The reference proposes several ways to compute the expectation of the absolute value of two random variables that follow same discrete positive distribution but with possibly different parameters. The case of Poisson processes is treated. Let’s describe the beginning of Katti’s paper [17]. Let’s note X 1 and X 2 two Poisson random variables of intensity λ 1 and λ 2 . We would like to compute the following number Δ 1 = E | X 1 − X 2 | . One can write the following:

Δ 1 = ∑ i , k   k P ( X 2 − X 1 = k | X 1 = i ) P ( X 1 = i )     + ∑ i , k   k P ( X 1 − X 2 = k | X 2 = i ) P ( X 2 = i ) = ∑ i , k   k P i 1 P i + k 2 + ∑ i , k   k P i 2 P i + k 1 ,

where the summations are over 1 , 2 , 3 , ⋯ and P i 1 = e − λ 1 λ 1 i i ! and P i 2 = e − λ 2 λ 2 i i ! . Then, one can develop it as follows:

Δ 1 = e − λ 1 − λ 2 ( ∑ i = 0 ∞ ∑ i = 0 ∞ k ( λ 1 λ 2 ) i λ 2 k i ! ( i + k ) ! + ∑ i = 0 ∞ ∑ i = 0 ∞ k ( λ 1 λ 2 ) i λ 1 k i ! ( i + k ) ! ) = e − λ 1 − λ 2 ( A 1 + B 2 ) ,

with A 1 and B 1 the two different sums. The author, in order to simplify the calculus and use a trick, makes the following assumptions: λ 1 λ 2 = ν , where ν is a constant not linked anymore to λ 2 nor λ 1 . It implies thus a relation between the two variables (for example λ 1 = 1 λ 2 ). Thanks to this assumption he can do the following:

A 1 = ( λ 2 δ δ λ 2 ) ( ∑ i = 0 ∞ ∑ k = 0 ∞ ν i λ 2 k i ! ( i + k ) ! ) = ( λ 2 δ δ λ 2 ) A 0 ,

say, with:

A 0 = ∑ i = 0 ∞ ν i ( i ! ) 2 F 1 ( 1 ; i + 1 ; λ 2 ) ,

where F 1 ( α , γ , x ) is a confluent hypergeometric function. Operating by ( λ 2 δ δ λ 2 ) it finally leads to:

A 1 = ( λ 2 − ν λ 2 ) A 0 + ν e 2 ν F 1 ( 3 2 ; 3 ; − 4 ν ) + ν λ 2 F 1 ( 1 2 ; 1 ; − 4 ν ) .

The particular case of λ 1 = λ 2 = λ cannot be treated with this trick because it would imply equal numbers are linked by an inverse relation, so that the product is independant of λ 2 . But λ 1 = λ 2 = ν , ν is not anymore a constant of the main parameters λ 1 and λ 2 , so applying the operator does not give the previous results. One may use here another reference, one cited by Katti [18]. We will detail later the same ideas for our precise the VPIN framework. Anyway, this case leads to the following:

Δ 1 = 2 λ e − 2 λ ( I 0 ( 2 λ ) + I 1 ( 2 λ ) ) ,

where I n ( x ) = ∑ i = 0 ∞ ( x 2 ) n + 2 i i ! ( n + i ) ! is a modified Bessel function of first kind.

First, let’s put ourselves in the context where we have the differences of only Poisson processes. It’s pretty simple, one just have to condition the expectation of E ( | K | ) for each case:

E ( | K | ) = α ( 1 − δ ) E ( | K | | g ) + α δ E ( | K | | b ) + ( 1 − α ) E ( | K | | n ) .

Then, remind K = S − B . S and B are, under the model assumption, Poisson processes describing the number of sells and buys in one day of trade. We only need two different kinds of Poisson processes to describe the mixture of Poisson processes resulting of informed and uninformed traders in each case (“good event”, “bad event” and “no event”). Let’s note them as follows X ϵ S ~ P ( ϵ ) , X ϵ B ~ P ( ϵ ) , Y μ B ~ P ( μ ) and Y μ S ~ P ( μ ) , S and B labelling buys or sells. One finds then:

E ( | K | ) = α ( 1 − δ ) E ( | X ϵ S − ( Y μ B + X ϵ B ) | ) + α δ E ( | Y μ B + X ϵ B − X ϵ S | )     + ( 1 − α ) E ( | X ϵ S − X ϵ B | ) .

As all Poisson processes are independant one can sum them to produce new Poisson processes, as follows³:

E ( | K | ) = α ( 1 − δ ) E ( | X ϵ − Y μ + ϵ | ) + α δ E ( | Y μ + ϵ − X ϵ | ) + ( 1 − α ) E ( | X ϵ ( 1 ) − X ϵ ( 2 ) | ) .

One can thus sum the two first terms and obtain the following:

E ( | K | ) = α E ( | X ϵ − Y μ + ϵ | ) + ( 1 − α ) E ( | X ϵ ( 1 ) − X ϵ ( 2 ) | ) .

One has to treat finally two different cases:

· different intensities: first term

· same intensities: second term

We have seen that Katti’s reference use the assumption that Poisson intensities are linked by a relation of the form λ 1 = ν λ 2 where ν is independent of these parameters. Here the respective parameters would be μ + ϵ and ϵ . The product ϵ ( ϵ + μ ) has clearly no single reason to be a constant. One could create some tricky cases, but it does not seem that the model would like to be limited to these cases (indeed, one may consider for example to fit the PIN parameters maximising likelihood, like in [1] ). Thus the assumptions are not met and the reference [17] cannot be invoked to say E | K | ≈ α μ at first order, as it was done in [1] for example.

Anyway, let’s do nevertheless calculations to compute E ( | S − B | ) . We follow the same natural ideas of T. A. Ramasubban in this paper which treats only the case of same Poisson intensities [18]. We begin with:

Δ 1 = α E ( | X ϵ − Y μ + ϵ | ) + ( 1 − α ) E ( | X ϵ 1 − X ϵ 2 | ) .

Let’s start with the easier calculation: the case where Poisson intensities are equal.

E ( | X ϵ 1 − X ϵ 2 | ) = ∑ i , j ∈ ℕ P ( X ϵ 1 = i ) P ( X ϵ 2 = j ) | i − j | = 2 ∑ i ∈ ℕ ∑ j = 0 i P ( X ϵ 1 = i ) P ( X ϵ 2 = j ) ( i − j ) = 2 e − 2 ϵ ∑ i ∈ ℕ ∑ j = 0 i ( i − j ) ϵ i i ! ϵ j j ! = 2 ϵ e − 2 ϵ ∑ i ∈ ℕ * ∑ j = 0 i ϵ i − 1 ( i − 1 ) ! ϵ j j ! − ∑ j = 1 i ϵ i i ! ϵ j − 1 ( j − 1 ) ! .

All the sums separately exist, we can split them in two different ones:

E ( | X ϵ 1 − X ϵ 2 | ) = 2 ϵ e − 2 ϵ ( ∑ i ∈ ℕ ∑ j = 0 i + 1 ϵ i i ! ϵ j j ! − ∑ i ∈ ℕ * ∑ j = 0 i − 1 ϵ i i ! ϵ j j ! ) = 2 ϵ e − 2 ϵ ( ∑ i ∈ ℕ ϵ i i ! ( ϵ i + 1 ( i + 1 ) ! + ϵ i i ! ) ) = 2 ϵ e − 2 ϵ ( ∑ i ∈ ℕ ϵ 2 i + 1 i ! ( i + 1 ) ! + ∑ i ∈ ℕ ϵ 2 i i ! 2 ) .

One recognizes here a modified Bessel functions of first kind: for an integer n and, say scalar x, I n ( x ) = ∑ i ∈ ℕ ( x 2 ) 2 i + n i ! ( n + i ) ! . Here we obtain:

E ( | Y B − Y S | ) = 2 ϵ e − 2 ϵ ( I 0 ( 2 ϵ ) + I 1 ( 2 ϵ ) ) .

which is the result of Ramasubban’s quoted paper. The computation with different intensities follow the same idea, expect that the symmetry of the two initial sums is broken, so we have to compute them separately.

E | X ϵ − Y μ + ϵ | = ∑ i ∈ ℕ ∑ j = 0 i P ( X ϵ = i ) P ( Y ϵ + μ = j ) ( i − j )     + ∑ i ∈ ℕ ∑ j = 0 i P ( X ϵ = j ) P ( Y ϵ + μ = i ) ( i − j ) .

Let’s calculate the first sum and then the second:

∑ i ∈ ℕ ∑ j = 0 i   P ( X ϵ = i ) P ( Y ϵ + μ = j ) ( i − j ) = e − 2 ϵ − μ ( ∑ i = 1 + ∞ ∑ j = 0 i ϵ i ( i − 1 ) ! ( ϵ + μ ) j j ! − ∑ i = 1 + ∞ ∑ j = 1 i ϵ i i ! ( ϵ + μ ) j ( j − 1 ) ! ) = e − 2 ϵ − μ ( ∑ i = 0 + ∞ ∑ j = 0 i + 1 ϵ i + 1 i ! ( ϵ + μ ) j j ! − ∑ i = 1 + ∞ ∑ j = 0 i − 1 ϵ i i ! ( ϵ + μ ) j + 1 j ! ) ,

which separates as follows as all sums exist separately:

∑ i ∈ ℕ ∑ j = 0 i   P ( X ϵ = i ) P ( Y ϵ + μ = j ) ( i − j ) = e − 2 ϵ − μ ϵ ∑ i = 0 + ∞ ϵ i i ! ( ( ϵ + μ ) i + 1 ( i + 1 ) ! + ( ϵ + μ ) i i ! ) − e − 2 ϵ − μ μ ∑ i = 0 + ∞ ∑ j = 0 i ϵ i + 1 ( i + 1 ) ! ( ϵ + μ ) j j ! .

Replacing first sum of the rigth hand side by Bessel functions of second kind, we finally find:

∑ i ∈ ℕ ∑ j = 0 i   P ( X ϵ = j ) P ( Y ϵ + μ = i ) ( i − j ) = e − 2 ϵ − μ ϵ ( ϵ + μ ) I 1 ( 2 ϵ ( ϵ + μ ) ) + e ϵ ( ϵ + μ ) ϵ I 0 ( 2 ϵ ( ϵ + μ ) )       − e − 2 ϵ − μ μ ∑ i = 0 + ∞ ∑ j = 0 i ϵ i + 1 ( i + 1 ) ! ( ϵ + μ ) j j ! .

For the second sum, we do an equivalent calculus and find the following:

∑ i ∈ ℕ ∑ j = 0 i   P ( X ϵ = j ) P ( Y ϵ + μ = i ) ( i − j ) = e − 2 ϵ − μ ϵ ∑ i = 0 + ∞ ( ϵ + μ ) i i ! ( ϵ i + 1 ( i + 1 ) ! + ϵ i i ! ) − e − 2 ϵ − μ μ ∑ i = 0 + ∞ ∑ j = 0 i ϵ i + 1 ( i + 1 ) ! ( ϵ + μ ) j j ! ,

thus:

∑ i ∈ ℕ ∑ j = 0 i   P ( X ϵ = j ) P ( Y ϵ + μ = i ) ( i − j ) = e − 2 ϵ − μ ϵ 2 ϵ ( ϵ + μ ) I 1 ( 2 ϵ ( ϵ + μ ) ) + e − 2 ϵ − μ ϵ I 0 ( 2 ϵ ( ϵ + μ ) )       + e − 2 ϵ − μ μ ∑ i = 0 + ∞ ( ϵ + μ ) i i ! ∑ j = 0 i + 1 ϵ j j ! .

If we put together all the terms we find:

E | X ϵ − Y ϵ + μ | = e − 2 ϵ − μ ( 2 ϵ 2 + ϵ μ ϵ ( ϵ + μ ) I 1 ( 2 ϵ ( ϵ + μ ) ) + 2 ϵ I 0 ( 2 ϵ ( ϵ + μ ) )     − μ ∑ i = 0 + ∞ ∑ j = 0 i ϵ i + 1 ( i + 1 ) ! ( ϵ + μ ) j j ! + μ ∑ i = 0 + ∞ ( ϵ + μ ) i i ! ∑ j = 0 i + 1 ϵ j j ! ) .

Arranging the last two sums of the left hand side of the equality we finnaly get:

E | X ϵ − Y ϵ + μ | = e − 2 ϵ − μ ( 2 ϵ 2 + ϵ μ ϵ ( ϵ + μ ) I 1 ( 2 ϵ ( ϵ + μ ) ) + 2 ϵ I 0 ( 2 ϵ ( ϵ + μ ) )     + μ ∑ i = 0 + ∞   P ( Y ϵ + μ = i ) ( P ( X ϵ ≤ i + 1 ) − P ( X ϵ ≥ i + 1 ) ) .

Thus E | K | equals:

E | K | = 2 ϵ ( 1 − α ) e − 2 ϵ [ I 0 ( 2 ϵ ) + I 1 ( 2 ϵ ) ]     + α e − 2 ϵ − μ [ 2 ϵ 2 + ϵ μ ϵ ( ϵ + μ ) I 1 ( 2 ϵ ( ϵ + μ ) ) + 2 ϵ I 0 ( 2 ϵ ( ϵ + μ ) ) ]     + α μ ∑ i = 0 + ∞   P ( Y ϵ + μ = i ) ( P ( X ϵ ≤ i + 1 ) − P ( X ϵ ≥ i + 1 ) ) .

With an arbitrarily time length t for a trading period, we find:

E | K | = 2 ϵ t ( 1 − α ) e − 2 ϵ t [ I 0 ( 2 ϵ t ) + I 1 ( 2 ϵ t ) ]     + α e − 2 ϵ t − μ t [ 2 ( ϵ t ) 2 + ϵ μ t 2 ϵ ( ϵ + μ ) t I 1 ( 2 ϵ ( ϵ + μ ) t ) + 2 ϵ t I 0 ( 2 ϵ ( ϵ + μ ) t ) ]     + α μ t ∑ i = 0 + ∞   P ( Y ( ϵ + μ ) t = i ) ( P ( X ϵ t ≤ i + 1 ) − P ( X ϵ t ≥ i + 1 ) ) .

Recall that ϵ and μ are rates of uninformed and informed traders per day (in the original the PIN model). Thus, these parameters are pretty high integers: this is the first intuition behind first order approximate. Moreover, Hankel [19] derived an asymptotic expansion of modified Bessel function of first kind as follows:

I α ( z ) ~ e z 2 π z ( 1 − 4 α 2 − 1 8 z + ( 4 α 2 − 1 ) ( 4 α 2 − 9 ) 2 ! ( 8 z ) 2     − ( 4 α 2 − 1 ) ( 4 α 2 − 9 ) ( 4 α 2 − 25 ) 3 ! ( 8 z ) 3 + ⋯ )

for | z | ≫ 1 and | arg z | < π 2

We first apply this expansion to E | K | with the condition μ ≫ 1 and ϵ ≫ 1 , as we consider there are a lot of informed and uninformed traders per day (compared to 1). We find the following:

E | K | ~ 2 ϵ π ( 1 − α ) + α 2 π ( ϵ + ϵ ( ϵ + μ ) ) 2 ( ϵ ( ϵ + μ ) ) 3 4 e 2 ϵ ( ϵ + μ ) − 2 ϵ − μ     + α μ ∑ i = 0 + ∞   P ( Y ϵ + μ = i ) ( P ( X ϵ ≤ i + 1 ) − P ( X ϵ ≥ i + 1 ) ) .

Let’s now distinguish these three cases:

· μ and ϵ are of same order,

· μ = o ( ϵ ) ,

· ϵ = o ( μ ) ,

If μ and ϵ are of same order,

in this case: ϵ ( ϵ + μ ) < 2 ϵ − μ , thus one can neglect the corresponding term. We obtain:

E | K | ~ 2 ϵ π ( 1 − α ) + α μ ∑ i = 0 + ∞   P ( Y ϵ + μ = i ) ( P ( X ϵ ≤ i + 1 ) − P ( X ϵ ≥ i + 1 ) ) .

Thus PIN ~ α μ α μ + 2 ϵ and

E | S − B | E ( S + B ) ~ 2 ϵ π ( 1 − α ) + α μ ∑ i = 0 + ∞   P ( Y ϵ + μ = i ) ( P ( X ϵ ≤ i + 1 ) − P ( X ϵ ≥ i + 1 ) ) α μ + 2 ϵ .

if ϵ ≪ μ then it reduces to:

E | K | ~ α μ ∑ i = 0 + ∞   P ( Y ϵ + μ = i ) ( P ( X ϵ ≤ i + 1 ) − P ( X ϵ ≥ i + 1 ) ) .

Thus:

E | S − B | E ( S + B ) ~ α μ α μ + 2 ϵ ∑ i = 0 + ∞   P ( Y ϵ + μ = i ) ( P ( X ϵ ≤ i + 1 ) − P ( X ϵ ≥ i + 1 ) ) .

If μ = o ( ϵ ) ,

we find:

E | S − B | E ( S + B ) ~ 2 ϵ π + α μ ∑ i = 0 + ∞   P ( Y ϵ + μ = i ) ( P ( X ϵ ≤ i + 1 ) − P ( X ϵ ≥ i + 1 ) ) 2 ϵ .

And: PIN ~ α μ 2 ϵ ≪ 1 .

If ϵ ≫ μ , then:

E | S − B | E ( S + B ) ~ 1 ϵ π .

If ϵ = o ( μ ) ,

we find:

E | S − B | E ( S + B ) ~ ∑ i = 0 + ∞ P ( Y ϵ + μ = i ) ( P ( X ϵ ≤ i + 1 ) − P ( X ϵ ≥ i + 1 ) ) .

and P I N ~ 1

Thus, we can see that first order approximation depends a lot of:

· the respective values of μ and ϵ ,

· and in a lot of cases of the weighted average of a given Poisson distribution of the difference between cumulative density functions from opposite parts of the tail of another Poisson distribution, i.e.: ∑ i = 0 + ∞   P ( Y ϵ + μ = i ) ( P ( X ϵ ≤ i + 1 ) − P ( X ϵ ≥ i + 1 ) )

The first order approximation E | S − B | ~ α μ proposed in [1] is not incorrect as we will see in the simulations, but sometimes, imprecise.

In [3] D. Easley, M. de Prado and M. O’Hara describe a new model to compute easily the VPIN and therefore the PIN using the above previous results:

· E ( | S − B | ) = α μ , supposedly at first order,

· E ( S + B ) = α μ + 2 ϵ

They introduce the paradigm of volume clock and time bars. Let’s first describe it and see that the assumptions are implicitly changed, but ignored. The idea is pretty simple. Consider a trade described by a time serie of price, say p t , labelled with time t. First, They package trades in objects called “bars” that have a fixed time volume, i.e.: they aggregate the time serie in, for example, one-minute time bars. It is equivalent to a sampling of the time serie. Each bar is a kind of new trade with several rules to guess its price. Second they agreggate these time bars to form fixed in volum “buckets”. Say these buckets have a volume V.

· Remark 1: nothing can ensure us that buckets will have a fixed volume size. Indeed, each time bar is sensitive to trading intensity. The last time bar can often be too big to be aggregated to a fixed size bucket. Which mean, that if one wants to force bucket size to be constant then, a lot of time bar won’t be of one minute lenght. If one on the contrary wants to preserve time size to be constant, a lot of buckets might not be of constant volume size.

Suppose anyway that everything is ideal and that each bucket is of constant volume. Authors note τ the label of a bucket of volume V, V τ , and V τ S and V τ B respectively the total number of sells and buys that occured in this bucket. They then refer to their previous work [1] result: E ( | V τ S − V τ B | ) ≈ α μ . But even if the result does not hold as previously shown, one must note the following:

· First: here the bucket is constant in volume, thus filling volume time is random, it is a really strong hypothesis, as we have then: V τ S + V τ B = V that holds almost surely,

· Second: they use the result indeed to say that as V τ S + V τ B = V then the expectation equals V = 2 ϵ + α μ .

· But finally, one should remark that this equality lacks a time, as we are talking of rates of traders. In the first model, the time was one day, and implicitly one would multiply within the time-clock framework, rates by one day. Here, in the volume-clock framework, one does not control anymore time. One should take into account filling bucket time which is a new random variable. At first glance, the expression is inhomogenous and even if right, it is far from being trivial.

Indeed the authors preciss us “recall that we divide the trading day into equal-sized volume buckets and treat each volume bucket as equivalent to a period for information arrival”. It’s misleading. Recall that in the initial model time is fixed (one day) and thus volum is random. Here one has the contrary, volume is fixed and time is thus random. Let’s detail a bit more the calculus with the new assumptions. To do so let’s precise a bit more the new implicit framework.

In fact we want to compute now Δ 1 = E ( | V τ S − V τ B | | V ) , as bucket volume is fixed. Note t − t ′ the filling time of the bucket τ and then note the following:

Δ 1 = E ( | S t t o t − S t ′ t o t − ( B t t o t − B t ′ t o t ) | | V ) ,

with S t t o t , S t ′ t o t , B t t o t and B t ′ t o t the Poisson processes of the total sell up to t and t ′ and the total buys up to t and t ′ . One has in distribution the following:

· S t t o t − S t ′ t o t = N S , t − t ′ t o t

· B t t o t − B t ′ t o t = N B , t − t ′ t o t

where N S , t − t ′ t o t and N B , t − t ′ t o t are Poisson processes describing total sells and buys in the bucket labelled with τ . The variables being independent, we can thus write the following:

Δ 1 = E ( | N S , t − t ′ t o t − N B , t − t ′ t o t | | V ) .

One must note that there is still the constraint of the volume of a bucket:

N S , t − t ′ t o t + N B , t − t ′ t o t = V .

Thus, imposing one value, imposes the other. Let’s calculate Δ 1 . First, one can condition the events: “good event” (g), “bad event” (b) and “no event” (n):

Δ 1 = α ( 1 − δ ) E ( | N S , t − t ′ t o t − N B , t − t ′ t o t | | g , V ) + α δ E ( | N S , t − t ′ t o t − N B , t − t ′ t o t | | b , V )     + ( 1 − α ) E ( | N S , t − t ′ t o t − N B , t − t ′ t o t | | n , V ) .

On each event, one knows the distribution of N S , t − t ′ t o t and N B , t − t ′ t o t . One can then re-write it the following way⁴:

Δ 1 = α ( 1 − δ ) E ( | N ϵ ( t − t ′ ) t o t ,1 − N ( ϵ + μ ) ( t − t ′ ) t o t ,1 | | V ) + α δ E ( | N ( μ + ϵ ) ( t − t ′ ) t o t ,2 − N ϵ ( t − t ′ ) t o t ,2 | | V )     + ( 1 − α ) E ( | N ϵ ( t − t ′ ) t o t ,3 − N ϵ ( t − t ′ ) t o t ,4 | | V ) .

The two first terms corresponding to “good” or “bad events” are equal in distribution, that’s why we have:

Δ 1 = α E ( | N ϵ ( t − t ′ ) t o t , 1 − N ( ϵ + μ ) ( t − t ′ ) t o t | | V ) + ( 1 − α ) E ( | N ϵ ( t − t ′ ) t o t , 2 − N ϵ ( t − t ′ ) t o t , 3 | | V ) .

Before going further, let’s implement the joint probability density function of for example, sells and buys and respective filling bucket time t-t’ in the case of a bad event. Let’s note it f ( S , B , t − t ′ | V , b ) . Now, we synthetise and refer to the great ideas of the proof of Kin and Le [14]. Remark first the following:

f ( S , B , t − t ′ | V , b ) = f ( S , B | V , t − t ′ , b ) f ( t − t ′ | V , b ) ,

and as f ( V | t − t ′ , b ) follows a Poisson law of intensity ( t − t ′ ) ( 2 ϵ + μ ) , then f ( t − t ′ | V , b ) classically follows an Erlang law with the following parameters Γ ( t − t ′ ; V ,2 ϵ + μ ) . Second, as S + B = V almost surely, we have the following equalities:

f ( S , B | t − t ′ , V , b ) = f ( S | t − t ′ , V , b ) = f ( B | t − t ′ , V , b ) ,

and:

f ( S , V | t − t ′ , b ) = f ( B | V , t − t ′ , b ) f ( V | t − t ′ , b ) .

We know f ( V | t − t ′ , b ) ~ P ( ( 2 ϵ + μ ) ( t − t ′ ) ) and considering for example a continuous bounded function g, one can guess easily f ( B | V , t − t ′ , b ) computing E ( g ( S , V | t − t ′ , b ) ) using that E ( g ( S , V | t − t ′ , b ) ) = E ( g ( S , S + B | t − t ′ , b ) ) . We find a binomial law B ( S ; V , ϵ 2 ϵ + μ ) i.e.:

f ( S | V , t − t ′ , b ) = f ( B | V , t − t ′ , b ) = B ( S ; V , ϵ 2 ϵ + μ ) = B ( B ; V , ϵ + μ 2 ϵ + μ ) .

So finally:

f ( S , B , t − t ′ | V , b ) = B ( S ; V , ϵ 2 ϵ + μ ) Γ ( V ,2 ϵ + μ ) .

The “no event” case is similar. We thus find the following:

f ( S , B , t − t ′ | V ) = α B ( S ; V , 1 2 ) Γ ( t − t ′ ; V ,2 ϵ ) + ( 1 − α ) B ( S ; V , ϵ 2 ϵ + μ ) Γ ( t − t ′ ; V ,2 ϵ + μ ) ,

And after an integration on the random variable t-t’:

f ( S , B | V ) = α B ( S ; V 1 2 ) + ( 1 − α ) B ( S ; V , ϵ 2 ϵ + μ ) .

Taking the previous joint probability into account we are thus computing the following expectations of let say X and Y in fact:

Δ 1 = α E ( | V − 2 X | ) + ( 1 − α ) E ( | V − 2 Y | )

with Y ~ B ( S ; V , 1 2 ) and X ~ B ( V , S , ϵ 2 ϵ + μ )

Moreover, if x follows the binomail distribution of which p.d.f is B ( x ; m , p ) , then using Jensen inequality for the concave function y → y we have:

· E ( | m − 2 x | ) m = E ( [ ( m − 2 x ) 2 ] 1 2 ) / m ≤ ( 2 p − 1 ) 2 + 4 p ( 1 − p ) m ≈ | 2 p − 1 | for large enough m and p differing from 1 2

· E ( | m − 2 x | ) m ≥ | E ( m − 2 x ) | m = | 2 p − 1 | .

Thus, for large enough V:

Δ 1 ≈ V α | 2 ϵ 2 ϵ + μ − 1 | + 0,

i.e.

Δ 1 ≈ V α μ 2 ϵ + μ .

Thus the VPIN metric approximates the following for large enough n as shown by Kin and Le [14] :

VPIN = ∑ τ = 1 n | V τ S − V τ B | n V ≈ E ( | V τ S − V τ B | | V ) V ≈ α μ 2 ϵ + μ ,

which is indeed different of PIN = α μ α μ + 2 ϵ .

For purpose of illustration, we compare the empirical form of E ( | B − S | ) E ( S + B ) with the PIN and the asymptotic limit5 found within the time clock framework for different cases of ϵ and μ . It is pretty easy to do, as controlling ex-ante all the parameters of the model one then just has to generate the appropriate Poisson processes to obtain all the values. We illustrate the results with three examples:

· ϵ = o ( μ ) and μ of same order than ϵ : we took ϵ = 100 and μ ∈ { 10000,20000,30000 } ,

· μ = o ( ϵ ) and ϵ of same order than μ : we took μ = 100 and ϵ ∈ { 10000,20000,30000 } ,

· ϵ of same order than μ : we took6 ϵ = 10000 and μ ∈ { 10000,2000,30000 } .

Remarks:

· We compute 20 values for each choice of ϵ and μ in the three cases above,

· For each of the 20 values, the empirical expectations are computed with an average of 10,000 values,

· To compute the sum ∑ i = 0 + ∞   P ( Y ϵ + μ = i ) ( P ( X ϵ ≤ i + 1 ) − P ( X ϵ ≥ i + 1 ) ) , considering the values of ϵ and ϵ + μ , we have bounded the sum to i = 100000 , when probability values starts to be then very little.

On each case, we plot first the empirical numerator E ( | S − B | ) , α μ , and the asymptotic limit found (Figure 2, Figure 4 and Figure 6). Second, we plot E ( | B − S | ) E ( S + B ) , the PIN (i.e. α μ α μ + 2 ϵ and the asymptotic limit divided by α μ + 2 ϵ (Figure 3, Figure 5 and Figure 7).

Case1: ϵ = 100 , μ ∈ { 10000 , 20000 , 30000 }

On Figure 2, first order and asymptotic estimations are very close.

Case 2: μ = 100 , ϵ ∈ { 10000 , 20000 , 30000 }

On Figure 3 and Figure 4, one can see better the difference when one does not change μ anymore.

Case 3: ϵ = 10000 , μ ∈ { 10000 , 2000 , 30000 }

This last case on Figure 6 and Figure 7 illustrates a market where the number of informed and uninformed traders are of same order.

The probability to obtain S sells during a time period of length t is the following:

P ( z t = S ) = α 2 e − ϵ t ( ϵ t ) S S ! + ( 1 − α ) e − ϵ t ( ϵ t ) S S ! + α 2 e − ( μ + ϵ ) t ( ( ϵ + μ ) t ) S S ! = ( 1 − α 2 ) e − ϵ t ( ϵ t ) S S ! + α 2 e − ( μ + ϵ ) t ( ( ϵ + μ ) t ) S S ! .

Let’s compute the moment-generating function of this process. We will estimate the numerator using relations between moments. Let u be a real value, let V S be the random variable representing the volume of sells and t the fixed time period associated. We have:

E ( e V S u ) = ( 1 − α 2 ) e ϵ t ( e u − 1 ) + α 2 e ( ϵ + μ ) t ( e u − 1 ) .

Let’s compute the first three moments of V S :

· First moment:

E ( V S e V S u ) = ( 1 − α 2 ) ϵ t e u e ϵ t ( e u − 1 ) + α 2 ( ϵ + μ ) t e u e ( ϵ + μ ) t ( e u − 1 ) ,

so:

E ( V S ) = ( ϵ + α μ 2 ) t .

· Second moment:

E ( V S 2 e V S u ) = ( 1 − α 2 ) ϵ t e u e ϵ t ( e u − 1 ) + ( 1 − α 2 ) ( ϵ t ) 2 e 2 u e ϵ t ( e u − 1 )     + α 2 ( ϵ + μ ) t e u e ( ϵ + μ ) t ( e u − 1 ) + α 2 ( ( ϵ + μ ) t ) 2 e 2 u e ( ϵ + μ ) t ( e u − 1 ) ,

so:

E ( V S 2 ) = ( ϵ + α μ 2 ) t + ( α μ ϵ + ϵ 2 + α μ 2 2 ) t 2 ,

i.e. we have the classic decomposition:

E ( V S 2 ) = E ( V S ) + E ( V S ( V S − 1 ) ) .

· Third moment:

E ( V S 3 e V S u ) = ( 1 − α 2 ) ϵ t e u e ϵ t ( e u − 1 ) + ( 1 − α 2 ) ( ϵ t ) 2 e 2 u e ϵ t ( e u − 1 )     + 2 ( 1 − α 2 ) ( ϵ t ) 2 e 2 u e ϵ t ( e u − 1 ) + ( 1 − α 2 ) ( ϵ t ) 3 e 3 u e ϵ t ( e u − 1 )     + α 2 ( ϵ + μ ) t e u e ( ϵ + μ ) t ( e u − 1 ) + α 2 ( ( ϵ + μ ) t ) 2 e 2 u e ( ϵ + μ ) t ( e u − 1 )     + 2 α 2 ( ( ϵ + μ ) t ) 2 e 2 u e ( ϵ + μ ) t ( e u − 1 ) + α 2 ( ( ϵ + μ ) t ) 3 e 3 u e ( ϵ + μ ) t ( e u − 1 ) ,

so:

E ( V S 3 ) = ( ϵ + α μ 2 ) t + 3 ( α μ ϵ + ϵ 2 + α μ 2 2 ) t 2 + ( ϵ 3 + 3 2 α μ ϵ 2 + 3 2 α ϵ μ 2 + μ 3 2 α ) t 3 ,

i.e. we just wrote:

E ( V S 3 ) = E ( V S ) + 3 E ( V S ( V S − 1 ) ) + E ( V S ( V S − 1 ) ( V S − 2 ) ) .

Remark the following:

E ( V S ( V S − 1 ) ) − ( E ( V S ) ) 2 = ( α μ ϵ + ϵ 2 + α μ 2 2 − ( ϵ + α μ 2 ) 2 ) t 2 = ( μ 2 α 2 − α 2 μ 2 4 ) t 2 = ( α μ t 2 ) 2 [ 2 − α α ] .

Then with the same idea let’s compute the following:

E ( V S ( V S − 1 ) ( V S − 2 ) ) − ( E ( V S ) ) 3 = [ ( ϵ 3 + 3 2 α μ ϵ 2 + 3 2 α ϵ μ 2 + μ 3 2 α ) − ( ϵ + α μ 2 ) 3 ] t 3 = 3 ϵ t ( α μ t 2 ) 2 [ 2 − α α ] + ( α μ t 2 ) 3 [ 4 − α 2 α 2 ] = 3 ϵ t ( α μ t 2 ) 2 [ 2 − α α ] + ( α μ t 2 ) 2 2 − α α α μ t 2 α + 2 α ,

and we know that ϵ t = E ( V S ) − α μ t 2 and that ( α μ t 2 ) 2 2 − α α = E ( V S ( V S − 1 ) ) − ( E ( V S ) ) 2 , so:

E ( V S ( V S − 1 ) ( V S − 2 ) ) − ( E ( V S ) ) 3 = ( E ( V S ( V S − 1 ) ) − ( E ( V S ) ) 2 ) ( 3 E ( V S ) − 3 α μ t 2 + μ t 2 ( 2 + α ) ) ,

i.e.

E ( V S ( V S − 1 ) ( V S − 2 ) ) − ( E ( V S ) ) 3 E ( V S ( V S − 1 ) ) − ( E ( V S ) ) 2 = 3 E ( V S ) + α μ t 2 2 ( 1 − α ) α ,

If we use again the formula, we can then replace α μ t 2 by E ( V S ( V S − 1 ) ) − ( E ( V S ) ) 2 2 − α α :

E ( V S ( V S − 1 ) ( V S − 2 ) ) − ( E ( V S ) ) 3 ( E ( V S ( V S − 1 ) ) − ( E ( V S ) ) 2 ) = 3 E ( V S ) + E ( V S ( V S − 1 ) ) − ( E ( V S ) ) 2 2 − α α 2 ( 1 − α ) α ,

so:

E ( V S ( V S − 1 ) ( V S − 2 ) ) − ( E ( V S ) ) 3 E ( V S ( V S − 1 ) ) − ( E ( V S ) ) 2 − 3 E ( V S ) E ( V S ( V S − 1 ) ) − ( E ( V S ) ) 2 = 2 ( 1 − α ) α ( 2 − α ) .

If we arrange a bit the expression on denominator and numerator on the left hand side of the equation, we remark the following:

· Remark 1:

E ( V S ( V S − 1 ) ) − ( E ( V S ) ) 2 = V a r ( V S ) − E ( V S ) ,

· Remark 2:

E ( V S ( V S − 1 ) ( V S − 2 ) ) − ( E ( V S ) ) 3 − 3 E ( V S ) E ( V S 2 ) + 3 E ( V S ) 2 + 3 ( E ( V S ) ) 3 = E ( ( V S − E ( V S ) ) 3 ) − 3 V a r ( V S ) + 2 E ( V S ) .

Thus:

E ( ( V S − E ( V S ) ) 3 ) − 3 V a r ( V S ) + 2 E ( V S ) ( V a r ( V S ) − E ( V S ) ) 3 2 = 2 ( 1 − α ) α ( 2 − α ) .

Introducing the skewness γ and the following notations: σ = V a r ( V S ) , γ = E ( ( V S − E ( V S ) σ ) 3 ) and m = E ( V S ) , we obtain finally:

γ σ 3 − 3 σ 2 + 2 m ( σ 2 − m ) 3 2 = 2 ( 1 − α ) α ( 2 − α ) .

Skewness, standard deviation and expectation are measured from data. To estimate α we thus just have to solve the following second order equation on α :

α 2 − 2 α + 4 4 + ( γ σ 3 − σ 2 + 2 m ) 2 ( σ 2 − m ) 3 = 0.

The discriminant is positive: Δ = 4 ( 1 − 4 4 + ( γ σ 3 − σ 2 + 2 m ) 2 ( σ 2 − m ) 3 ) . As α is a probability, we finally find:

α = 1 − 1 − 4 4 + ( γ σ 3 − σ 2 + 2 m ) 2 ( σ 2 − m ) 3 ,

which is indeed between 0 and 1.

We know that E ( V S ( V S − 1 ) ) − ( E ( V S ) ) 2 = ( α μ t 2 ) 2 [ 2 − α α ] , so let’s replace the α of the right hand side of the equality (not with μ t ) by previous expresion. We then estimate α μ t 2 . We finally obtain the following as E ( V S ( V S − 1 ) ) − ( E ( V S ) ) 2 = σ 2 − m with the previous notations:

α μ t 2 = 1 − 1 − 4 4 + ( γ σ 3 − σ 2 + 2 m ) 2 ( σ 2 − m ) 3 1 + 1 − 4 4 + ( γ σ 3 − σ 2 + 2 m ) 2 ( σ 2 − m ) 3 ( σ 2 − m )

Finally we obtain the following equivalent exact formula:

PIN = α μ t 2 ϵ t + α μ t 2 ,

i.e.:

PIN = 1 m 1 − 1 − 4 4 + ( γ σ 3 − σ 2 + 2 m ) 2 ( σ 2 − m ) 3 1 + 1 − 4 4 + ( γ σ 3 − σ 2 + 2 m ) 2 ( σ 2 − m ) 3 ( σ 2 − m ) ,

or after simplifying a bit:

PIN = 2 m ( σ 2 − m ) 2 4 ( σ 2 − m ) 3 + ( γ σ 3 − 3 σ 2 + 2 m ) 2 + γ σ 3 − 3 σ 2 + 2 m .

One then just have to estimate on a arbitrary time lenght t, m, σ and γ to estimate the PIN number. The difficulty is then put on estimating on this time period the volume of direction of trades. We describe further a possible framework to compute this number. One can verify numerically that these two formula give the exact same numbers of the PIN.

For purpose of illustration, we compare the empirical form E ( | B − S | ) E ( S + B ) with the PIN and the new formula8 found within the time clock framework for different cases of ϵ and μ . It is pretty easy to do, as controlling ex-ante all the parameters of the model one then just has to generate the appropriate Poisson processes to obtain all the values. We illustrate the results with three examples:

· ϵ = o ( μ ) and μ of same order than ϵ : we took ϵ = 100 and μ ∈ { 10000,20000,30000 } ,

· μ = o ( ϵ ) and ϵ of same order than μ : we took μ = 100 and ϵ ∈ { 10000,20000,30000 } ,

· ϵ of same order than μ : we took9 ϵ = 10000 and μ ∈ { 10000,2000,30000 } .

Remarks:

· We compute 20 values for each choice of ϵ and μ in the three cases above,

· For each of the 20 values, for a choice of ϵ and μ , we generate 1,000,000 Poisson processes, we divide them in 100 consecutive intervals of 10,000 values. For each of the 100 intervals we compute empirical average to approximate mean m, standard deviation σ and skewness γ . We then compute an approximation of the PIN with an average of these 100 values1⁰.

On each case (Figure 8, Figure 9 and Figure 10), we plot E ( | B − S | ) E ( S + B ) (VPIN), the PIN ( α μ α μ + 2 ϵ ) and the new PIN value (labelled as NPIN).

Case1: ϵ = 100, μ ∈ { 10000,20000,30000 }

On Figure 8, new formula (NPIN) and PIN are very close.

Case 2: μ = 100, ϵ ∈ { 10000,20000,30000 }

Here on Figure 9 one can see better the difference when one does not change μ anymore.

Case 3: ϵ = 10000, μ ∈ { 10000,2000,30000 }

This last case on Figure 10 illustrates a market where the number of informed and uninformed traders are of same order. The VPIN really slightly over-estimates the true PIN value.

In any case one sees that new formula estimated is closer than the VPIN one. By the way, we have checked that new PIN formula obviously equals true PIN formula for any parameter ϵ , μ and α of the model.