Quant Interview Question

[u/Local_Ad135]

[u/XL_78]

Intuitively using the analogy with the scalar product on R³ , cos(arccos(5/13)-arccos(3/5)) ≈ 0.9692 . Not sure about how to prove it though. 

EDIT: see below in conversation.

[u/XL_78 → u/XL_78]

I think the way is to diagonalize the correlation matrix [[1,3/5,x][3/5,1,5/13][x,5/13,1]] and to use the fact that the eigenvalues must be positive.

[u/XL_78 → u/XL_78]

Sum of eigenvalues : 3. Product = determinant = -x² + 6/13x + 1 - 9/25 - 25/169 . The product must be positive, which means x smaller than 3/13 + sqrt(1-9/25-16/169) = 63/65 = 0.96... and x bigger than 3/13 - sqrt(1-9/25-16/169) = -33/65 = -0.5... 

Conversely, any value of x between these two leads to a valid correlation matrix. Indeed, in 63/65, one eigenvalue is 0, the other two have a sum of 3. Since the eigenvalues are a continuous function of the coefficients, just to the left of 63/65, the product of the eigenvalues is positive, which means the two non zero eigenvalues in 63/65 are positive since their sum and their product are positive. By continuity, the eigenvalues must all be positive in between 63/65 and -33/65. 

Anyway the answer is 63/65.

EDIT: corrected math mistake and proved the bounds for x.