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

#### Apashanka

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

#### Doc Al

Mentor
Note that while $x$ may be negative, $x^2 \geq 0$.

#### Dale

Mentor
Let $x=\{-1,1\}$.

Then $\bar x=0$ but $\bar{x^2}=1$

#### Apashanka

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

Mentor
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?

#### Apashanka

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

Gold Member
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.

Gold Member
Just to be clear, do you mean $\overline{A^2_i}$? The rendering for \bar{} is terrible in $\LaTeX$. Consider using \overline{} instead.

#### Apashanka

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

Homework Helper
Dearly Missed
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

Gold Member
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 $A_i$?

#### Apashanka

What kind of Mathematical object is your $A_i$?
$A_i$ are components of a vector

#### WWGD

Gold Member
$A_i$ 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?

#### Apashanka

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

Gold Member
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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#### Apashanka

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

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.

#### Apashanka

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

Gold Member
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

Gold Member
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.

#### Nik_2213

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...

#### Master1022

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)

#### Michael Hardy

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

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

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