• We are pleased to announce that the winner of our Feedback Prize Draw for the Winter 2024-25 session and winning £150 of gift vouchers is Zhao Liang Tay. Congratulations to Zhao Liang. If you fancy winning £150 worth of gift vouchers (from a major UK store) for the Summer 2025 exam sitting for just a few minutes of your time throughout the session, please see our website at https://www.acted.co.uk/further-info.html?pat=feedback#feedback-prize for more information on how you can make sure your name is included in the draw at the end of the session.
  • Please be advised that the SP1, SP5 and SP7 X1 deadline is the 14th July and not the 17th June as first stated. Please accept out apologies for any confusion caused.

Efficient Portfolio (2007 Sept)

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:

E
p = 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 E
p will be equal
c) the values of σp will be equal.

From equation (4), we know the gradient is:

m = (E
p – 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


 
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:

E
p = 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 E
p will be equal
c) the values of σp will be equal.

From equation (4), we know the gradient is:

m = (E
p – 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!
 
Back
Top