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

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In summary: Variance 0 implies all values are...0?No, it would imply that there is a value for which the sum of the squares of all the values is 0.
  • #1
Apashanka
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if ##\bar x =\frac{\int x dv}{\int dv}=0##
Is necessarily ##\bar {x^2}=0##??
 
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  • #2
Note that while ##x## may be negative, ##x^2 \geq 0##.
 
  • #3
Let ##x=\{-1,1\}##.

Then ##\bar x=0## but ##\bar{x^2}=1##
 
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  • #4
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}##
 
  • #5
Apashanka said:
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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  • #6
Nugatory said:
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
 
  • #7
Apashanka said:
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.
 
  • #8
Just to be clear, do you mean ##\overline{A^2_i}##? The rendering for \bar{} is terrible in ##\LaTeX##. Consider using \overline{} instead.
 
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  • #9
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}##)
 
  • #10
Apashanka said:
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.
 
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  • #11
Apashanka said:
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##?
 
  • #12
WWGD said:
What kind of Mathematical object is your ##A_i##?
##A_i## are components of a vector
 
  • #13
Apashanka said:
##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?
 
  • #14
WWGD said:
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)##
 
  • #15
Apashanka said:
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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  • #16
WWGD said:
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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  • #17
Apashanka said:
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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  • #18
haushofer said:
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
 
  • #19
Apashanka said:
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)
Apashanka said:
Yes it's variance is then coming 0
Variance 0 implies all values are equal.
 
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  • #20
Apashanka said:
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)
WWGD said:
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.
 
  • #21
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...
 
  • #22
Apashanka said:
if ##\bar x =\frac{\int x dv}{\int dv}=0##
Is necessarily ##\bar {x^2}=0##??
Not necessarily, think about sin(x) and cos(x)
 
  • #23
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##.
 
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  • #24
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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  • #25
For the initial question isn't this one of those cases where tn suffices to think of a simple example? For example for the mean of two quantities to be zero they have to be equal and opposite sign. The squares are also equal but they are not opposite. Proved except for unineresting qualification where everything is zero.
 
  • #26
epenguin said:
For the initial question isn't this one of those cases where tn suffices to think of a simple example? For example for the mean of two quantities to be zero they have to be equal and opposite sign. The squares are also equal but they are not opposite. Proved except for unineresting qualification where everything is zero.
Yes, that is what we did. Take X~N(0,1). Then ##E[X]=0, E[X^2]=E[(X-0)^2]=1 \neq E[X]^2##
 

1. What does it mean for the mean of x to be 0?

The mean of x being 0 means that when all the values of x are added together and divided by the number of values, the result is 0. In other words, the average value of x is 0.

2. Does the mean of x being 0 guarantee that the mean of x-squared is also 0?

No, the mean of x being 0 does not guarantee that the mean of x-squared will also be 0. This is because the mean of x-squared is calculated by squaring each value of x and then finding the mean, which can result in a non-zero value even if the mean of x is 0.

3. What is the relationship between the mean of x and the mean of x-squared?

The mean of x-squared is related to the mean of x through the variance of x. The variance of x is equal to the mean of x-squared minus the square of the mean of x. So if the mean of x is 0, the variance of x will also be 0, but the mean of x-squared may still be non-zero.

4. Can the mean of x and the mean of x-squared be equal if the mean of x is not 0?

Yes, it is possible for the mean of x and the mean of x-squared to be equal even if the mean of x is not 0. This can occur if the values of x are symmetrically distributed around a non-zero mean.

5. How does the mean of x being 0 affect the distribution of x-squared?

If the mean of x is 0, the distribution of x-squared will have a smaller spread compared to the distribution of x. This is because squaring the values of x will result in all positive values, which will reduce the variability of the data. However, the shape of the distribution may still be affected by other factors such as the sample size and the underlying distribution of x.

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