$1-\rho $ is proportional to the variance abut the
$\rho =1$ line and
$1+\rho $ is proportional
to the variance about the
$\rho =-1$ line.
$\rho =0$ iff the variances about both are the same.
Uncorrelated but not independent
Suppose the joint density for
$\{X,\phantom{\rule{0.166667em}{0ex}}Y\}$ is constant on the unit circle about the origin. By the
rectangle test, the pair cannot be independent. By symmetry, the
$\rho =1$ line is
$u=t$ and the
$\rho =-1$ line is
$u=-t$ . By symmetry, also, the variance about each of these
lines is the same. Thus
$\rho =0$ , which is true iff
$\mathrm{Cov}\phantom{\rule{0.166667em}{0ex}}[X,Y]=0$ . This fact can be
verified by calculation, if desired.
Consider the three distributions in
[link] . In case (a), the distribution is
uniform over the square centered at the origin with vertices at (1,1), (-1,1), (-1,-1),(1,-1). In case (b), the distribution is uniform over two squares, in the first and
third quadrants with vertices (0,0), (1,0), (1,1), (0,1) and (0,0),
(-1,0), (-1,-1), (0,-1).
In case (c) the two squares are in the second and fourth quadrants. The marginalsare uniform on (-1,1) in each case, so that in each case
This means the
$\rho =1$ line is
$u=t$ and the
$\rho =-1$ line is
$u=-t$ .
By symmetry,
$E\left[XY\right]=0$ (in fact the pair is independent) and
$\rho =0$ .
For every pair of possible values, the two signs must be the same, so
$E\left[XY\right]>0$ which
implies
$\rho >0$ . The actual value may be calculated to give
$\rho =3/4$ . Since
$1-\rho <1+\rho $ , the variance about the
$\rho =1$ line
is less than that about the
$\rho =-1$ line. This is evident from the figure.
$E\left[XY\right]<0$ and
$\rho <0$ . Since
$1+\rho <1-\rho $ , the variance about the
$\rho =-1$ line is less than that about the
$\rho =1$ line. Again, examination of the figure
confirms this.
With the aid of m-functions and MATLAB we can easily caluclate the covariance
and the correlation coefficient. We use the joint distribution for
Example 9 in "Variance." In
that example calculations show
The pair
$\{X,Y\}$ has joint density function
${f}_{XY}(t,u)=\frac{6}{5}(t+2u)$ on the triangular region bounded by
$t=0,u=t$ , and
$u=1$ . By the usual integration techniques, we have
From this we obtain
$E\left[X\right]=2/5,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\mathrm{Var}\phantom{\rule{0.166667em}{0ex}}\left[X\right]=3/50,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}E\left[Y\right]=3/4$ , and
$\mathrm{Var}\phantom{\rule{0.166667em}{0ex}}\left[Y\right]=3/80$ . To complete the picture we need
The parameter
ρ is usually called the correlation coefficient. A more descriptive
name would be
coefficient of linear correlation . The following example shows that
all probability mass may be on a curve, so that
$Y=g\left(X\right)$ (i.e., the value of
Y is
completely determined by the value of
X ), yet
$\rho =0$ .
$Y=g\left(X\right)$ But
$\rho =0$
Suppose
$X\sim $ uniform (-1,1), so that
${f}_{X}\left(t\right)=1/2,\phantom{\rule{0.277778em}{0ex}}-1<t<1$ and
$E\left[X\right]=0$ . Let
$Y=g\left(X\right)=cosX$ . Then
Thus
$\rho =0$ . Note that
g could be any even function defined on (-1,1). In this case
the integrand
$tg\left(t\right)$ is odd, so that the value of the integral is zero.
We wish to determine
$\mathrm{Cov}\phantom{\rule{0.166667em}{0ex}}[X,Y]$ and
$\mathrm{Var}\phantom{\rule{0.166667em}{0ex}}\left[X\right]$ . It is convenient to work with the centered random variables
${X}^{\text{'}}=X-{\mu}_{X}$ and
${Y}^{\text{'}}=Y-{\mu}_{y}$ . Since by linearity of expectation,
Note that
a
_{i}^{2} does not depend upon the sign of
a
_{i} . If the
X
_{i} form an independent class, or are
otherwise uncorrelated, the expression for variance reduces to
In this morden time nanotechnology used in many field .
1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc
2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc
3- Atomobile -MEMS, Coating on car etc.
and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change .
maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.