R7.P1.T1.Elton_Topic: Minimum_Variance_Portfolio

gargi.adhikari

Active Member
Hi,
In reference to R7.P1.T1.Elton_Topic: Minimum_Variance_Portfolio:
I am trying to get a feel of the fact that the Minimum Variance of a Portfolio = First Partial Derivative of the Standard Deviation of the Portfolio set to zero. Can anyone explain why we are doing that in order to derive the Minimum Variance of a Portfolio formula..?
I did notice that point of Minimum Variance is the point where the slope of the tangent =0. What is the first partial derivative wrt to..? What is the X..? in the dX...?
 
Last edited:

ShaktiRathore

Well-Known Member
Subscriber
Hi,
We are in fact finding the point of minima by setting the First Partial Derivative of the Standard Deviation of the Portfolio to zero.
wA+wB=1=>wB=1-wA

σp²=wA²*σA²+wB²*σB²+2*ρ*σA* σB *wA*wB

σp²=wA²*σA²+(1-wA)²*σB²+2*ρ*σA* σB *wA*(1-wA)
Variance of portfolio σp² is a function of wA and wB we can find the point of minima that is the point where σp² attains its minimum value by setting the First Partial Derivative of the Standard Deviation of the Portfolio to zero. Infact First Partial Derivative is the slope only of the function σp² at any point therefore at minimum the function should have 0 slope thus we set Partial Derivative to 0 and find the corresponding value of w1/w2 at which the function σp² is minimum.
Thus For min. variance,
X are the weights only.(differentiate variance σp² w.r.t X where X=w1/w2)
dσp²/dwA=0=>

d/dwA[wA²*σA²+(1-wA)²*σB²+ 2 *ρ*σA* σB *wA*(1-wA)]=0

2wA*σA²-2(1-wA)*σB²+2 *ρ*σA* σB *(1-2wA)=0

2wA*σA²-2(1-wA)*σB²+2 *ρ*σA* σB *(1-2wA)=0

wA*σA²-(1-wA)*σB²+ρ*σA* σB *(1-2wA)=0…cancelling 2

wA(σA²+σB²-2*ρ*σA* σB )-σB²+ρ*σA* σB =0

wA(σA²+σB²-2*ρ*σA* σB )=σB²-ρ*σA* σB

wA = (σB²-ρ*σA* σB)/(σA²+σB²-2*ρ*σA* σB )

To be certain that variance is min,

d²σp²/dwA²

= d/dσA[2wA*σA²-2(1-wA)*σB²+2 *ρ*σA* σB *(1-2wA)]

= [2σA²+2σB²+2 *ρ*σA* σB *(-2)]

= 2[σA²+σB²-2ρ*σA* σB]

= 2[σA²+σB²-2*σA* σB+2*σA* σB-2ρ*σA* σB]

= 2[(σA-σB) ²+2*σA* σB-2ρ*σA* σB]

= 2[(σA-σB) ²+2(1-ρ)*σA* σB]>0 since at point of minimum the slope of function σp² shall increase from negative to 0 and then increase again from 0 to positive thus rate of change of slope at point of minimum is positive that is d²σp/dwA²>0 (slope increases with increase in value of wA)

Hence variance is minimum.
thanks
 

David Harper CFA FRM

David Harper CFA FRM
Subscriber
Hi @gargi.adhikari

Hopefully you can see the first step, which is a fundamental calculus idea. You can find local minimums and maximums by taking the first (partial) derivative and setting the value of this first derivative function equal to zero. Why? The first derivative is the slope of the tangent line, so it will "go flat" (i.e, equal zero) at a local min/max. For example, consider the function: y = 0.5*x^2 - 2x + 3. Where is (y) the lowest? at the local minimum, where dy/dx = 0 = x - 2, or at x = 2. As the variance is a function of weights, @ShaktiRathore 's first step is to setup d/dwA = 0. Here is https://en.wikipedia.org/wiki/Maxima_and_minima (first derivative rule at https://en.wikipedia.org/wiki/Derivative_test#First_derivative_test, although IMO this is just not a great explainer at all). I hope that helps!
 
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