Minimum / Maximum and other Advanced Properties of the Covariance of Two Random Variables Are there any advanced results established regarding the behavior of the Covariance of two random variables other than the bounds on the correlation and independence when it is zero etc. which are usually summarized in introductory notes on this topic such as at this link? http://www.stat.yale.edu/~pollard/Courses/241.fall97/Variance.pdf
Some Possibilities for Advanced Properties
1)
For example, whether it is a convex or concave function and so on and under what restrictions on the density or joint density functions etc.?
---- It seems to be both convex or concave since it is linear in the variables.
2)
Where do the maximum and the minimum occur etc.?
3)
Also, to get the joint density (for example, the bivariate normal) we need the correlation coefficient which is based on the covariance. And to get the covariance we need the joint density? Seems like a cyclical; which came first - the chicken or the egg problem?
Please let me know if anything is not clear or if this question is too trivial or incorrect in some sense etc.
Steps Tried to get Maximum / Minimum
$$
Cov\left(X,Y\right)=\int\int\left(t-\mu_{X}\right)\left(u-\mu_{Y}\right)f_{XY}\left(t,u\right)\:dt\:du
$$
$$
=\int\int\left(t\:u\right)f_{XY}\left(t,u\right)\:dt\:du-\mu_{X}\:\mu_{Y}
$$
Taking derivatives and First Order Conditions,
$$
\frac{\partial Cov\left(X,Y\right)}{\partial\mu_{X}}=\int\int ut\left[\frac{\partial\left\{ f_{XY}\left(t,u\right)\right\} }{\partial\mu_{X}}\right]\:dt\:du-\mu_{Y}
$$
$$
\Rightarrow\mu_{Y}=\int\int ut\left[\frac{\partial\left\{ f_{XY}\left(t,u\right)\right\} }{\partial\mu_{X}}\right]\:dt\:du
$$
$$
\frac{\partial Cov\left(X,Y\right)}{\partial\mu_{Y}}=\int\int ut\left[\frac{\partial\left\{ f_{XY}\left(t,u\right)\right\} }{\partial\mu_{Y}}\right]\:dt\:du-\mu_{X}
$$
$$
\Rightarrow\mu_{X}=\int\int ut\left[\frac{\partial\left\{ f_{XY}\left(t,u\right)\right\} }{\partial\mu_{Y}}\right]\:dt\:du
$$
Can we take derivatives as above, assuming the densities are differentiable? Is another approach advisable? Can we simplify this further?
The only material I could find was this. (related but not the same)
http://www.math.tu-dresden.de/sto/schmidt/dsvm/dsvm2003-4.pdf
Related Questions
Joint Density and Covariance between Two Random Variables with the same Mean and Variance
https://math.stackexchange.com/questions/74677/covariance-of-increasing-functions/74681
 A: The only useful bound I'm aware of is the Cauchy-Schwarz inequality which says that
$$
|\text{Cov}(X, Y)| \leq \sqrt{\text{Var}(X) \text{Var}(Y)} ,
$$
but in general covariances are unbounded.  One way to see this is to observe that the Cauchy-Schwarz inequality is tight (consider when $X$ and $Y$ are perfectly correlated) and variances can be arbitrarily large.
A: This question is closely related to this previous question but also includes the incorrect belief that random variables are independent if they have
zero correlation.

... on the correlation, and independence when it is zero etc.

I will answer only a simpler version of the OP's "chicken and egg"
question:

to get the joint density (for example, the bivariate normal) we need the correlation coefficient which is based on the covariance. And to get the covariance we need the joint density? Seems like a cyclical; which came first - the chicken or the egg problem?

The simpler version relates to the variance of a normal random variable
and asks

To find the density of a (zero-mean) normal random variable, we need to know the variance, and to get the variance we need the density. Seems like a cyclical; which came first - the chicken or the egg?

Answer: 
If we are told that a zero-mean normal random variable $X$ has variance
$\sigma^2$, then we can write down the density function as
$$f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}},
~-\infty < x < \infty. \tag{1}$$
It is not necessary to use the density to compute the variance
since we know the value of the variance already, but if we do
choose to proceed with the calculation -- preferably without asking on
stats.SE or math.SE for help in calculating the value of 
$\operatorname{var}(X) = E[X^2] = \int_{-\infty}^\infty x^2f(x)\,\mathrm dx$ -- we come up with
the result that the variance is indeed $\sigma^2$.  So, in this
case, the egg (value of the variance) hatched into the chicken
(the density) and then the chicken laid the egg when we went through
all the rigmarole of finding the variance. We can, if we like,
say to the density (as
Dorothy Parker did to a friend who had just
had a baby): "Good job! We all knew you had it in you!"
On the other hand, if we simply told that a zero-mean random variable
has density function
$$f(x) = e^{-\pi x^2}, ~ -\infty < x < \infty,\tag{2}$$
and are asked to find its variance, we can proceed to try and
calculate the variance of $X$ as in the previous paragraph,
again hopefully without asking for help in computing the value
of $\int_{-\infty}^\infty x^2e^{-\pi x^2}\,\mathrm dx$,
or we can muse that the density function in $(2)$ looks
a heck of a lot like the density function in $(1)$.
The constant term $\frac{1}{\sigma\sqrt{2\pi}}$ in $(1)$ equals $1$
if $\sigma\sqrt{2\pi} = 1$, that is, if $\frac{1}{2\sigma^2} = \pi$,
and by golly, that makes the exponent $-\frac{x^2}{2\sigma^2}$ equal
$-\pi x^2$. So, $(2)$ is in fact the density of a zero-mean
normal random variable whose variance is $\frac{1}{2\pi}$.  At this
time, we run wet and naked through campus streets crying "Eureka!"
and burst into our professor's office to tell him that we have not
just found the value of the variance as the professor had asked for
but also discovered that the random variable
in question is a normal random variable. So in this case, the
chicken (density) came first and laid the egg of the variance.
Knowing the value of the variance, we can, if we wish, write down
the density in $(2)$ as
$$f(x) = \frac{1}{\sqrt{\frac{1}{2\pi}}\sqrt{2\pi}}
e^{-\frac{x^2}{2\left(\sqrt{\frac{1}{2\pi}}\right)^2}},
~-\infty < x < \infty \tag{3}$$
which is a one-to-one match with the form written in $(1)$
so that everybody can tell immediately what the value of
the variance is. In short, the egg hatches into the density
stated exactly as one should write normal densities instead
of the dubious form $(2)$ which is clearly designed to conceal
rather than reveal.

All right, then. Suppose $X$ and $Y$ have joint density function
$$f_{X,Y}(x,y) = \frac{1}{\pi}e^{-2x^2-2y^2+2\sqrt{3} xy},~
-\infty < x, y < \infty.$$
What are the means and variances of $X$ and $Y$ and what is their
covariance? Which came first? the chicken or the egg?
Now let the down-votes begin....
