How to determine the range of values ​of A(i,j) in Covariance matrix A? Let $A(i,j), i,j=0,1,2$ be the covariance matrix of three random variables. If we know all the entries except $A(2,0)$ and $A(0,2)$, how to determine the range of possible values of $A(2,0)$?
 A: Let us write 
$$A=\left(
\begin{array}{ccc}
 a & b & c \\
 b & d & e \\
 c & e & f \\
\end{array}
\right).
$$
Then $A$ will be a covariance matrix iff it is positive semidefinite ($A\ge0$), that is, iff 
$$\text{$a\ge0$, $d\ge0$, $f\ge0$, $ad\ge b^2$, $d f\ge e^2$, }\tag{1}
$$
$a f\ge c^2$, and
$$\det A=-c^2 d + 2 b c e - a e^2 - b^2 f + a d f\ge0. 
$$
Of these seven inequalities, only the last two involve $c=A(2,0)$, and they are quadratic inequalities for $c$. Note also that the inequalities $a d\ge b^2$ and $d f\ge e^2$ imply $D:=\left(a d-b^2\right) \left(d f-e^2\right)\ge0$. 
Also, if $d=0$ and $A\ge0$, then $b=d=e=0$. 
So, if $d=0$, then 
$A$ is a covariance matrix iff $c\in[-\sqrt{af},\sqrt{af}]$, provided that conditions (1) hold. 
Finally, if $d>0$, then $A$ is a covariance matrix iff $c\in[c_1,c_2]\cap[-\sqrt{af},\sqrt{af}]$, provided that conditions (1) hold, where 
$$c_1:=\frac{-\sqrt{D}+b e}{d}\quad\text{and}\quad
c_2:=\frac{\sqrt{D}+b e}{d}.
$$

In fact, the answer in the case $d>0$ can be simplified: if $d>0$, then $A$ is a covariance matrix iff $c\in[c_1,c_2]$, provided that conditions (1) hold. Indeed, suppose that $d>0$ and let 
\begin{equation}
 M_1:=\left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 -{b}/{d} & 1 & 0 \\
 0 & 0 & 1 \\
\end{array}
\right),
\quad
M_2:=\left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 1 & -{e}/{d} \\
 0 & 0 & 1 \\
\end{array}
\right). 
\end{equation}
Then $A\ge0$ iff 
\begin{equation}
 0\le B:=M_2^T M_1^T AM_1M_2
 =\left(
\begin{array}{ccc}
 a-{b^2}/{d} & 0 & c-{b e}/{d} \\
 0 & d & 0 \\
 c-{b e}/{d} & 0 & f-{e^2}/{d} \\
\end{array}
\right) 
\end{equation}
iff $(c-{b e}/{d})^2\le(a-{b^2}/{d})(f-{e^2}/{d})[=D/d^2]$ iff $c\in[c_1,c_2]$, as claimed. 
A: I am guessing, but I think in this case when det(A) > 0.   Therefore look at det(A) = 0 as a quadratic equation in x = A(0,2), and find it's zeros, and it ought ot be the area between the zeros.   Here is the circumstantial evidence:  det(A) = 0 will give  a semidefinite matrix, which should be the boundary between  positive definite and indefinite, and det(A) > 0 means the product of the eigenvalues is > 0, which probably means the smallest is also, since the structure of the rest of the matrix ought to  be good for 2 positive eigenvalues.   This non-argument might have to be modified a bit in case the first 2 r.v.s are actually the same, but that should be a simple special case.
