# Calculate Upper Lower bound in $4\times 4$ correlation matrix

Let $X_i$, $i = 1,2,3 4$, be random variables on the same probability space such that \begin{align*} \mathrm{corr}(X_1,X_3) &= 0.3;\\ \mathrm{corr}(X_2,X_3) &= 0.1;\\ \mathrm{corr}(X_1,X_4) &= 0.2;\\ \mathrm{corr}(X_2,X_4) &= −0.1;\\ \mathrm{corr}(X_3,X_4) &= −0.2. \end{align*} Find upper and lower bounds for $\mathrm{corr}(X_1,X_2)$.

Any help on how to approach this will be appreciated.

I have been able to create the partial matrix, but not sure how to proceed from here.

$$\mathrm{corr}(x_1,x_2)=\mathrm{corr}(x_2,x_1)=x$$ $$\left(\begin{array}{rrrr} 1 & x & 0.3 & 0.2\\ x & 1 & 0.1 & -0.1\\ 0.3& 0.1& 1 & -0.2\\ 0.2 &-0.1 &-0.2& 1 \end{array}\right)$$

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A correlation matrix $C$ must be positive definite, which means that it must satisfy $\det(C)>0$. For your case you have

$$C = \left[ \begin{array}{rrrr} 1 & x & 0.3 & 0.2 \\ x & 1 & 0.1 & -0.1 \\ 0.3 & 0.1 & 1 & -0.2 \\ 0.2 & -0.1& -0.2 & 1 \end{array}\right]$$

and you can use Wolfram Alpha to calculate that

$$\det(C) = 0.7925+0.016 x-0.96 x^2$$

You can now find the roots of this quadratic, which will tell you where the upper and lower bounds on $x$ are.

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Thank you sir!!!..Roots came out to be -0.900287 and 0.916953. Thanks again for your help. – MathBoy Mar 7 '12 at 8:48
@MathBoy If this has answered your question, you should click the 'accept' button (the tick mark to the left of the answer). Apart from being nice for me, it will prevent this question from repeatedly popping up on the front page. – Chris Taylor Mar 7 '12 at 9:10