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Efficient Portfolio (2007 Sept)
- Thread starter Jia Syuen
- Start date
Hi Jia
It's the equation for the expected return on the portfolio:
Ep = (2/9 + 4c) x 0.09 + (2/9 + c) x 0.06 + (5/9 - 5c) x 0.03
Ep = 0.05 + 0.27c (1)
We also have an equation for the variance of the return on the portfolio from part (iii)(a) of the question:
σp^2 = 0.001111 + 1.35c^2 (2)
Rearranging (2) to make c the subject and substituting into (1) gives:
Ep = 0.05 + 0.27 √[(σp^2 – 0.001111) / 1.35] (3)
This is the equation of the original efficient frontier in terms of Ep and σp.
The question is asking us to work out the equation of the tangent to the efficient frontier, which passes through the point
Ep = 0.04, σp = 0.
This tangent will have a (straight line) equation:
Ep = 0.04 + mσp (4)
where 0.04 is the intercept and m is the gradient.
At the point where the tangent intercepts the original efficient frontier:
a) the gradients of the two lines, (3) and (4) will be the same
b) the values of Ep will be equal
c) the values of σp will be equal.
From equation (4), we know the gradient is:
m = (Ep – 0.04)/σp (5)
And we can differentiate (3) to get the gradient of the original efficient frontier:
m = dEp / dσp (6)
The idea is that we set the two values of m equal to each other and then solve. However, this all gets very messy algebraically.
To make things easier, the examiners keep everything in terms of c so that there ends up being just a single unknown. We have:
Ep = 0.05 + 0.27c from (1)
σp = √[0.001111 + 1.35c^2] from (2)
Then, equation (5) becomes:
m = (0.01 + 0.27c)/√[0.001111 + 1.35c^2] (7)
and we use the chain rule to work out (6):
m = dEp / dσp
= (dEp / dc) x (dc / dσp)
= (dEp / dc) divided by (dσp / dc)
= 0.27 divided by 1.35 x 2c x 0.5 / √[0.001111 + 1.35c^2]
= (0.27/1.35c) x √[0.001111 + 1.35c^2] (8)
Finally, (if you are still following, very well done), equate (7) and (8) together and solve to get c = 2/90.
then m = 0.3795 from equation (7)
And, from Equation (4), the tangential line is Ep = 0.04 + 0.3795σp
What a difficult question!
Anna
It's the equation for the expected return on the portfolio:
Ep = (2/9 + 4c) x 0.09 + (2/9 + c) x 0.06 + (5/9 - 5c) x 0.03
Ep = 0.05 + 0.27c (1)
We also have an equation for the variance of the return on the portfolio from part (iii)(a) of the question:
σp^2 = 0.001111 + 1.35c^2 (2)
Rearranging (2) to make c the subject and substituting into (1) gives:
Ep = 0.05 + 0.27 √[(σp^2 – 0.001111) / 1.35] (3)
This is the equation of the original efficient frontier in terms of Ep and σp.
The question is asking us to work out the equation of the tangent to the efficient frontier, which passes through the point
Ep = 0.04, σp = 0.
This tangent will have a (straight line) equation:
Ep = 0.04 + mσp (4)
where 0.04 is the intercept and m is the gradient.
At the point where the tangent intercepts the original efficient frontier:
a) the gradients of the two lines, (3) and (4) will be the same
b) the values of Ep will be equal
c) the values of σp will be equal.
From equation (4), we know the gradient is:
m = (Ep – 0.04)/σp (5)
And we can differentiate (3) to get the gradient of the original efficient frontier:
m = dEp / dσp (6)
The idea is that we set the two values of m equal to each other and then solve. However, this all gets very messy algebraically.
To make things easier, the examiners keep everything in terms of c so that there ends up being just a single unknown. We have:
Ep = 0.05 + 0.27c from (1)
σp = √[0.001111 + 1.35c^2] from (2)
Then, equation (5) becomes:
m = (0.01 + 0.27c)/√[0.001111 + 1.35c^2] (7)
and we use the chain rule to work out (6):
m = dEp / dσp
= (dEp / dc) x (dc / dσp)
= (dEp / dc) divided by (dσp / dc)
= 0.27 divided by 1.35 x 2c x 0.5 / √[0.001111 + 1.35c^2]
= (0.27/1.35c) x √[0.001111 + 1.35c^2] (8)
Finally, (if you are still following, very well done), equate (7) and (8) together and solve to get c = 2/90.
then m = 0.3795 from equation (7)
And, from Equation (4), the tangential line is Ep = 0.04 + 0.3795σp
What a difficult question!
Anna
Thanks for your reply! I really appreciate it!
You're welcome Jia 