I have some covariance matrix $$A = \begin{bmatrix}121 & c\\c & 81\end{bmatrix} $$

The problem is to determine the possible values of $c$.

Now I know that the elements of this matrix are given by the usual definition of the covariance,

$$ \frac{1}{N-1} \sum_{i=1}^N (X_i - \bar{x})(Y_i - \bar{y})$$

and so e.g.

$$ \frac{1}{N-1} \sum_{i=1}^N (X_i - \bar{x})^2 = 121$$

$$ \frac{1}{N-1} \sum_{i=1}^N (Y_i - \bar{y})^2 = 81$$

But I can't see how to go from here to determining $c$?

  • 2
    $\begingroup$ is it always 2x2 matrix? $\endgroup$
    – Aksakal
    Commented Apr 16, 2021 at 13:58

4 Answers 4


You might find it instructive to start with a basic idea: the variance of any random variable cannot be negative. (This is clear, since the variance is the expectation of the square of something and squares cannot be negative.)

Any $2\times 2$ covariance matrix $\mathbb A$ explicitly presents the variances and covariances of a pair of random variables $(X,Y),$ but it also tells you how to find the variance of any linear combination of those variables. This is because whenever $a$ and $b$ are numbers,

$$\operatorname{Var}(aX+bY) = a^2\operatorname{Var}(X) + b^2\operatorname{Var}(Y) + 2ab\operatorname{Cov}(X,Y) = \pmatrix{a&b}\mathbb A\pmatrix{a\\b}.$$

Applying this to your problem we may compute

$$\begin{aligned} 0 \le \operatorname{Var}(aX+bY) &= \pmatrix{a&b}\pmatrix{121&c\\c&81}\pmatrix{a\\b}\\ &= 121 a^2 + 81 b^2 + 2c^2 ab\\ &=(11a)^2+(9b)^2+\frac{2c}{(11)(9)}(11a)(9b)\\ &= \alpha^2 + \beta^2 + \frac{2c}{(11)(9)} \alpha\beta. \end{aligned}$$

The last few steps in which $\alpha=11a$ and $\beta=9b$ were introduced weren't necessary, but they help to simplify the algebra. In particular, what we need to do next (in order to find bounds for $c$) is complete the square: this is the process emulating the derivation of the quadratic formula to which everyone is introduced in grade school. Writing

$$C = \frac{c}{(11)(9)},\tag{*}$$

we find

$$\alpha^2 + \beta^2 + \frac{2c^2}{(11)(9)} \alpha\beta = \alpha^2 + 2C\alpha\beta + \beta^2 = (\alpha+C\beta)^2+(1-C^2)\beta^2.$$

Because $(\alpha+C\beta)^2$ and $\beta^2$ are both squares, they are not negative. Therefore if $1-C^2$ also is non-negative, the entire right side is not negative and can be a valid variance. Conversely, if $1-C^2$ is negative, you could set $\alpha=-c\beta$ to obtain the value $(1-C^2)\beta^2\lt 0$ on the right hand side, which is invalid.

You therefore deduce (from these perfectly elementary algebraic considerations) that

If $A$ is a valid covariance matrix, then $1-C^2$ cannot be negative.

Equivalently, $|C|\le 1,$ which by $(*)$ means $-(11)(9) \le c \le (11)(9).$

There remains the question whether any such $c$ does correspond to an actual variance matrix. One way to show this is true is to find a random variable $(X,Y)$ with $\mathbb A$ as its covariance matrix. Here is one way (out of many).

I take it as given that you can construct independent random variables $A$ and $B$ having unit variances: that is, $\operatorname{Var}(A)=\operatorname{Var}(B) = 1.$ (For example, let $(A,B)$ take on the four values $(\pm 1, \pm 1)$ with equal probabilities of $1/4$ each.)

The independence implies $\operatorname{Cov}(A,B)=0.$ Given a number $c$ in the range $-(11)(9)$ to $(11)(9),$ define random variables

$$X = \sqrt{11^2-c^2/9^2}A + (c/9)B,\quad Y = 9B$$

(which is possible because $11^2 - c^2/9^2\ge 0$) and compute that the covariance matrix of $(X,Y)$ is precisely $\mathbb A.$

Finally, if you carry out the same analysis for any symmetric matrix $$\mathbb A = \pmatrix{a & b \\ b & d},$$ you will conclude three things:

  1. $a \ge 0.$

  2. $d \ge 0.$

  3. $ad - b^2 \ge 0.$

These conditions characterize symmetric, positive semi-definite matrices. Any $2\times 2$ matrix satisfying these conditions indeed is a variance matrix. (Emulate the preceding construction.)

  • 4
    $\begingroup$ It might be worth mentioning that $C$ here is the correlation and, as shown, is always between $-1$ and $+1$ $\endgroup$
    – Henry
    Commented Apr 17, 2021 at 12:00

An intuitive method to determine this answer quickly is to just remember that covariance matrices may be interpreted in the form

\begin{equation} A = \begin{pmatrix} \sigma_1^2 & \rho_{12}\sigma_1\sigma_2 &\rho_{13}\sigma_1\sigma_3 & \cdots & \rho_{1n}\sigma_1 \sigma_n \\ & \sigma_2^2 & \rho_{23}\sigma_2\sigma_3 & \cdots & \rho_{2n}\sigma_2 \sigma_n \\ & & \sigma_3^2 & \cdots & \rho_{3n}\sigma_3 \sigma_n \\ & & & \ddots & \vdots \\ & & & & \sigma_n^2 \end{pmatrix} \end{equation}

where $\rho_{ab} \in [-1,1]$ is a Pearson Correlation Coefficient. In your case you have

\begin{align} \sigma_1^2 = 121 ,~~~ \sigma_2^2 = 81 ~\Longrightarrow ~ |c| \leq \sqrt{121\cdot 81} = 99 \end{align}

i.e. $c \in [-99, 99]$.

  • 5
    $\begingroup$ +1 This is a great answer for readers familiar with this representation of covariance matrices. I don't think it would be remiss, though, to point out that when $n\gt 2$ the correlation coefficients are subject to additional restrictions: it doesn't suffice for them all just to lie between $-1$ and $1.$ The case $n=3$ is discussed at stats.stackexchange.com/questions/72790. $\endgroup$
    – whuber
    Commented Apr 17, 2021 at 18:05

$A$ is posdef so, by Sylvesters criterion, $det(A) = 121 \cdot 81 - c^2 \geq 0$. Thus, any $c \in [-99, 99]$ will produce a valid covariance matrix.

  • $\begingroup$ Covariance matrices can be positive semi-definite right? Does the semi here change anything? $\endgroup$ Commented Apr 16, 2021 at 15:23
  • 3
    $\begingroup$ semi is the case when c is exactly 99 and det = 0. $\endgroup$
    – Hunaphu
    Commented Apr 16, 2021 at 16:16
  • $\begingroup$ Constructive comments about your voting are always welcome. $\endgroup$
    – whuber
    Commented Apr 16, 2021 at 17:14
  • 6
    $\begingroup$ "posdef" might be a bit misleading here ... I think it's worth writing "positive semi-definite" as @user1887919 says $\endgroup$
    – Silverfish
    Commented Apr 16, 2021 at 23:39
  • 3
    $\begingroup$ @Hunaphu $c=+99$ and $c=-99$ both lead to zero determinant. If $-99 < c < +99$ then you would have a strictly positive definite matrix $\endgroup$
    – Henry
    Commented Apr 17, 2021 at 12:03

There are three main possibilities of note. One is that the variable are uncorrelated, in which case the off-diagonal entries are easy to calculate as 0. Another possibility is that you don't really have two different variables. $y$ is simply a scalar multiple of $x$ (i.e. perfect correlation). If $y= c x$, then $\sigma_{xy} =\sigma_{x}\sigma_{xy}=99$. We get a third possibility in noting that the above assumes $c>0$. For $c<0$, we get $\sigma_{xy} =-99$.

Geometrically, the covariance between two vectors is the product of their lengths times the cosine of the angle between them. Since the cosine varies from $-1$ to $1$, the covariance ranges from the product of their lengths to the negative of the product.

Another approach is to consider $z_1 = \frac{x-\mu_{x}}{\sigma_{x}}$ and $z_2 = \frac{y-\mu_y}{\sigma_{y}}$. $\sigma_{xy} = \sigma_{(\sigma_x z_1)(\sigma_y z_2)}=\sigma_x \sigma_y \sigma_{z_1z_2}=99\sigma_{z_1z_2}$ and $\sigma_{z_1z_2}$ is simply the correlation between $x$ and $y$, which ranges from $-1$ to $1$.


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