Hi,
•Standard deviation of two asset A and B portfolio,
σp=√wA²*σA²+wB²*σB²+2*Cov(A,B)*wA*wB=√wA²*σA²+wB²*σB²+2*ρ(A,B)*wA*wB*σA*σB
for perfect negative correlation,ρ(A,B)=-1
σp=√wA²*σA²+wB²*σB²+2*(-1)*wA*wB*σA*σB
σp=√wA²*σA²+wB²*σB²-2*wA*wB*σA*σB
σp=√(wA*σA-wB*σB)²
σp=wA*σA-wB*σB( case1) or σp= wB*σB-wA*σA (case 2)
we first take the case σp=wA*σA-wB*σB
case 1) σp+wB*σB=wA*σA-wB*σB+wB*σB
σp+wB*σB=wA*σA
σp+(1-wA)*σB=wA*σA (wA+wB=1 =>wB=1-wA)
σp+σB-wA*σB=wA*σA
σp+σB-wA*σB+wA*σB=wA*σA+wA*σB
σp+σB= wA*σA+wA*σB=wA(σA+σB)
(σp+σB)/(σA+σB)=wA(σA+σB)/(σA+σB)
(σp+σB)/(σA+σB)=wA
wA=(σp+σB)/(σA+σB) ...1)
•Expected return of two asset portfolio A and B
E(Rp)=wA*E(RA)+wB*E(RB) ...2)
E(Rp)-wB*E(RB)=wA*E(RA)+wB*E(RB)-wB*E(RB)
E(Rp)-wB*E(RB)=wA*E(RA)
E(Rp)-(1-wA)*E(RB)=wA*E(RA)
E(Rp)-E(RB)+wA*E(RB)=wA*E(RA)
E(Rp)-E(RB)+wA*E(RB)-wA*E(RB)=wA*E(RA)-wA*E(RB)
E(Rp)-E(RB)=wA*E(RA)-wA*E(RB)=wA(E(RA)-E(RB))
E(Rp)-E(RB)=wA(E(RA)-E(RB))
(E(Rp)-E(RB))/(E(RA)-E(RB))=wA(E(RA)-E(RB))/(E(RA)-E(RB))
(E(Rp)-E(RB))/(E(RA)-E(RB))=wA
wA=(E(Rp)-E(RB))/(E(RA)-E(RB)) ...2)
equating equations 1 and 2,
wA=(σp+σB)/(σA+σB)=(E(Rp)-E(RB))/(E(RA)-E(RB))
(E(Rp)-E(RB))=((E(RA)-E(RB)) /(σA+σB))*(σp+σB)
E(Rp)=E(RB)+[(E(RA)-E(RB)) /(σA+σB)]*(σp+σB) is the equation of the case 1 straight line
Let m=slope=[(E(RA)-E(RB)) /(σA+σB)]
Thus equation becomes,
E(Rp)=E(RB)+m*(σp+σB)=m*(σp+σB)+E(RB)
This line is upward sloping as is evident from the slope m=[(E(RA)-E(RB)) /(σA+σB)]>0 (we assume for both cases that E(RA)>E(RB))
case 2)
σp= wB*σB-wA*σA
σp+wA*σA= wB*σB-wA*σA+wA*σA
σp+wA*σA=wB*σB
σp+wA*σA-wB*σB=wB*σB-wB*σB
σp+wA*σA-wB*σB=0
σp+wA*σA-wB*σB-wA*σA=0-wA*σA
σp-wB*σB=-wA*σA
σp-(1-wA)*σB=-wA*σA (wA+wB=1 =>wB=1-wA)
σp-σB+wA*σB=-wA*σA
σp-σB+wA*σB-wA*σB=-wA*σA-wA*σB
σp-σB=-wA*σA-wA*σB
σp-σB=-wA(σA+σB)
(σp-σB)/(σA+σB)=-wA(σA+σB)/(σA+σB)
-(σp-σB)/(σA+σB)=wA
wA=-(σp-σB)/(σA+σB) ...1)
•Expected return of two asset portfolio A and B
Also we know from case 1 equation 2,
wA=(E(Rp)-E(RB))/(E(RA)-E(RB)) ...2)
equating equations 1 and 2,
wA=-(σp-σB)/(σA+σB)=(E(Rp)-E(RB))/(E(RA)-E(RB))
(E(Rp)-E(RB))=-((E(RA)-E(RB)) /(σA+σB))*(σp-σB)
E(Rp)=E(RB)-((E(RA)-E(RB)) /(σA+σB))*(σp-σB) is the equation of the case 2 straight line
Let m=slope=-[(E(RA)-E(RB)) /(σA+σB)]
Thus equation becomes,
E(Rp)=E(RB)+m*(σp-σB)=m*(σp-σB)+E(RB) ( we see from relation case 2 ,σp= wB*σB-wA*σA such that σp>=σB)
This line is downward sloping as is evident from the negative slope m=-[(E(RA)-E(RB)) /(σA+σB)]<0 (we assume for both cases that E(RA)>E(RB))
Thus there are two straight lines
1) E(Rp)=m*(σp+σB)+E(RB) where m=[(E(RA)-E(RB)) /(σA+σB)] =positive slope(upward sloping) and
2) E(Rp)=m*(σp-σB)+E(RB) where m=-[(E(RA)-E(RB)) /(σA+σB)]=negative slope (downward sloping)
If we had assumed that E(RA)<E(RB) in both cases then
we would have the following pair of straight lines ,
1) E(Rp)=m*(σp+σB)+E(RB) where m=[(E(RA)-E(RB)) /(σA+σB)]=negative slope (downward sloping) and
2) E(Rp)=m*(σp-σB)+E(RB) where m=-[(E(RA)-E(RB)) /(σA+σB)]=positive slope (upward sloping)
Thus its proved that for perfect negative correlation there are two pair of straight lines one with a negative slope (downward sloping) and other with a positive slope (upward sloping) (as can be seen from the graph also such that there are two straight lines one downward sloping and the other upward sloping)depending on the relation whether E(RA)>E(RB) or E(RA)<E(RB).
Thanks
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