# Correlation -1 and standard deviation [closed]

My book says that for a portfolio of two stocks:

$\sigma_p = \sqrt{w_A^2 \sigma_A^2 + (1-w_A)^2 \sigma_B^2 + 2 w_A (1 - w_A) \rho_{AB} \sigma_A \sigma_B}$

Elsewhere it says that if the correlation is -1, then the standard deviation is 0.

However, when I substitute $\rho_{AB}$ with $-1$ clearly $\sigma_p \neq 0$.

What am I missing here?

• You need $\sigma_A^2=\sigma_B^2$, otherwise you will need a specific weight $w_A$ to get $\sigma_P^2=0$. – Richard Hardy Mar 6 '16 at 21:04
• Let's close this one .. this is pure algebra. Plug in the values of the variables. – Ric Mar 7 '16 at 9:16
• @ChrisDegnen, algebraically you are wrong. $\sqrt{x^2}=|x|$ and never $-|x|$; that is in the definition of square root. However, it is true that $(-x)^2=x^2$ just as well as $x^2=x^2$, which you likely wanted to emphasize. – Richard Hardy Mar 7 '16 at 19:05
• @RichardHardy Thanks. Indeed, the text I was referring to states that the term under the square root has two equivalent forms. – Chris Degnen Mar 9 '16 at 15:29
• Amending my first comment: What you are missing is that with ρAB = -1 the minimum σP = 0. – Chris Degnen Mar 9 '16 at 16:48

$\sigma_p=\sqrt{\omega_a^2 \sigma_a^2+(1-\omega_a)^2 \sigma_b^2+2 \omega_a (1-\omega_a) \rho_{ab} \sigma_a \sigma_b}$

with

$\rho_{ab}=-1$

the term under the square root simplifies to

$(\omega_a \sigma_a-(1-\omega_a) \sigma_b)^2$

which is equivalent to $(-\omega_a \sigma_a+(1-\omega_a) \sigma_b)^2$

therefore

$\sigma_p=\omega_a \sigma_a-(1-\omega_a) \sigma_b$

or $\sigma_p=-\omega_a \sigma_a+(1-\omega_a) \sigma_b$

"Each equation is only valid when the right-hand side is positive. Since one is always positive when the other is negative (except when both equations equal zero), there is a unique solution for the risk and return of any combination of securities A and B."

Ref. Modern Portfolio Theory & Investment Analysis, page 72 (Case 2)

Running some test data, with perfect negative correlation the minimum portfolio s.d. is zero.

Test data

a = {0.9624, 1.6462, -0.0378, -4.0397, 0.2045}
b = {-3.6569, -4.5494, -2.2938, 3.1099, -2.6359}


$\sigma_a=2.21804$

$\sigma_b=2.99359$

$\omega_a1=\frac{\sigma_b+\sigma_p}{\sigma_a+\sigma_b}$

$\omega_a2=\frac{\sigma_b-\sigma_p}{\sigma_a+\sigma_b}$

with $\sigma_p=0$

$\omega_a1=\omega_a2=0.574406$ • Here you use $b=-a$ which does not follow from $\text{corr}(a,b)=-1$ (but it is a special case, of course). – Richard Hardy Mar 7 '16 at 19:08
• Added some more interesting test data and revised my answer. – Chris Degnen Mar 9 '16 at 16:50
• I recommend changing the png files to latex. – John Mar 9 '16 at 19:28