# Constructing a long futures hedge

I'm taking a financial engineering course through coursera and on a slide one of the lecturers talks about how a long hedge is used in the futures market. Here is the text:

Today is Sept 1st. A baker needs 500,000 bushels of wheat on December 1st. So, the baker faces the risk of an uncertain price on Dec. 1st.

Hedging strategy: buy 100 futures contracts maturing on Dec 1st - each for 5000 bushels

It goes on to talk about the cash flows on Dec 1st. (numbering mine):

1. Futures position at maturity: $$F_T - F_0 = S_T - F_0$$
2. Buy in the spot market: $$S_T$$
3. Effective cash flow: $$S_T - F_0 - S_T = -F_0$$

However, the lecturer glosses over how (3) is arrived at.

So my questions are as follows:

for (2) - why are we buying in the spot market? Shouldn't the baker be taking delivery of the underlying and therefore already own the underlying?

How is (3) arrived at algebraically? I can see how (1) is mutated into it, but I'm wondering the reasoning behind how the $$-F_0$$ got there, as well as the second $$S_T$$ (which I'm assuming comes from (2).

Thank you! It's been a while since I've messed with this stuff and it sure shows.

• Agree with you regarding (2). It seems like the course is assuming that the futures contract is cash-settled - i.e. it does not deliver the underlying but simply settles the cash difference. (3) is just (1) - (2). Keep in mind that these are cash-flows. So buying in the spot market means that you have a cash-outflow of $S_T$. Feb 20, 2018 at 7:46
• @localvolatility Thank you. Wouldn't the cash flow associated with the purchase in the spot market be negative?
– user20664
Feb 20, 2018 at 7:48
• Yes - that what I mean by "outflow". Feb 20, 2018 at 7:54
• Oh of course. I read too fast. Haha
– user20664
Feb 20, 2018 at 7:54

1) Futures position at maturity $$F_T-F_0=\underbrace{S_T-F_0}_{\text{Futures converge to spot at maturity}}$$
2) Buy in the spot market $$\underbrace{-S_T}_{\text{negative number denotes a cash outflow to purchase spot}}$$
3) Adding 1 and 2 together yields $$\require{cancel}\underbrace{\cancel{S_T}-F_0\cancel{-S_T}}_{\text{simultaneous inflow and outflow of S_T cancel}}=\underbrace{-F_0}_{\text{effective outflow/price paid by baker}}$$
Which shows your ending outflow is simply equal to the futures price at time 0, that is, the baker was hedged at this price. The same holds under physical delivery, or I guess that instead of thinking of the $S_T$ terms as cancelling, they are simply not there.