Expected value:
μ = ∑ x ⋅ P ( x ) \mu=\sum x\cdot P(x) μ = ∑ x ⋅ P ( x )
μ X = 0.5 ⋅ 120 + 0.3 ⋅ 100 + 0.2 ⋅ 250 = 140 \mu_X=0.5\cdot120+0.3\cdot100+0.2\cdot250=140 μ X = 0.5 ⋅ 120 + 0.3 ⋅ 100 + 0.2 ⋅ 250 = 140
μ Y = 0.5 ⋅ 200 + 0.3 ⋅ 150 + 0.2 ⋅ 100 = 165 \mu_Y=0.5\cdot200+0.3\cdot150+0.2\cdot100=165 μ Y = 0.5 ⋅ 200 + 0.3 ⋅ 150 + 0.2 ⋅ 100 = 165
Standard deviation:
σ = ∑ ( x − μ ) 2 ⋅ P ( x ) \sigma=\sqrt{\sum(x-\mu)^2\cdot P(x)} σ = ∑ ( x − μ ) 2 ⋅ P ( x )
σ X = 0.5 ⋅ ( 120 − 140 ) 2 + 0.3 ⋅ ( 100 − 140 ) 2 + 0.2 ⋅ ( 250 − 140 ) 2 = 55.68 \sigma_X=\sqrt{0.5\cdot(120-140)^2+0.3\cdot(100-140)^2+0.2\cdot(250-140)^2}=55.68 σ X = 0.5 ⋅ ( 120 − 140 ) 2 + 0.3 ⋅ ( 100 − 140 ) 2 + 0.2 ⋅ ( 250 − 140 ) 2 = 55.68
σ Y = 0.5 ⋅ ( 200 − 165 ) 2 + 0.3 ⋅ ( 150 − 165 ) 2 + 0.2 ⋅ ( 100 − 165 ) 2 = 39.05 \sigma_Y=\sqrt{0.5\cdot(200-165)^2+0.3\cdot(150-165)^2+0.2\cdot(100-165)^2}=39.05 σ Y = 0.5 ⋅ ( 200 − 165 ) 2 + 0.3 ⋅ ( 150 − 165 ) 2 + 0.2 ⋅ ( 100 − 165 ) 2 = 39.05
Coefficient of variation:
C V = σ μ CV=\frac{\sigma}{\mu} C V = μ σ C V X = 55.68 140 = 0.4 CV_X=\frac{55.68}{140}=0.4 C V X = 140 55.68 = 0.4
C V Y = 39.05 165 = 0.27 CV_Y=\frac{39.05}{165}=0.27 C V Y = 165 39.05 = 0.27
Correlation between the two investments:
r = ∑ ( x − μ X ) ( y − μ Y ) ∑ ( x − μ X ) 2 ∑ ( y − μ Y ) 2 r=\frac{\sum (x-\mu_X)(y-\mu_Y)}{\sqrt{\sum (x-\mu_X)^2\sum (y-\mu_Y)^2}} r = ∑ ( x − μ X ) 2 ∑ ( y − μ Y ) 2 ∑ ( x − μ X ) ( y − μ Y ) r = ( 120 − 140 ) ( 200 − 165 ) + ( 100 − 140 ) ( 150 − 165 ) + ( 250 − 140 ) ( 100 − 165 ) ( ( 120 − 140 ) 2 + ( 100 − 140 ) 2 + ( 250 − 140 ) 2 ) ( ( 200 − 165 ) 2 + ( 150 − 165 ) 2 + ( 100 − 165 ) 2 ) = r=\frac{(120-140)(200-165)+(100-140)(150-165)+(250-140)(100-165)}{\sqrt{((120-140)^2+(100-140)^2+(250-140)^2)((200-165)^2+(150-165)^2+(100-165)^2)}}= r = (( 120 − 140 ) 2 + ( 100 − 140 ) 2 + ( 250 − 140 ) 2 ) (( 200 − 165 ) 2 + ( 150 − 165 ) 2 + ( 100 − 165 ) 2 ) ( 120 − 140 ) ( 200 − 165 ) + ( 100 − 140 ) ( 150 − 165 ) + ( 250 − 140 ) ( 100 − 165 ) =
= − 20 ⋅ 35 + 40 ⋅ 15 − 110 ⋅ 65 ( ( − 20 ) 2 + ( − 40 ) 2 + 11 0 2 ) ( 3 5 2 + ( − 15 ) 2 + ( − 65 ) 2 ) = − 7250 14100 ⋅ 5675 = − 0.81 =\frac{-20\cdot35+40\cdot15-110\cdot65}{\sqrt{((-20)^2+(-40)^2+110^2)(35^2+(-15)^2+(-65)^2)}}=\frac{-7250}{\sqrt{14100\cdot 5675}}=-0.81 = (( − 20 ) 2 + ( − 40 ) 2 + 11 0 2 ) ( 3 5 2 + ( − 15 ) 2 + ( − 65 ) 2 ) − 20 ⋅ 35 + 40 ⋅ 15 − 110 ⋅ 65 = 14100 ⋅ 5675 − 7250 = − 0.81
Answer:
μ X = 140 \mu_X=140 μ X = 140 μ Y = 165 \mu_Y=165 μ Y = 165
σ X = 55.68 \sigma_X=55.68 σ X = 55.68 σ Y = 39.05 \sigma_Y=39.05 σ Y = 39.05
C V X = 0.4 CV_X=0.4 C V X = 0.4 C V Y = 0.27 CV_Y=0.27 C V Y = 0.27
r = − 0.81 r=-0.81 r = − 0.81
The coefficient of variation shows the risk per unit of return. The smaller the coefficient of variation, the lower risk factor occurs. Therefore choose Y.
The two investment has fairly strong negative relationship.
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