Futures: Linearity and Equivalence

Introduction

Futures contracts are a common financial instrument used in crypto for hedging, speculation, or other purposes. Futures are linear, meaning their payoff (aka P/L) is linear w.r.t. the underlying asset (e.g. BTC in a BTC-USDT future).

A future position is defined by:

Size $s$ (measured in the underlying asset, e.g. BTC)
Entry price $p$ (measured in quote per underlying units, e.g. 100k USDT/BTC)

For example, a long position of 1 BTC entered at 100k would have $(s, p) = (1, 100 k)$ . Conversely, a short position with same size and entry will have negative size, i.e. $(s, p) = (- 1, 100 k)$ .

Time Value

The value of a position $(s, p)$ , given the underlying price $p_{t}$ at a time $t$ is

V (t) = s (p_{t} - p)

For example, our 1BTC long @ 100k, if the current price is 110k, will have a value of 10k:

V (t) = 1 \cdot (110 k - 100 k) = 10 k

Conversely, the short will have negative value:

V (t) = - 1 \cdot (110 k - 100 k) = - 10 k

Linearity

With the value definition, we can easily see how it’s linear w.r.t. the underlying price $p_{t}$ :

Δ = \frac{d V}{d p _{t}} = s

I.e., linearity means that the delta is constant. The change in value of the position is directly proportional to the underlying.

Equivalence

A nice thing of future contracts is their multiple dualities. Opening a long is equivalent to closing a short; closing a long equivalent to opening a short. In essence, only two operations are possible: going long and going short—buying and selling.

Further, multiple positions are always equivalent to a single position. Say for example you do these operations:

Long 2BTC @ 100k
Short 1BTC @ 110k
Long 1BTC @ 90k

What’s the value of the resulting position? Can it be described as a single future position, or do we require a composite equation? Turns out there’s an equivalent, single position. In this case, the resulting position will be $(s, p) = (2, 90 k)$ .

Let’s prove it:

We have a set of positions $(s_{1}, p_{1}), \dots, (s_{n}, p_{n})$ , each with value $V_{i} (t) = s_{i} (p_{t} - p_{i})$ . Thus the total value is:

V (t) = i = 1 \sum n V_{i} (t) = i = 1 \sum n s_{i} (p_{t} - p_{i})

We can (1) split the sum and (2) take out $p_{t}$ , since it’s a constant within the sum:

V (t) = p_{t} i = 1 \sum n s_{i} - i = 1 \sum n s_{i} p_{i}

Now, we’ll call $s = \sum_{i = 1}^{n} s_{i}$ the “total size”, fair enough:

V (t) = p_{t} s - i = 1 \sum n s_{i} p_{i}

And, why not, let’s factor it out:

V (t) = s (p_{t} - \frac{\sum _{i = 1}^{n} s _{i} p _{i}}{s})

What’s that thing on the right? Turns out it’s the “average entry price”—a weighted average of the entry prices (weighted by their size):

p = \frac{\sum _{i = 1}^{n} s _{i} p _{i}}{s}

Putting it all together:

V (t) = s (p_{t} - p)

And what’s that? That’s the value of a future position of size $s$ , entry price $p$ . I.e., our set of positions $(s_{1}, p_{1}), \dots, (s_{n}, p_{n})$ is equivalent to a single position $(s, p)$ , defined as:

$s = \sum_{i = 1}^{n} s_{i}$ , the total size
$p = \sum_{i = 1}^{n} s_{i} p_{i} / s$ , the average entry price

Let’s compute our example above:

$s = 2 - 1 + 1 = 2$
$p = (2 \cdot 100 k - 1 \cdot 110 k + 1 \cdot 90 k) /2 = 90 k$

as promised.

Summary

We’ve seen some insightful things about future instruments:

A future position is entirely described by its size $s$ and entry price $p$
The value of a position is linear w.r.t. the underlying price
A set of future positions is equivalent to a single position with total size and average entry price