HULL Ch 5 Practice Question 5.23

[gargi.adhikari]

Hi,
In Reference to FIN_PRODS_HULL_CH5_Determination_Of_Forward_and_Futures_Prices_Practice_Question_5.23 :-

I have the following Practice Questions Hull 5.23:-
1) It states that Ft = St * e ^ (R- Rf ) (T-t)
Isn't Ft = S0 * e ^ (R- Rf ) (T-t) ......?

If we were to expressFt in terms of St, then, Ft= St * e ^ (R-k) *T

2) If the Hedge Ratio is h, how is the Pricing with hedging h( F0- Ft) + St ...?

3) The third I have is in Part b of this question. It says if t= 1 day, then h= e^(R-Rf) *(T-t) and so h reduces to h= e^(R-Rf) *T

If t=1, shouldn't then h= e^(R-Rf)*(T-1) or h= e^(R-Rf)* ( 1 ) if this is interpreted as (T-t) = 1 ..?

 

[David Harper CFA FRM]

Hi @gargi.adhikari As the last question in Hull's chatper 5, this is not an easy question. Briefly is all the time I have today for this, I cannot guarantee I will have the time to go deep in supporting Hull's question, my workload is just too much:

1. I think the confusion is related to a point I made here https://forum.bionicturtle.com/threads/hull-ch-5-practice-question-5-20.13811/. Our basic COC is given by F(0) = S(0)*exp[(r-q)*T], where q = Rf under IRP. Normally, we are retrieving today's theoretical price of the futures contract, and notationally this can alternatively be represented by: F(0,T) = S(0)*exp[(r-q)*T]; i.e., the price today which is the predetermined delivery price for a futures contract that matures in (T) years. But Hull has generalized this to any point in time. If it's true today, T(0), then it's true at any point in time T(t): F(t) = S(t)*exp[(r-q)*T]
2. He means the net price (net value) of the position including the hedge. The hedge ratio is simply the ratio of units hedged (to the underlying exposure units). If the hedge ratio is 0.50, and your underlying exposure is long the commodity (e.g., you are producer who plans to sell in the future at the then-prevailing price. Say, corn farmer), then in the future, you will receive +S(t) which is the price of corn. Your hedge will be short futures contract, so your gain on the hedge will be F(0) - F(t); e.g., a gain (loss) if the future price decreases (increases). But you are hedging only 0.5 units (as a ratio), so your gain is 0.5*[F(0) - F(t)]. Hence the value of your overall position, in the future at time (t) is given by 0.5*[F(0) - F(t)] + S(t); i.e., hedge + exposure
3. Under h= e^(R-Rf)*(T-t), when t = 1, we have h= e^(R-Rf)*(T-1) --> h ~ e^(R-Rf)*(T) because T-1 is approximately equal to T. I personally think his notation is confusing here. I think his question is basically about a hedge that you would put on tomorrow (0 + 1 day) rather than today, for a contract expiring in the future at time T. I hope that's helpful,

 

[gargi.adhikari]

Thanks so much @David Harper CFA FRM for the above explanation- crystal clear now - Thank You so much for taking the time to explain the above

 

[ankit4685]

Hi @David Harper CFA FRM
Can you kindly help me to understand the logic of "Zero Variance Hedge" in context of this question. I'm bit confussed on Zero Variance hedge condition so not sure how we drived zero Variance hedge over here.

Price with Hedging : hF(0) + S(t) - hS(t)e^(r-rf)(T-t).
we have h = e^( rf-r )(T-t)

As per the given example, post substitiion price of hedging reduce to hF(0).
Not sure why we substitied "e^( rf-r )(T-t)" only for hS(t)e^(r-rf)(T-t) not hf(0)

 

[David Harper CFA FRM]

Hi @ankit4685 Hull could also have substituted into the first (h), but he didn't need to, as he only substitutes into the second to neutralize the exponent and simplify. FWIW, this is intermediate/advanced and not essential for the exam.

Hull uses cost of carry to replace Ft as function of St: Ft= St*exp[(r-rf)(T-t)]
Then: h*F0 + St - h*St*exp[(r-rf)(T-t)]

We are selecting (reverse engineering) the hedge ratio to achieve zero variance: h = exp[(rf-r)(T-t)], where rf is AUD riskfree rate; this hedge ratio is "solved for" precisely because it neutralizes in the exponent, so that:

• h*F0 + St - St*exp[(r-rf)(T-t)]*exp[(rf-r)(T-t)]
• = h*F0 + St - St*exp[(r-rf + rf -r)(T-t)]
• = h*F0 + St - St*exp[(0)(T-t)]
• = h*F0 + St - St*1.0
• = h*F0 + St - St
• = h*F0; you could still substitute into the other h, but Hull had no purpose for doing that:
• h*F0 = exp[(rf-r)(T-t)]*F0. I hope that's helpful,