# Computing covariance matrix given partial elements

I have a random vector $$\mathbf{X}$$ consisting of three random variables, $$\mathbf{X} = [X_1 X_2 X_3]^T$$. I am given a partial covariance matrix C for $$\mathbf{X}$$, which is $$C=\begin{bmatrix} 6 & 0 & ? \\ ? & ? & ? \\ 0 & 1 & 4 \end{bmatrix}$$ where the $$?$$ entry implies we do not have a value. There are three parts I must compute. I have given my approaches below. Are they correct?

a) What are the values of the $$?$$ entries?

Since $$\operatorname{cov}(X_a, X_b) = \operatorname{cov}(X_b, X_a)$$, we know right away the off diagonal terms: $$C= \begin{bmatrix} 6 & 0 & 0 \\ 0 & ? & 1 \\ 0 & 1 & 4 \end{bmatrix}$$ I do not believe we can compute $$\operatorname{cov}(X_2,X_2)=\operatorname{var}[X_2]$$ since $$\operatorname{cov}(X_1,X_2) = E[X_1\cdot X_2] - E[X_1]E[X_2],\quad \operatorname{var}[X_2] = E[X_2^2]-E^2[X^2]$$ (inside outside rule).

I know that covariance matrices are positive semidefinite but I do not see how I can exploit that fact to determine the center term.

b) Are any of the random variables correlated?

Yes, $$X_3$$ and $$X_2$$ are correlated because $$\operatorname{cov}(X_2,X_3)>0$$, otherwise they would be orthogonal (and thus uncorrelated).

c) What is $$\operatorname{var}[X_3]$$?

$$\operatorname{var}[X_3] = 4$$ based on the definition of a covariance matrix and $$C_{3,3}=4$$.

• thanks for the cleanup. I am new here. – Avedis Dec 23 '18 at 14:03
• $C_{2,3}$ should be $1$ to match $C_{3,2}$. At the moment you've entered $0$. – Alex Dec 23 '18 at 14:15
• @Alex you are right. that was a type. – Avedis Dec 23 '18 at 17:06
• The only constraint on $Var(X_2)$ that I can see is that $Var(X_2) \geq 1/4$. This is because $Cov(X_2, X_3)^2 \leq Var(X_2) Var(X_3)$ by Cauchy-Schwarz and so $1 \leq Var(X_2) \cdot 4$ so $1/4 \leq Var(X_2)$. I tried using the fact that $C$ must be positive semidefinite but it didn't yield any other bound. Cauchy-Schwarz: en.wikipedia.org/wiki/… – Alex Dec 23 '18 at 18:19