I If the mean of x is 0, is the mean of x-squared also 0?

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if ##\bar x =\frac{\int x dv}{\int dv}=0##
Is necessarily ##\bar {x^2}=0##??
 

Doc Al

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Note that while ##x## may be negative, ##x^2 \geq 0##.
 
387
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Is the reverse true, e.g if ##\bar {x^2}=0## then ##\bar x=0##??, given ##x>0## and
##\bar x=\frac{\int x dv}{\int dv}##
 

Nugatory

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Is the reverse true, e.g if ##\bar {x^2}=0## then ##\bar x=0##??, given ##x>0## and
##\bar x=\frac{\int x dv}{\int dv}##
What do you think and why?
 
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What do you think and why?
Because I came across a situation where it is given that as ##\bar {A^2}##=0 ,so ##\bar {\vec A}##=0,that why I am asking
 

sophiecentaur

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Because I came across a situation where it is given that as ##\bar {A^2}##=0 ,so ##\bar {\vec A}##=0,that why I am asking
"Situation"? That could only be true if all values of A are zero or complex, I think.
 
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For ##\vec A## if ##\overline {A^2}=0## implies that ##\overline{A_i^2}=0## using this ##\frac{A_i \int_V A_i dv}{\int_V dv}-\frac{\int_V (\frac{dA_i}{dv}\int_V(A_i dv))dv}{\int_V dv}(=\bar A_i \int_V dA_i)=0## can this be used to proof that ##\bar A_i=0??##
(Average is taken over a volume V and ##\bar A_i=\frac{\int_V A_i dv}{\int_V dv}##)
 

Ray Vickson

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For ##\vec A## if ##\overline {A^2}=0## implies that ##\overline{A_i^2}=0## using this ##\frac{A_i \int_V A_i dv}{\int_V dv}-\frac{\int_V (\frac{dA_i}{dv}\int_V(A_i dv))dv}{\int_V dv}(=\bar A_i \int_V dA_i)=0## can this be used to proof that ##\bar A_i=0??##
(Average is taken over a volume V and ##\bar A_i=\frac{\int_V A_i dv}{\int_V dv}##)
The equation ##\overline{A_i^2}=0## means that
$$\frac{\int_V A_i^2 \, dv}{\int_V \, dv} = 0.$$ So, if the ##A_i## are real numbers, this implies ##A_i = 0## almost everywhere, and so ##\bar{A_i} = 0## also.
 

WWGD

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For ##\vec A## if ##\overline {A^2}=0## implies that ##\overline{A_i^2}=0## using this ##\frac{A_i \int_V A_i dv}{\int_V dv}-\frac{\int_V (\frac{dA_i}{dv}\int_V(A_i dv))dv}{\int_V dv}(=\bar A_i \int_V dA_i)=0## can this be used to proof that ##\bar A_i=0??##
(Average is taken over a volume V and ##\bar A_i=\frac{\int_V A_i dv}{\int_V dv}##)
What kind of Mathematical object is your ##A_i##?
 

WWGD

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##A_i## are components of a vector
But how do you define the integral in this case? Is it given as a Real- or Otehrwise- valued function?
 
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But how do you define the integral in this case? Is it given as a Real- or Otehrwise- valued function?
It's real and of course positive which are functions of ##f(x,y,z,t)##
 

WWGD

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It's real and of course positive which are functions of ##f(x,y,z,t)##
EDIT: Well, if the (Riemann) integral of a positive Real-valued function is 0, then the function must be identically 0. Otherwise, if ## f \geq \epsilon \geq 0 ## then ##\int_{[a,b]} f \geq \epsilon(b-a) >0 ##

EDIT2: At most f can be non-zero at a set of measure 0.
 
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387
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EDIT: Well, if the (Riemann) integral of a positive Real-valued function is 0, then the function must be identically 0. Otherwise, if ## f \geq \epsilon \geq 0 ## then ##\int_{[a,b]} f \geq \epsilon(b-a) >0 ##

EDIT2: At most f can be non-zero at a set of measure 0.
What I am just thinking( say in cartesian coordinates) is ##\int_v f(x,y,z)dv=\int_v f(x,y,z)dxdydz=\int_x f_x(x)dx\int_y f_y(y)\int_z f_z(z)dz##this expression is zero if each of the area under the curve on the ##x,y,z## axis is 0 ,or if it is having a point of zero crossing.
For the function to be positive valued the first condition above is to be satisfied,isn't it??
 
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haushofer

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if ##\bar x =\frac{\int x dv}{\int dv}=0##
Is necessarily ##\bar {x^2}=0##??
Besides Dale's example: Wouldn't that imply that a mean of 0 would imply a variance of 0? That would be really odd, I'd say.
 
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Besides Dale's example: Wouldn't that imply that a mean of 0 would imply a variance of 0? That would be really odd, I'd say.
Yes it's variance is then coming 0
 

WWGD

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What I am just thinking( say in cartesian coordinates) is ##\int_v f(x,y,z)dv=\int_v f(x,y,z)dxdydz=\int_x f_x(x)dx\int_y f_y(y)\int_z f_z(z)dz##this expression is zero if each of the area under the curve on the ##x,y,z## axis is 0 ,or if it is having a point of zero crossing.
For the function to be positive valued the first condition above is to be satisfied,isn't it??
I don't think you can always find a primitive. But if you did, since f (x,y,z) is Real valued, it would evaluate as
F(x_1,y_1,z_1)-F(x_0,y_0,z_0)
Yes it's variance is then coming 0
Variance 0 implies all values are equal.
 

WWGD

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What I am just thinking( say in cartesian coordinates) is ##\int_v f(x,y,z)dv=\int_v f(x,y,z)dxdydz=\int_x f_x(x)dx\int_y f_y(y)\int_z f_z(z)dz##this expression is zero if each of the area under the curve on the ##x,y,z## axis is 0 ,or if it is having a point of zero crossing.
For the function to be positive valued the first condition above is to be satisfied,isn't it??
I don't think you can always find a primitive. But if you did, since f (x,y,z) is Real valued, it would evaluate as
F(x_1,y_1,z_1)-F(x_0,y_0,z_0)
I don't think you can always find a primitive. But if you did, since f (x,y,z) is Real valued, it would evaluate as
F(x_1,y_1,z_1)-F(x_0,y_0,z_0)
Variance 0 implies all values are equal.
Expanding on this, yes, if your volume is zero, you can see it as having "degenerate" volume, i.e., a figure that is not strictly 3D ( under the same assumption that f(x,y,z) is non-negative), and that allows for degeneracy along different dimensions.
 
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Without wandering into calculus, I'm reminded of the statistics 'standard deviation' algorithm where you sum a list items, and also sum the squares of those list items...
 
The variance of a real-valued random variable is the mean of the square of something whose mean is ##0##. If the proposed proposition were true, then every variance would be ##0,## and that is nonsense. The example already pointed out, where the two possible values of ##x## are ##\pm1,## each with probability ##1/2,## may be the simplest example.

However, it is true that if the average value of the square of ##x## is ##0## and ##x## is real, then the average value of ##x## is ##0,## and in fact the probability that ##x\ne0## in that case is ##0##.
 

WWGD

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Ultimately there is the counter of the standard normal. ##E[x]=0; \sigma=E[(x-0)^2]=E[x^2]=1\neq (E[x])^2=0^2=0##.
 
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