Chebyshev variant
$begingroup$
Show that $P(a<X<b)geq 1-frac{sigma^2+(mu-frac{a+b}{2})^2}{(frac{b-a}{2})^2}$ where $X$ is a random variable with mean $mu$ and variance $sigma^2$.
I'm having a hard time with this question. I was trying to think of a new variable $X=Y+k$ to so that on $P(a-k<Y<b-k)$ you can just directly apply chebyshev, but I now realize that just shifts everything over so it doesn't help. I was thinking of splitting the interval into $a,mu+mu-a$ and $mu+mu-a,b$, but then I have no idea what to do with the second interval.
It kind of looks to me like the right side somehow combines the inequality for two different variables, but I have no idea how I would get a variable with variance $(mu-frac{a+b}{2})^2$.
Any ideas for how this problem is supposed to be solved?
probability
asked Nov 30 '18 at 5:05
Miles JohnsonMiles Johnson
1928
$endgroup$
add a comment |
$begingroup$
Show that $P(a<X<b)geq 1-frac{sigma^2+(mu-frac{a+b}{2})^2}{(frac{b-a}{2})^2}$ where $X$ is a random variable with mean $mu$ and variance $sigma^2$.
I'm having a hard time with this question. I was trying to think of a new variable $X=Y+k$ to so that on $P(a-k<Y<b-k)$ you can just directly apply chebyshev, but I now realize that just shifts everything over so it doesn't help. I was thinking of splitting the interval into $a,mu+mu-a$ and $mu+mu-a,b$, but then I have no idea what to do with the second interval.
It kind of looks to me like the right side somehow combines the inequality for two different variables, but I have no idea how I would get a variable with variance $(mu-frac{a+b}{2})^2$.
Any ideas for how this problem is supposed to be solved?
probability
asked Nov 30 '18 at 5:05
Miles JohnsonMiles Johnson
1928
$endgroup$
add a comment |
$begingroup$
Show that $P(a<X<b)geq 1-frac{sigma^2+(mu-frac{a+b}{2})^2}{(frac{b-a}{2})^2}$ where $X$ is a random variable with mean $mu$ and variance $sigma^2$.
I'm having a hard time with this question. I was trying to think of a new variable $X=Y+k$ to so that on $P(a-k<Y<b-k)$ you can just directly apply chebyshev, but I now realize that just shifts everything over so it doesn't help. I was thinking of splitting the interval into $a,mu+mu-a$ and $mu+mu-a,b$, but then I have no idea what to do with the second interval.
It kind of looks to me like the right side somehow combines the inequality for two different variables, but I have no idea how I would get a variable with variance $(mu-frac{a+b}{2})^2$.
Any ideas for how this problem is supposed to be solved?
probability
asked Nov 30 '18 at 5:05
Miles JohnsonMiles Johnson
1928
$endgroup$
Show that $P(a<X<b)geq 1-frac{sigma^2+(mu-frac{a+b}{2})^2}{(frac{b-a}{2})^2}$ where $X$ is a random variable with mean $mu$ and variance $sigma^2$.
I'm having a hard time with this question. I was trying to think of a new variable $X=Y+k$ to so that on $P(a-k<Y<b-k)$ you can just directly apply chebyshev, but I now realize that just shifts everything over so it doesn't help. I was thinking of splitting the interval into $a,mu+mu-a$ and $mu+mu-a,b$, but then I have no idea what to do with the second interval.
It kind of looks to me like the right side somehow combines the inequality for two different variables, but I have no idea how I would get a variable with variance $(mu-frac{a+b}{2})^2$.
Any ideas for how this problem is supposed to be solved?
probability
probability
asked Nov 30 '18 at 5:05
Miles JohnsonMiles Johnson
1928
asked Nov 30 '18 at 5:05
Miles JohnsonMiles Johnson
1928
asked Nov 30 '18 at 5:05
Miles JohnsonMiles Johnson
1928
asked Nov 30 '18 at 5:05
Miles JohnsonMiles Johnson
1928
asked Nov 30 '18 at 5:05
Miles JohnsonMiles Johnson
1928
1928
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You're on the right track with the approach $X=Y+k$. Try setting $k$ to be the midpoint of the interval $(a,b)$, so $k:=frac12(a+b)$, and then the event ${a<X<b}$ is the same as the event ${|Y|<frac12(b-a)}$. For brevity write $c:=frac12(b-a)$.The complement of this last event is then ${|Y|ge c}$, and its probability is
$$
Pleft(|Y|ge cright)=P(Y^2ge c^2)le frac{E(Y^2)}{c^2}
$$
by Markov's inequality. Finish off by computing $E(Y^2)=operatorname{Var}(Y)+[E(Y)]^2$.
answered Nov 30 '18 at 5:23
grand_chatgrand_chat
20.2k11226
$endgroup$
add a comment |
$begingroup$
For $a<b$,
$$
mathsf{P}(a<X<b)=mathsf{P}(|X-(a+b)/2|<(b-a)/2).
$$
answered Nov 30 '18 at 5:20
d.k.o.d.k.o.
9,325628
$endgroup$
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%2f3019678%2fchebyshev-variant%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You're on the right track with the approach $X=Y+k$. Try setting $k$ to be the midpoint of the interval $(a,b)$, so $k:=frac12(a+b)$, and then the event ${a<X<b}$ is the same as the event ${|Y|<frac12(b-a)}$. For brevity write $c:=frac12(b-a)$.The complement of this last event is then ${|Y|ge c}$, and its probability is
$$
Pleft(|Y|ge cright)=P(Y^2ge c^2)le frac{E(Y^2)}{c^2}
$$
by Markov's inequality. Finish off by computing $E(Y^2)=operatorname{Var}(Y)+[E(Y)]^2$.
answered Nov 30 '18 at 5:23
grand_chatgrand_chat
20.2k11226
$endgroup$
add a comment |
$begingroup$
You're on the right track with the approach $X=Y+k$. Try setting $k$ to be the midpoint of the interval $(a,b)$, so $k:=frac12(a+b)$, and then the event ${a<X<b}$ is the same as the event ${|Y|<frac12(b-a)}$. For brevity write $c:=frac12(b-a)$.The complement of this last event is then ${|Y|ge c}$, and its probability is
$$
Pleft(|Y|ge cright)=P(Y^2ge c^2)le frac{E(Y^2)}{c^2}
$$
by Markov's inequality. Finish off by computing $E(Y^2)=operatorname{Var}(Y)+[E(Y)]^2$.
answered Nov 30 '18 at 5:23
grand_chatgrand_chat
20.2k11226
$endgroup$
add a comment |
$begingroup$
You're on the right track with the approach $X=Y+k$. Try setting $k$ to be the midpoint of the interval $(a,b)$, so $k:=frac12(a+b)$, and then the event ${a<X<b}$ is the same as the event ${|Y|<frac12(b-a)}$. For brevity write $c:=frac12(b-a)$.The complement of this last event is then ${|Y|ge c}$, and its probability is
$$
Pleft(|Y|ge cright)=P(Y^2ge c^2)le frac{E(Y^2)}{c^2}
$$
by Markov's inequality. Finish off by computing $E(Y^2)=operatorname{Var}(Y)+[E(Y)]^2$.
answered Nov 30 '18 at 5:23
grand_chatgrand_chat
20.2k11226
$endgroup$
You're on the right track with the approach $X=Y+k$. Try setting $k$ to be the midpoint of the interval $(a,b)$, so $k:=frac12(a+b)$, and then the event ${a<X<b}$ is the same as the event ${|Y|<frac12(b-a)}$. For brevity write $c:=frac12(b-a)$.The complement of this last event is then ${|Y|ge c}$, and its probability is
$$
Pleft(|Y|ge cright)=P(Y^2ge c^2)le frac{E(Y^2)}{c^2}
$$
by Markov's inequality. Finish off by computing $E(Y^2)=operatorname{Var}(Y)+[E(Y)]^2$.
answered Nov 30 '18 at 5:23
grand_chatgrand_chat
20.2k11226
edited Nov 30 '18 at 5:31
answered Nov 30 '18 at 5:23
grand_chatgrand_chat
20.2k11226
answered Nov 30 '18 at 5:23
grand_chatgrand_chat
20.2k11226
answered Nov 30 '18 at 5:23
grand_chatgrand_chat
20.2k11226
20.2k11226
add a comment |
add a comment |
$begingroup$
For $a<b$,
$$
mathsf{P}(a<X<b)=mathsf{P}(|X-(a+b)/2|<(b-a)/2).
$$
answered Nov 30 '18 at 5:20
d.k.o.d.k.o.
9,325628
$endgroup$
add a comment |
$begingroup$
For $a<b$,
$$
mathsf{P}(a<X<b)=mathsf{P}(|X-(a+b)/2|<(b-a)/2).
$$
answered Nov 30 '18 at 5:20
d.k.o.d.k.o.
9,325628
$endgroup$
add a comment |
$begingroup$
For $a<b$,
$$
mathsf{P}(a<X<b)=mathsf{P}(|X-(a+b)/2|<(b-a)/2).
$$
answered Nov 30 '18 at 5:20
d.k.o.d.k.o.
9,325628
$endgroup$
For $a<b$,
$$
mathsf{P}(a<X<b)=mathsf{P}(|X-(a+b)/2|<(b-a)/2).
$$
answered Nov 30 '18 at 5:20
d.k.o.d.k.o.
9,325628
answered Nov 30 '18 at 5:20
d.k.o.d.k.o.
9,325628
answered Nov 30 '18 at 5:20
d.k.o.d.k.o.
9,325628
answered Nov 30 '18 at 5:20
d.k.o.d.k.o.
9,325628
9,325628
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.
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%2f3019678%2fchebyshev-variant%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
