# Minimum / Maximum and other Advanced Properties of the Covariance of Two Random Variables [closed]

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.?

---- Answered as convex and concave in the comments.

# 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

$$\operatorname{Cov}(X,Y)=\int\int(t-\mu_X)(u-\mu_Y) f_{XY} (t,u)\:dt\:du$$ $$=\int\int(t\:u)f_{XY}(t,u)\:dt\:du-\mu_X\:\mu_Y$$ Taking derivatives and First Order Conditions, $$\frac{\partial \operatorname{Cov}(X,Y)}{\partial\mu_X}=\int\int ut \left[ \frac{\partial\left\{ f_{XY}(t,u)\right\} }{\partial\mu_X}\right]\:dt\:du-\mu_Y$$ $$\Rightarrow\mu_Y=\int\int ut\left[\frac{\partial\left\{ f_{XY}(t,u)\right\} }{\partial\mu_X}\right]\:dt\:du$$ $$\frac{\partial \operatorname{Cov}(X,Y)}{\partial\mu_Y}=\int\int ut \left[ \frac{\partial\left\{ f_{XY}(t,u)\right\} }{\partial\mu_Y}\right]\:dt\:du-\mu_X$$ $$\Rightarrow\mu_X=\int\int ut\left[\frac{\partial\left\{ f_{XY}(t,u)\right\} }{\partial\mu_Y}\right]\:dt\:du$$

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

covariance of increasing functions

Proof that Derivative of Expected Value is Zero (Using Differentiation show Unconditional Expectation is Constant)

## closed as unclear what you're asking by Did, Harish Chandra Rajpoot, 3SAT, Daniel W. Farlow, SchrodingersCatFeb 20 '16 at 15:38

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• $u$ and $t$ are dummy variables, so you can't really differentiate w.r.p. them. Covariance is a linear in both of its variables, so you could probably say it's both convex and concave. – A.S. Feb 4 '16 at 5:16
• Thanks @A.S. I am unclear as to why we cannot diffentiate using the Leibniz Integral Rule and assuming the density is differentiable. Also, if differentiaion is ruled out how can be prove the covariance is convex or concave. – texmex Feb 4 '16 at 5:20
• 1. You can differentiate $g(\alpha)=\int f(x,\alpha)\,dx$ with respect to $\alpha$ but not with respect to $x$. 2. A function can be convex/linear/concave without any references to differentiability – A.S. Feb 4 '16 at 5:27
• Much appreciative of your time to clarify my basic questions, Why can we not differentiate with respect to x? What other method can be use to demonstrate convexity etc. in this case other than differntiation .. Could you please elaborate or point me to any other resources where this is established for the covariance? – texmex Feb 4 '16 at 5:57
• Because $g(\alpha)$ is not a function of $x$ which is simply a dummy variable of the integration. Here is a definition of a convex function. Covariance is linear in its arguments $X$ and $Y$ (bilinear in other word), so you can say it's both convex and concave. – A.S. Feb 4 '16 at 6:04