# Is there any measure that is a non-trivial combination of VWAP and TWAP?

Is there any measure that is a non-trivial combination of VWAP and TWAP? For example:

\begin{equation} \textrm{VTWAP} = \frac{\textrm{VWAP}+\textrm{TWAP}}{2} \end{equation}

I'm thinking about something like this:

\begin{equation} \textrm{VTWAP}_{\textrm{exp}}(\alpha,T) = \frac{\sum{P_i * V_i * e^{-i*\alpha}}}{\sum{V_i * e^{-i*\alpha}}} \end{equation}

where $P_i$ is the price at time $T-i+1$ and $V_i$ is the volume at time $T-i+1$.

Influence of past volumes is exponentially decayed with factor $\alpha$.

We can see that $\textrm{VTWAP}_{\textrm{exp}}(0,T)=\textrm{VWAP}(T)$.

I think that good point to start to analyse this problem is to find out types of existing TWAPs.

Second part of the question:

Are there any mathematical requirements or equations that measures like TWAP and VWAP should meet?

Something like that, but more advanced: $\textrm{VWAP}(T+1)=\textrm{VWAP}(T)$ for $V_T=0$ which state that there was no trade at time $T$.

• Note sure whether such a measure exist as when looking at the definition it are two different concepts. What kind of applications could such a measure have? – JohnAndrews Apr 13 '12 at 0:45

In fact if you make the time change $$t\rightarrow \int_{\tau\leq t} V_\tau d\tau$$ a TWAP is a VWAP.

So just define the FWAP associate to a transform F: (you should ask to F to be an adapted stochastic process if you want to use models) $$t\rightarrow \int_{\tau\leq t} F(\tau) d\tau$$

You will have a new benchmark.

The real question is "what do you want to capture?"

You can also see a VWAP as the expectation of a volume at price density $d\mu(P)$: $${\rm VWAP} = \mathbb{E}_\mu (P)$$

In such a case just define a GWAP (associating a measure to a measure) as: $${\rm GWAP} = \mathbb{E}_G(\mu) (P)$$

For $G$ transforming a measure into the uniform one over its support: a GWAP is a TWAP (and for G being identity, it is a VWAP).

• Nice contributions last couple of days! @lehalle – Ram Ahluwalia Apr 19 '12 at 5:47
• You are welcome, I am surprised taht this stackexchange is not more active. It is far more useful than a forum or wikipedia. – lehalle Apr 19 '12 at 10:12