Probability of selecting positive integer with odd number of digits (Grinstead and Snell Ch 1.2 Q 28 (d))
$begingroup$
If $A$ is the set of all positive integers with an odd number of digits, let $A,(N)$ be the number of elements of $A$ which are less than or equal to $N$. Show that $P,(A) = lim_{Ntoinfty} A,(N)/N$ does not exist.
My rusty knowledge of limits says $P,(A) = 0$, since the denominator dominates. Any help?
probability limits
asked Dec 13 '18 at 14:20
fenfen
334
$endgroup$
add a comment |
$begingroup$
If $A$ is the set of all positive integers with an odd number of digits, let $A,(N)$ be the number of elements of $A$ which are less than or equal to $N$. Show that $P,(A) = lim_{Ntoinfty} A,(N)/N$ does not exist.
My rusty knowledge of limits says $P,(A) = 0$, since the denominator dominates. Any help?
probability limits
asked Dec 13 '18 at 14:20
fenfen
334
$endgroup$
$begingroup$
Why do you think the denominator dominates?
$endgroup$
– jgon
Dec 13 '18 at 14:30
$begingroup$
Just an intuitive guess. The difference between $A,(N)$ and $N$ increases as $N$ increases.
$endgroup$
– fen
Dec 14 '18 at 11:47
add a comment |
$begingroup$
If $A$ is the set of all positive integers with an odd number of digits, let $A,(N)$ be the number of elements of $A$ which are less than or equal to $N$. Show that $P,(A) = lim_{Ntoinfty} A,(N)/N$ does not exist.
My rusty knowledge of limits says $P,(A) = 0$, since the denominator dominates. Any help?
probability limits
asked Dec 13 '18 at 14:20
fenfen
334
$endgroup$
If $A$ is the set of all positive integers with an odd number of digits, let $A,(N)$ be the number of elements of $A$ which are less than or equal to $N$. Show that $P,(A) = lim_{Ntoinfty} A,(N)/N$ does not exist.
My rusty knowledge of limits says $P,(A) = 0$, since the denominator dominates. Any help?
probability limits
probability limits
asked Dec 13 '18 at 14:20
fenfen
334
asked Dec 13 '18 at 14:20
fenfen
334
asked Dec 13 '18 at 14:20
fenfen
334
asked Dec 13 '18 at 14:20
fenfen
334
asked Dec 13 '18 at 14:20
fenfen
334
334
$begingroup$
Why do you think the denominator dominates?
$endgroup$
– jgon
Dec 13 '18 at 14:30
$begingroup$
Just an intuitive guess. The difference between $A,(N)$ and $N$ increases as $N$ increases.
$endgroup$
– fen
Dec 14 '18 at 11:47
add a comment |
$begingroup$
Why do you think the denominator dominates?
$endgroup$
– jgon
Dec 13 '18 at 14:30
$begingroup$
Just an intuitive guess. The difference between $A,(N)$ and $N$ increases as $N$ increases.
$endgroup$
– fen
Dec 14 '18 at 11:47
$begingroup$
Why do you think the denominator dominates?
$endgroup$
– jgon
Dec 13 '18 at 14:30
$begingroup$
Why do you think the denominator dominates?
$endgroup$
– jgon
Dec 13 '18 at 14:30
$begingroup$
Just an intuitive guess. The difference between $A,(N)$ and $N$ increases as $N$ increases.
$endgroup$
– fen
Dec 14 '18 at 11:47
$begingroup$
Just an intuitive guess. The difference between $A,(N)$ and $N$ increases as $N$ increases.
$endgroup$
– fen
Dec 14 '18 at 11:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you look at the beginning of the even set with $N_k=10^{2k+1}$ where k is an Integer then the last $10^{2k+1}-10^{2k}$ have been elements with an odd number of digits. Therefore we get $P(N_k) = A(N_k)/N_k geq frac{10^{2k+1}-10^{2k}}{10^{2k+1}} $ which goes to 0.9 for k to infinity. But we can also look at the $M_k=10^{2k}$. For this we can show that $P(M_k)$ goes to 0.1 . Therefore the limit can‘t converge since we found 2 different accumulation points.
answered Dec 13 '18 at 14:38
BörgeBörge
1,037415
$endgroup$
$begingroup$
Thanks, I'm not sure your answer is completely correct? but the principle makes sense. Is this standard methodology? i.e. is this method applicable to a large number of situations?
$endgroup$
– fen
Dec 14 '18 at 11:50
$begingroup$
Proving something diverges is normally (in a compact banach set) done by showing that there is more then one accumulation point. And this can be shown by finding a subsequence (existence of this is guaranteed due to the compact banach set) which converges to this accumulation point.
$endgroup$
– Börge
Dec 14 '18 at 11:55
$begingroup$
Could be that I made a mistake up there but I don‘t see it... which part, do you think is not correct and why?
$endgroup$
– Börge
Dec 14 '18 at 11:56
1
$begingroup$
$A(N)$ is the number of positive integers with an odd number of digits less than or equal to $N$. For what value of $k$ is $A(N_k)=10^{2k}-10^{2k-1}-1$? e.g. If $k=1$, $N_k = 100$ and $A,(N_k) = 10$, ${1,2,3,4,5,6,7,8,9,100}$ but your definition gives 89.
$endgroup$
– fen
Dec 14 '18 at 12:18
1
$begingroup$
This has the right idea, but it's wrong. The lim sup isn't 1, and the thing you claimed goes to 1 clearly equals $frac{9}{10}$ and therefore can't possibly go to 1.
$endgroup$
– jgon
Dec 14 '18 at 14:05
|
show 3 more comments
$begingroup$
Hint:
Compute $A(10^{2n})/10^{2n}$ and $A(10^{2n+1})/10^{2n+1}$.
answered Dec 13 '18 at 14:30
ajotatxeajotatxe
53.8k23890
$endgroup$
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%2f3038057%2fprobability-of-selecting-positive-integer-with-odd-number-of-digits-grinstead-a%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
$begingroup$
If you look at the beginning of the even set with $N_k=10^{2k+1}$ where k is an Integer then the last $10^{2k+1}-10^{2k}$ have been elements with an odd number of digits. Therefore we get $P(N_k) = A(N_k)/N_k geq frac{10^{2k+1}-10^{2k}}{10^{2k+1}} $ which goes to 0.9 for k to infinity. But we can also look at the $M_k=10^{2k}$. For this we can show that $P(M_k)$ goes to 0.1 . Therefore the limit can‘t converge since we found 2 different accumulation points.
answered Dec 13 '18 at 14:38
BörgeBörge
1,037415
$endgroup$
$begingroup$
Thanks, I'm not sure your answer is completely correct? but the principle makes sense. Is this standard methodology? i.e. is this method applicable to a large number of situations?
$endgroup$
– fen
Dec 14 '18 at 11:50
$begingroup$
Proving something diverges is normally (in a compact banach set) done by showing that there is more then one accumulation point. And this can be shown by finding a subsequence (existence of this is guaranteed due to the compact banach set) which converges to this accumulation point.
$endgroup$
– Börge
Dec 14 '18 at 11:55
$begingroup$
Could be that I made a mistake up there but I don‘t see it... which part, do you think is not correct and why?
$endgroup$
– Börge
Dec 14 '18 at 11:56
1
$begingroup$
$A(N)$ is the number of positive integers with an odd number of digits less than or equal to $N$. For what value of $k$ is $A(N_k)=10^{2k}-10^{2k-1}-1$? e.g. If $k=1$, $N_k = 100$ and $A,(N_k) = 10$, ${1,2,3,4,5,6,7,8,9,100}$ but your definition gives 89.
$endgroup$
– fen
Dec 14 '18 at 12:18
1
$begingroup$
This has the right idea, but it's wrong. The lim sup isn't 1, and the thing you claimed goes to 1 clearly equals $frac{9}{10}$ and therefore can't possibly go to 1.
$endgroup$
– jgon
Dec 14 '18 at 14:05
|
show 3 more comments
$begingroup$
If you look at the beginning of the even set with $N_k=10^{2k+1}$ where k is an Integer then the last $10^{2k+1}-10^{2k}$ have been elements with an odd number of digits. Therefore we get $P(N_k) = A(N_k)/N_k geq frac{10^{2k+1}-10^{2k}}{10^{2k+1}} $ which goes to 0.9 for k to infinity. But we can also look at the $M_k=10^{2k}$. For this we can show that $P(M_k)$ goes to 0.1 . Therefore the limit can‘t converge since we found 2 different accumulation points.
answered Dec 13 '18 at 14:38
BörgeBörge
1,037415
$endgroup$
$begingroup$
Thanks, I'm not sure your answer is completely correct? but the principle makes sense. Is this standard methodology? i.e. is this method applicable to a large number of situations?
$endgroup$
– fen
Dec 14 '18 at 11:50
$begingroup$
Proving something diverges is normally (in a compact banach set) done by showing that there is more then one accumulation point. And this can be shown by finding a subsequence (existence of this is guaranteed due to the compact banach set) which converges to this accumulation point.
$endgroup$
– Börge
Dec 14 '18 at 11:55
$begingroup$
Could be that I made a mistake up there but I don‘t see it... which part, do you think is not correct and why?
$endgroup$
– Börge
Dec 14 '18 at 11:56
1
$begingroup$
$A(N)$ is the number of positive integers with an odd number of digits less than or equal to $N$. For what value of $k$ is $A(N_k)=10^{2k}-10^{2k-1}-1$? e.g. If $k=1$, $N_k = 100$ and $A,(N_k) = 10$, ${1,2,3,4,5,6,7,8,9,100}$ but your definition gives 89.
$endgroup$
– fen
Dec 14 '18 at 12:18
1
$begingroup$
This has the right idea, but it's wrong. The lim sup isn't 1, and the thing you claimed goes to 1 clearly equals $frac{9}{10}$ and therefore can't possibly go to 1.
$endgroup$
– jgon
Dec 14 '18 at 14:05
|
show 3 more comments
$begingroup$
If you look at the beginning of the even set with $N_k=10^{2k+1}$ where k is an Integer then the last $10^{2k+1}-10^{2k}$ have been elements with an odd number of digits. Therefore we get $P(N_k) = A(N_k)/N_k geq frac{10^{2k+1}-10^{2k}}{10^{2k+1}} $ which goes to 0.9 for k to infinity. But we can also look at the $M_k=10^{2k}$. For this we can show that $P(M_k)$ goes to 0.1 . Therefore the limit can‘t converge since we found 2 different accumulation points.
answered Dec 13 '18 at 14:38
BörgeBörge
1,037415
$endgroup$
If you look at the beginning of the even set with $N_k=10^{2k+1}$ where k is an Integer then the last $10^{2k+1}-10^{2k}$ have been elements with an odd number of digits. Therefore we get $P(N_k) = A(N_k)/N_k geq frac{10^{2k+1}-10^{2k}}{10^{2k+1}} $ which goes to 0.9 for k to infinity. But we can also look at the $M_k=10^{2k}$. For this we can show that $P(M_k)$ goes to 0.1 . Therefore the limit can‘t converge since we found 2 different accumulation points.
answered Dec 13 '18 at 14:38
BörgeBörge
1,037415
edited Dec 14 '18 at 14:19
answered Dec 13 '18 at 14:38
BörgeBörge
1,037415
answered Dec 13 '18 at 14:38
BörgeBörge
1,037415
answered Dec 13 '18 at 14:38
BörgeBörge
1,037415
1,037415
$begingroup$
Thanks, I'm not sure your answer is completely correct? but the principle makes sense. Is this standard methodology? i.e. is this method applicable to a large number of situations?
$endgroup$
– fen
Dec 14 '18 at 11:50
$begingroup$
Proving something diverges is normally (in a compact banach set) done by showing that there is more then one accumulation point. And this can be shown by finding a subsequence (existence of this is guaranteed due to the compact banach set) which converges to this accumulation point.
$endgroup$
– Börge
Dec 14 '18 at 11:55
$begingroup$
Could be that I made a mistake up there but I don‘t see it... which part, do you think is not correct and why?
$endgroup$
– Börge
Dec 14 '18 at 11:56
1
$begingroup$
$A(N)$ is the number of positive integers with an odd number of digits less than or equal to $N$. For what value of $k$ is $A(N_k)=10^{2k}-10^{2k-1}-1$? e.g. If $k=1$, $N_k = 100$ and $A,(N_k) = 10$, ${1,2,3,4,5,6,7,8,9,100}$ but your definition gives 89.
$endgroup$
– fen
Dec 14 '18 at 12:18
1
$begingroup$
This has the right idea, but it's wrong. The lim sup isn't 1, and the thing you claimed goes to 1 clearly equals $frac{9}{10}$ and therefore can't possibly go to 1.
$endgroup$
– jgon
Dec 14 '18 at 14:05
|
show 3 more comments
$begingroup$
Thanks, I'm not sure your answer is completely correct? but the principle makes sense. Is this standard methodology? i.e. is this method applicable to a large number of situations?
$endgroup$
– fen
Dec 14 '18 at 11:50
$begingroup$
Proving something diverges is normally (in a compact banach set) done by showing that there is more then one accumulation point. And this can be shown by finding a subsequence (existence of this is guaranteed due to the compact banach set) which converges to this accumulation point.
$endgroup$
– Börge
Dec 14 '18 at 11:55
$begingroup$
Could be that I made a mistake up there but I don‘t see it... which part, do you think is not correct and why?
$endgroup$
– Börge
Dec 14 '18 at 11:56
1
$begingroup$
$A(N)$ is the number of positive integers with an odd number of digits less than or equal to $N$. For what value of $k$ is $A(N_k)=10^{2k}-10^{2k-1}-1$? e.g. If $k=1$, $N_k = 100$ and $A,(N_k) = 10$, ${1,2,3,4,5,6,7,8,9,100}$ but your definition gives 89.
$endgroup$
– fen
Dec 14 '18 at 12:18
1
$begingroup$
This has the right idea, but it's wrong. The lim sup isn't 1, and the thing you claimed goes to 1 clearly equals $frac{9}{10}$ and therefore can't possibly go to 1.
$endgroup$
– jgon
Dec 14 '18 at 14:05
$begingroup$
Thanks, I'm not sure your answer is completely correct? but the principle makes sense. Is this standard methodology? i.e. is this method applicable to a large number of situations?
$endgroup$
– fen
Dec 14 '18 at 11:50
$begingroup$
Thanks, I'm not sure your answer is completely correct? but the principle makes sense. Is this standard methodology? i.e. is this method applicable to a large number of situations?
$endgroup$
– fen
Dec 14 '18 at 11:50
$begingroup$
Proving something diverges is normally (in a compact banach set) done by showing that there is more then one accumulation point. And this can be shown by finding a subsequence (existence of this is guaranteed due to the compact banach set) which converges to this accumulation point.
$endgroup$
– Börge
Dec 14 '18 at 11:55
$begingroup$
Proving something diverges is normally (in a compact banach set) done by showing that there is more then one accumulation point. And this can be shown by finding a subsequence (existence of this is guaranteed due to the compact banach set) which converges to this accumulation point.
$endgroup$
– Börge
Dec 14 '18 at 11:55
$begingroup$
Could be that I made a mistake up there but I don‘t see it... which part, do you think is not correct and why?
$endgroup$
– Börge
Dec 14 '18 at 11:56
$begingroup$
Could be that I made a mistake up there but I don‘t see it... which part, do you think is not correct and why?
$endgroup$
– Börge
Dec 14 '18 at 11:56
1
1
$begingroup$
$A(N)$ is the number of positive integers with an odd number of digits less than or equal to $N$. For what value of $k$ is $A(N_k)=10^{2k}-10^{2k-1}-1$? e.g. If $k=1$, $N_k = 100$ and $A,(N_k) = 10$, ${1,2,3,4,5,6,7,8,9,100}$ but your definition gives 89.
$endgroup$
– fen
Dec 14 '18 at 12:18
$begingroup$
$A(N)$ is the number of positive integers with an odd number of digits less than or equal to $N$. For what value of $k$ is $A(N_k)=10^{2k}-10^{2k-1}-1$? e.g. If $k=1$, $N_k = 100$ and $A,(N_k) = 10$, ${1,2,3,4,5,6,7,8,9,100}$ but your definition gives 89.
$endgroup$
– fen
Dec 14 '18 at 12:18
1
1
$begingroup$
This has the right idea, but it's wrong. The lim sup isn't 1, and the thing you claimed goes to 1 clearly equals $frac{9}{10}$ and therefore can't possibly go to 1.
$endgroup$
– jgon
Dec 14 '18 at 14:05
$begingroup$
This has the right idea, but it's wrong. The lim sup isn't 1, and the thing you claimed goes to 1 clearly equals $frac{9}{10}$ and therefore can't possibly go to 1.
$endgroup$
– jgon
Dec 14 '18 at 14:05
|
show 3 more comments
$begingroup$
Hint:
Compute $A(10^{2n})/10^{2n}$ and $A(10^{2n+1})/10^{2n+1}$.
answered Dec 13 '18 at 14:30
ajotatxeajotatxe
53.8k23890
$endgroup$
add a comment |
$begingroup$
Hint:
Compute $A(10^{2n})/10^{2n}$ and $A(10^{2n+1})/10^{2n+1}$.
answered Dec 13 '18 at 14:30
ajotatxeajotatxe
53.8k23890
$endgroup$
add a comment |
$begingroup$
Hint:
Compute $A(10^{2n})/10^{2n}$ and $A(10^{2n+1})/10^{2n+1}$.
answered Dec 13 '18 at 14:30
ajotatxeajotatxe
53.8k23890
$endgroup$
Hint:
Compute $A(10^{2n})/10^{2n}$ and $A(10^{2n+1})/10^{2n+1}$.
answered Dec 13 '18 at 14:30
ajotatxeajotatxe
53.8k23890
answered Dec 13 '18 at 14:30
ajotatxeajotatxe
53.8k23890
answered Dec 13 '18 at 14:30
ajotatxeajotatxe
53.8k23890
answered Dec 13 '18 at 14:30
ajotatxeajotatxe
53.8k23890
53.8k23890
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%2f3038057%2fprobability-of-selecting-positive-integer-with-odd-number-of-digits-grinstead-a%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
$begingroup$
Why do you think the denominator dominates?
$endgroup$
– jgon
Dec 13 '18 at 14:30
$begingroup$
Just an intuitive guess. The difference between $A,(N)$ and $N$ increases as $N$ increases.
$endgroup$
– fen
Dec 14 '18 at 11:47