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Why the variance of x is 1/4 for a uniform distribution range in 0 to 1?
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$begingroup$
Pls read page 14
Why not Var$(x) = frac{1}{12}$?
geometry probability-distributions euclidean-geometry
edited yesterday
Anirban Niloy
8291218
asked yesterday
林大权林大权
62
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add a comment |
$begingroup$
Pls read page 14
Why not Var$(x) = frac{1}{12}$?
geometry probability-distributions euclidean-geometry
edited yesterday
Anirban Niloy
8291218
asked yesterday
林大权林大权
62
New contributor
林大权 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
$1/12$ is right.
$endgroup$
– Kavi Rama Murthy
yesterday
add a comment |
$begingroup$
Pls read page 14
Why not Var$(x) = frac{1}{12}$?
geometry probability-distributions euclidean-geometry
edited yesterday
Anirban Niloy
8291218
asked yesterday
林大权林大权
62
New contributor
林大权 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Pls read page 14
Why not Var$(x) = frac{1}{12}$?
geometry probability-distributions euclidean-geometry
geometry probability-distributions euclidean-geometry
edited yesterday
Anirban Niloy
8291218
asked yesterday
林大权林大权
62
New contributor
林大权 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Anirban Niloy
8291218
asked yesterday
林大权林大权
62
New contributor
林大权 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Anirban Niloy
8291218
edited yesterday
Anirban Niloy
8291218
edited yesterday
Anirban Niloy
8291218
8291218
asked yesterday
林大权林大权
62
New contributor
林大权 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
林大权林大权
62
asked yesterday
林大权林大权
62
62
New contributor
林大权 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
林大权 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
林大权 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
$1/12$ is right.
$endgroup$
– Kavi Rama Murthy
yesterday
add a comment |
2
$begingroup$
$1/12$ is right.
$endgroup$
– Kavi Rama Murthy
yesterday
2
2
$begingroup$
$1/12$ is right.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
$1/12$ is right.
$endgroup$
– Kavi Rama Murthy
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$E(X)=int_0^1 x dx = frac12$$
$$E(X^2)=int_0^1 x^2 dx = frac13$$
$$Var(X)=E(X^2)-(E(X))^2=frac13-(frac12)^2=frac{1}{12}$$
answered yesterday
Peter ForemanPeter Foreman
3,3051216
$endgroup$
$begingroup$
Looking a bit into the paper mentioned, it doesn't look like the error made there is material, it is even in the paper's disadvantage. The main conclusion is some probability being at least $1-frac1{4t^2}$ when the correct calculation gives the better $1-frac1{12t^2}$
$endgroup$
– Ingix
yesterday
$begingroup$
@Ingix Thank you for your explanation.
$endgroup$
– 林大权
22 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$E(X)=int_0^1 x dx = frac12$$
$$E(X^2)=int_0^1 x^2 dx = frac13$$
$$Var(X)=E(X^2)-(E(X))^2=frac13-(frac12)^2=frac{1}{12}$$
answered yesterday
Peter ForemanPeter Foreman
3,3051216
$endgroup$
$begingroup$
Looking a bit into the paper mentioned, it doesn't look like the error made there is material, it is even in the paper's disadvantage. The main conclusion is some probability being at least $1-frac1{4t^2}$ when the correct calculation gives the better $1-frac1{12t^2}$
$endgroup$
– Ingix
yesterday
$begingroup$
@Ingix Thank you for your explanation.
$endgroup$
– 林大权
22 hours ago
add a comment |
$begingroup$
$$E(X)=int_0^1 x dx = frac12$$
$$E(X^2)=int_0^1 x^2 dx = frac13$$
$$Var(X)=E(X^2)-(E(X))^2=frac13-(frac12)^2=frac{1}{12}$$
answered yesterday
Peter ForemanPeter Foreman
3,3051216
$endgroup$
$begingroup$
Looking a bit into the paper mentioned, it doesn't look like the error made there is material, it is even in the paper's disadvantage. The main conclusion is some probability being at least $1-frac1{4t^2}$ when the correct calculation gives the better $1-frac1{12t^2}$
$endgroup$
– Ingix
yesterday
$begingroup$
@Ingix Thank you for your explanation.
$endgroup$
– 林大权
22 hours ago
add a comment |
$begingroup$
$$E(X)=int_0^1 x dx = frac12$$
$$E(X^2)=int_0^1 x^2 dx = frac13$$
$$Var(X)=E(X^2)-(E(X))^2=frac13-(frac12)^2=frac{1}{12}$$
answered yesterday
Peter ForemanPeter Foreman
3,3051216
$endgroup$
$$E(X)=int_0^1 x dx = frac12$$
$$E(X^2)=int_0^1 x^2 dx = frac13$$
$$Var(X)=E(X^2)-(E(X))^2=frac13-(frac12)^2=frac{1}{12}$$
answered yesterday
Peter ForemanPeter Foreman
3,3051216
answered yesterday
Peter ForemanPeter Foreman
3,3051216
answered yesterday
Peter ForemanPeter Foreman
3,3051216
answered yesterday
Peter ForemanPeter Foreman
3,3051216
3,3051216
$begingroup$
Looking a bit into the paper mentioned, it doesn't look like the error made there is material, it is even in the paper's disadvantage. The main conclusion is some probability being at least $1-frac1{4t^2}$ when the correct calculation gives the better $1-frac1{12t^2}$
$endgroup$
– Ingix
yesterday
$begingroup$
@Ingix Thank you for your explanation.
$endgroup$
– 林大权
22 hours ago
add a comment |
$begingroup$
Looking a bit into the paper mentioned, it doesn't look like the error made there is material, it is even in the paper's disadvantage. The main conclusion is some probability being at least $1-frac1{4t^2}$ when the correct calculation gives the better $1-frac1{12t^2}$
$endgroup$
– Ingix
yesterday
$begingroup$
@Ingix Thank you for your explanation.
$endgroup$
– 林大权
22 hours ago
$begingroup$
Looking a bit into the paper mentioned, it doesn't look like the error made there is material, it is even in the paper's disadvantage. The main conclusion is some probability being at least $1-frac1{4t^2}$ when the correct calculation gives the better $1-frac1{12t^2}$
$endgroup$
– Ingix
yesterday
$begingroup$
Looking a bit into the paper mentioned, it doesn't look like the error made there is material, it is even in the paper's disadvantage. The main conclusion is some probability being at least $1-frac1{4t^2}$ when the correct calculation gives the better $1-frac1{12t^2}$
$endgroup$
– Ingix
yesterday
$begingroup$
@Ingix Thank you for your explanation.
$endgroup$
– 林大权
22 hours ago
$begingroup$
@Ingix Thank you for your explanation.
$endgroup$
– 林大权
22 hours ago
add a comment |
林大权 is a new contributor. Be nice, and check out our Code of Conduct.
林大权 is a new contributor. Be nice, and check out our Code of Conduct.
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2
$begingroup$
$1/12$ is right.
$endgroup$
– Kavi Rama Murthy
yesterday