# Previsibility in Binomial Representation Theorem

I'm working through Baxter and Rennie's "Financial Calculus: An Introduction to Derivative Pricing". It was going very well and I've actually found it an easy read up until the point where they introduce the Binomial Representation Theorem.

Here they ask us to consider the branching from tick i-1 to tick i i.e. $\Delta S_i = S_i - S_{i-1}$ and $\Delta N_i = N_i - N_{i-1}$ where S and N are our two martingales.

Q1: At this point they comment that the variability of these increments depends on the branch structure. Are they referring to the probabilities associated to each branch or to the value of the random variables at time i?

Moving on, they say that since there are only two places to go, any random variable dependent on the branch is fully determined by its width size and a constant offset dependent on the filtration $F_{i-1}$. So if we want to construct one random process out of another it will in general be a construction based on a scaling (to match the widths) and a shift (to match the offsets).

Q2: Am I correctly interpreting this when I rearrange the above eqns to give $S_i = \Delta S_i + S_{i-1}$ and $N_i = \Delta N_i + N_{i-1}$, and then view the widths as being the $\Delta S_i, \Delta N_i$ parts and the offsets as being the $S_{i-1}, N_{i-1}$ parts?

Q3: So he's basically saying that to transform one martingale into another we need to multiply it and then shift it? Is there some more intuitive way to see why this should be the case?

Now to the more confusing stuff....he next says that we'll consider the scaling first. The size of the difference between and up and down jump is $\delta s_i = s_u - s_d$ for S and $\delta n_i = n_u - n_d$ for N, and that both of these are dependent only on the filtration $F_{i-1}$. So we can define the ratio of these branch widths as $\phi_i = \frac{\delta n_i}{\delta s_i}$.

Q4: I cannot wrap my head around why $\delta s_i, \delta n_i$ are previsible - they clearly depend on future values of the random variable, namely $s_u,s_d,n_u,n_d$. This is glossed over in all the notes I've found which is a bit irritating. It seems important as only once we can confirm that this is true can we conclude that $\phi_i$ is indeed previsible as the Theorem states.

Q5: He is now referring to $\delta s_i$ as a branch width, whereas he was calling $\Delta S_i$ a width a few lines earlier (hence my second question above). In any case, $\phi_i$ is constructed as a ratio such that when it multiplies the $\Delta S_i$ increment it will bring it to the correct scale for $\Delta N_i$. Is this correct?

Thank you!

• I had very similar questions! Thanks for asking. – Dhruv Gupta Jul 8 '19 at 8:22

Are they referring to the probabilities associated to each branch or to the value of the random variables at time $i$?

They're referring to the variability in the values of the random processes.

Am I correctly interpreting this when I rearrange the above eqns to give $S_i = \Delta S_i + S_{i-1}$ and $N_i = \Delta N_i + N_{i-1}$, and then view the widths as being the $\Delta S_i, \Delta N_i$ parts and the offsets as being the $S_{i-1}, N_{i-1}$ parts?

The authors probably don't mean that. Later on they express $\Delta N_i = \phi_i \Delta S_i + k$, where $k$ is the offset and $\phi_i$ the scaling factor (I found the terminology used in the proof very tedious myself).

So he's basically saying that to transform one martingale into another we need to multiply it and then shift it? Is there some more intuitive way to see why this should be the case?

The intuitive analog that comes to my mind is that of a coordinate transformation. Suppose we have a 2-D Cartesian coordinate system and the coordinates of a point in it. If you change to a different coordinate system in the same plane, then the coordinates of that point will be scaled (depending on the scale of the new axes relative to old axes) and shifted (depending on location of new origin relative to old).

I cannot wrap my head around why $\delta s_i, \delta n_i$ are previsible - they clearly depend on future values of the random variable, namely $s_u,s_d,n_u,n_d$. This is glossed over in all the notes I've found which is a bit irritating. It seems important as only once we can confirm that this is true can we conclude that $\phi_i$ is indeed previsible as the Theorem states.

The key is that both $S$ and $N$ are $\mathbb{Q}-$martingales. Assuming that $S_{i+1}$ can either be $s_u$ or $s_d$ (similarly for $N_{i+1}$)

$S_i = qs_u + (1-q)s_d \implies (q-1)(s_d-S_i) = q(s_u-S_i)$

$N_i = qn_u + (1-q)n_d \implies (q-1)(n_d-N_i) = q(n_u-N_i)$

Dividing the first equation by the second, we get

$\frac{s_d-S_i}{n_d-N_i} = \frac{s_u-S_i}{n_u-N_i}$.

Again, tinkering with the original 2 equations:

$S_i = qs_u + (1-q)s_d \implies S_i-s_d = q(s_u-s_d)$

$N_i = qn_u + (1-q)n_d \implies N_i-n_d = q(n_u-n_d)$,

which gives us

$\frac{s_d-S_i}{n_d-N_i} = \frac{s_u-s_d}{n_u-n_d} = \frac{s_u-S_i}{n_u-N_i}$.

So irrespective of whether the next move is up or down, the ratio $\frac{S_{i+1}-S_i}{N_{i+1}-N_i} = \frac{s_u-s_d}{n_u-n_d}$ will remain the same and is therefore known with the information available till the current step $i$.

$\phi_i$ is constructed as a ratio such that when it multiplies the $\Delta S_i$ increment it will bring it to the correct scale for $\Delta N_i$. Is this correct?

That is correct.