How to compute the correlation between the min and max of two random variables? A joint bivariate function (PDF) with variables $A,B$ is given in the function:
$f_{A,B}(a,b) = Cab(a+b),  \quad 0<a<1, \quad 0<b<1.$
where $C$ is just a constant. Assume that $Q=\min(A,B)$ and $R = \max(A,B).$
Compute Corr($Q,R),$ ie the correlation.
From $f(a,b)= \int_0^1\int_0^1 Cab (a+b) dxdy = 1,$ $C = 18/5.$ so that
$$f_{AB}(a,b)= \frac{18}{5} ab (a+b) da db, \ 0 < a, b < 1.$$
UPDATE:
\begin{align*}
f_A(a) & = 
 1/5 \ (9a^2 + 6a), \ 0 < a < 1\\
f_B(b) &= 1/5 \ (9b^2 + 6b), \ 0 < b < 1
\end{align*}
$$E(A) = 1/5 \int_0^1 a \cdot(9a^2 + 6a) da =17/20 = E(B)$$
$$E(A^2)=1/5 \int _0^1 \left(9a^4 + 6a^3\right) da = 33/50 = E(B^2)$$
\begin{align*}
1 - F_Q(q) & =  P(A >q)P(B > q)\text{ (CDF of min)}\\
& = 
\int _q^1 1/5 (9a^2 + 6a) da \cdot \int _q^1 1/5 (9b^2 + 6b) db\\
& =
\left(\frac{1}{5}(-3q^3-3q^2+6)\right)^2, & \ 0 \le q < 1
\end{align*}
The expectation and variance of $Q$ are:
\begin{align*}
\mathbb{E}(Q) & = \int_0^1  \frac{1}{5}(-3q^3-3q^2+6)dq + 
\int_1^2  \frac{1}{25}(-3q^3-3q^2+6)^2 dq =
9.85234\\
\mathbb{E}(Q^2) & = \int_0^1 \frac{2q}{5}(-3q^3-3q^2+6)dq 
+ 
\int_1^2  \frac{2q}{25}(-3q^3-3q^2+6)^2 dq = 32.9597\\
\text{Var}(Q) & = \mathbb{E}(Q^2) - (\mathbb{E}(Q))^2   \\
& =  32.9597 - (9.85234)^2 \\
& = \text{negative number; but variance is nonnegative}
\end{align*}
I got stuck-in here because I got the variance to be negative which is wrong and I don't know where I went wrong. This is why this question is different to the other one. Thank you.

Because this is really the same as your previous question, changing only the details, permit me to outline a general approach to finding moments of the max and min of a bivariate random variable.
So, generally, let $(X,Y)$ be any bivariate random variable and write $M=\max(X,Y),$ $m=\min(X,Y)$ for its extrema.  For any numbers $(i,j)$ (usually natural numbers) the (raw) $(i,j)$ moment of $(M,m)$ is
$$\mu_{i,j}(M,m) = E[M^i m^j].$$
These are relevant because correlation is a function of some of these moments.  One formula for it is
$$\rho(M,m) = \frac{\mu_{1,1}(\bar M, \bar m)}{
\sqrt{\mu_{2,0}(\bar M, \bar m)\ \mu_{0,2}(\bar M, \bar m)}}$$
where $\bar M = M - \mu_{1,0}(M,m)$ and $\bar m = m - \mu_{0,1}(M,m)$ are the centered versions of $M$ and $m.$
The idea is to condition the calculation of the expectation according to which variable(s) $X$ and $Y$ correspond to $M$ and $m.$  There are three possibilities:

*

*When $X \gt Y,$ $X=M$ and $Y=m.$


*When $X \lt Y,$ $X=m$ and $Y=M.$


*When $X = Y,$ $X=Y=M=m.$
Thus, according to the basic properties of conditional expectation,
$$\begin{aligned}
\mu_{i,j}(M,m) &= E[X^i Y^j\mid X\gt Y]\Pr(X\gt Y)\\
& + E[Y^i X^j\mid X\lt Y]\Pr(X \lt Y)\\
& + E[X^{i+j}\mid X=Y]\Pr(X=Y) \\
\\
& = E[X^i Y^j\text{ and }X\gt Y] + E[Y^i X^j\text{ and }X\lt Y] +E[X^{i+j}\text{ and }X = Y].
\end{aligned}$$
When $\Pr(X=Y)=0,$ as when $(X,Y)$ has a continuous distribution, the third term drops out.
When $(X,Y)$ has a density function $f,$ it is continuous.  Fubini's Theorem usually applies and reduces the foregoing to

$$\begin{aligned}
&\mu_{i,j}(M,m) \\ &= \int_{-\infty}^\infty y^j \left(\int_y^\infty x^if(x,y)\mathrm dx\right)\mathrm dy + \int_{-\infty}^\infty y^i \left(\int_{-\infty}^yx^j f(x,y)\mathrm dx\right)\mathrm dy.
\end{aligned}\tag{*}$$


Example
Suppose, as in the present question,
$$f(x,y) = C xy(x+y)\ \mathcal{I}_{[0,1]\times [0,1]}.$$
The indicator function $\mathcal I$ explicitly states the density is zero unless $0\le x \le 1$ and $0 \le y \le 1.$
We plug this $f$ into $(*)$ and compute
$$\begin{aligned}
\mu_{i,j}(M,m) &= \int_{-\infty}^\infty\int_y^\infty x^i y^j  C xy(x+y)\mathcal{I}_{[0,1]\times [0,1]}\,\mathrm dx\mathrm dy \\
&+ \int_{-\infty}^\infty\int_{-\infty}^y y^i x^j  C xy(x+y)\mathcal{I}_{[0,1]\times [0,1]}\,\mathrm dx\mathrm dy\\
&= C\int_0^1 \int_y^1 x^i y^j   xy(x+y)\,\mathrm dx\mathrm dy 
+ C \int_0^1\int_{0}^y y^i x^j  xy(x+y)\,\mathrm dx\mathrm dy\\
&= C\int_0^1\left\{\int_y^1 x^{i+2}y^{j+1} + x^{i+1}y^{j+2}\,\mathrm dx+ \int_0^y x^{j+2}y^{i+1} + x^{j+1}y^{i+2}\,\mathrm dx\right\}\mathrm dy
\end{aligned}$$
The rest is just calculations.  For the values of $i$ and $j$ needed in the question (natural numbers), this formula involves the sum of four univariate integrals over $x.$  Just evaluate them (all four have such similar forms that you need only evaluate one and apply that result in all four cases), integrate that with respect to $y,$ and plug the answers into the formula for $\rho(M,m).$  (The constant $C$ can be found using the formula for $\mu_{0,0}.$)
