0.867$
– Anvit
Nov 25 at 7:11
add a comment |
1 Answer
1
active
oldest
votes
You misused Chebyshev's inequality. You computed $P(|X|> .2)$ not $P(|X-.6|>.2)$. Using Markov's inequality (i.e. Chebyshev's) we get $$P(|X-.6|>.2)=P((X-.6)^2>.04)lefrac{E(X-.6)^2}{.04}=frac{1-0+.36}{.04}$$
Still, this is greater than $1$. This is to be expected. Consider $X$ to be a coinflip with $50/50$ shot of $pm 1$. Then $E(X)=0, operatorname{var}(X)=1$. And for all $varepsilon<1$ we have $$P(|X|>varepsilon)=1$$
You can modify this for the case of $|X-.6|$.
So the best bound we can get is $1$ without further information.
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3012506%2fderiving-the-probability-on-the-probability-of-an-event-chebyshevs-theorem%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
You misused Chebyshev's inequality. You computed $P(|X|> .2)$ not $P(|X-.6|>.2)$. Using Markov's inequality (i.e. Chebyshev's) we get $$P(|X-.6|>.2)=P((X-.6)^2>.04)lefrac{E(X-.6)^2}{.04}=frac{1-0+.36}{.04}$$
Still, this is greater than $1$. This is to be expected. Consider $X$ to be a coinflip with $50/50$ shot of $pm 1$. Then $E(X)=0, operatorname{var}(X)=1$. And for all $varepsilon<1$ we have $$P(|X|>varepsilon)=1$$
You can modify this for the case of $|X-.6|$.
So the best bound we can get is $1$ without further information.
add a comment |
You misused Chebyshev's inequality. You computed $P(|X|> .2)$ not $P(|X-.6|>.2)$. Using Markov's inequality (i.e. Chebyshev's) we get $$P(|X-.6|>.2)=P((X-.6)^2>.04)lefrac{E(X-.6)^2}{.04}=frac{1-0+.36}{.04}$$
Still, this is greater than $1$. This is to be expected. Consider $X$ to be a coinflip with $50/50$ shot of $pm 1$. Then $E(X)=0, operatorname{var}(X)=1$. And for all $varepsilon<1$ we have $$P(|X|>varepsilon)=1$$
You can modify this for the case of $|X-.6|$.
So the best bound we can get is $1$ without further information.
add a comment |
You misused Chebyshev's inequality. You computed $P(|X|> .2)$ not $P(|X-.6|>.2)$. Using Markov's inequality (i.e. Chebyshev's) we get $$P(|X-.6|>.2)=P((X-.6)^2>.04)lefrac{E(X-.6)^2}{.04}=frac{1-0+.36}{.04}$$
Still, this is greater than $1$. This is to be expected. Consider $X$ to be a coinflip with $50/50$ shot of $pm 1$. Then $E(X)=0, operatorname{var}(X)=1$. And for all $varepsilon<1$ we have $$P(|X|>varepsilon)=1$$
You can modify this for the case of $|X-.6|$.
So the best bound we can get is $1$ without further information.
You misused Chebyshev's inequality. You computed $P(|X|> .2)$ not $P(|X-.6|>.2)$. Using Markov's inequality (i.e. Chebyshev's) we get $$P(|X-.6|>.2)=P((X-.6)^2>.04)lefrac{E(X-.6)^2}{.04}=frac{1-0+.36}{.04}$$
Still, this is greater than $1$. This is to be expected. Consider $X$ to be a coinflip with $50/50$ shot of $pm 1$. Then $E(X)=0, operatorname{var}(X)=1$. And for all $varepsilon<1$ we have $$P(|X|>varepsilon)=1$$
You can modify this for the case of $|X-.6|$.
So the best bound we can get is $1$ without further information.
edited Nov 25 at 7:24
answered Nov 25 at 7:19
Zachary Selk
532311
532311
add a comment |
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3012506%2fderiving-the-probability-on-the-probability-of-an-event-chebyshevs-theorem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
$E[X]=0$ not $0.6$. So the probablity you're calculating is $P(|X|ge.2)$
– Anvit
Nov 25 at 7:07
Also, assuming normal distribution, $P(|X-.6|>.2)=P(X>0.8 or X<0.4)$ which is equal to approx $0.867$
– Anvit
Nov 25 at 7:11