A feedback information-theoretic transmission scheme (FITTS) for modeling trajectory variability in aimed movements

Abstract

Trajectories in human aimed movements are inherently variable. Using the concept of positional variance profiles, such trajectories are shown to be decomposable into two phases: In a first phase, the variance of the limb position over many trajectories increases rapidly; in a second phase, it then decreases steadily. A new theoretical model, where the aiming task is seen as a Shannon-like communication problem, is developed to describe the second phase: Information is transmitted from a “source” (determined by the position at the end of the first phase) to a “destination” (the movement’s end-point) over a “channel” perturbed by Gaussian noise, with the presence of a noiseless feedback link. Information-theoretic considerations show that the positional variance decreases exponentially with a rate equal to the channel capacity C. Two existing datasets for simple pointing tasks are re-analyzed and observations on real data confirm our model. The first phase has constant duration, and C is found constant across instructions and task parameters, which thus characterizes the participant’s performance. Our model provides a clear understanding of the speed-accuracy tradeoff in aimed movements: Since the participant’s capacity is fixed, a higher prescribed accuracy necessarily requires a longer second phase resulting in an increased overall movement time. The well-known Fitts’ law is also recovered using this approach.

Appendix

1.1 Segmentation algorithm

The algorithm works in the following steps (using the preprocessed time series as presented in Sect. 4):

• 1. Identify time instants $\lbrace { t_{0,i} \rbrace }_{i= 1}^n$ when position crosses half of the distance between start and target while maintaining a positive speed, thereby identifying n movements¹⁶;

• 2. Compute the velocity profile (from the position profile), normalize it with respect to its maximum value, to determine the start of each movement via thresholding (go back in time from $t_{0,i}$ until the normalized velocity reaches, say, 1%);

• 3. Look for “dwell periods” after $t_{0,i}$, i.e., intervals where the absolute value of the normalized velocity is below $1\%$ and when the current position is above the one obtained at $t_{0,i}$. The latest instant of the last dwell period is the end of the movement.¹⁷

1.2 $\tau$ Box-plot for each participant

The Box plot for $\tau$ for each participant is displayed Fig. 8.

Fig. 8

$\tau$ (s) plotted for each participant of the PD-dataset

1.3 Mathematical proofs

In the proofs, we use the natural logarithm ($\log$) rather than its base-2 form ($\log _2$), for simplicity and consistency with information-theoretic textbooks and is equivalent barring the change of units from bits ($\log _2$) to nats ($\log$) (Cover and Thomas 2012).

Proof of Theorem 1

We use well-known ingredients from information-theory (Cover and Thomas 2012). For inequality (10a):

$$\begin{aligned} I({\mathbf {A}};\widehat{{\mathbf {A}}}_n)&= H({\mathbf {A}}) - H({\mathbf {A}} | \widehat{{\mathbf {A}}}_n) \end{aligned}$$

(35)

$$\begin{aligned}&= H({\mathbf {A}}) - H({\mathbf {A}}-\widehat{{\mathbf {A}}}_n | \widehat{{\mathbf {A}}}_n) \end{aligned}$$

(36)

$$\begin{aligned}&\ge H({\mathbf {A}}) - H({\mathbf {A}}-\widehat{{\mathbf {A}}}_n) \end{aligned}$$

(37)

$$\begin{aligned}&\ge H({\mathbf {A}}) - \frac{1}{2}\log \left( 2 \pi e {\mathbb {E}}[ ({\mathbf {A}} - \widehat{{\mathbf {A}}}_n)^2]\right) \end{aligned}$$

(38)

$$\begin{aligned}&= \frac{1}{2}\log \frac{\sigma ^2_0}{D_n}, \end{aligned}$$

(39)

where (35) is by definition of mutual information; (36) because of the conditioning by $\widehat{{\mathbf {A}}}_n$; (37) because conditioning reduces entropy; (38) because the Gaussian distribution maximizes entropy under power constraints and by the entropy formula for a Gaussian distribution; (39) by definition of the distortion and the entropy formula for a Gaussian distribution.

For inequality (10b):

$$\begin{aligned} I({\mathbf {A}};\widehat{{\mathbf {A}}}_n)&\le I({\mathbf {A}};{\mathbf {Y}}^n) \end{aligned}$$

(40)

$$\begin{aligned}&= H({\mathbf {Y}}^n) - H({\mathbf {Y}}^n|{\mathbf {A}}) \end{aligned}$$

(41)

$$\begin{aligned}&= \sum \nolimits _{i=1}^n \left[ H({\mathbf {Y}}_i|{\mathbf {Y}}^{i-1}) - H({\mathbf {Y}}_i|{\mathbf {Y}}^{i-1},{\mathbf {A}}) \right] \end{aligned}$$

(42)

$$\begin{aligned}&= \sum \left[ H({\mathbf {Y}}_i|{\mathbf {Y}}^{i-1}) - H({\mathbf {Y}}_i|{\mathbf {X}}_i) \right] \end{aligned}$$

(43)

$$\begin{aligned}&\le \sum \left[ H({\mathbf {Y}}_i) - H({\mathbf {Z}}_i)\right] \end{aligned}$$

(44)

$$\begin{aligned}&\le \sum \left[ \frac{1}{2} \log (2\pi e (P_i + N)) - \frac{1}{2} \log (2\pi e N)\right] \end{aligned}$$

(45)

$$\begin{aligned}&\le \sum \nolimits _{i=1}^n \left[ \frac{1}{2} \log ( 1 + P_i/N) \right] \le nC \end{aligned}$$

(46)

where (40) is by the data processing inequality (Cover and Thomas 2012) applied the Markov chain ${\mathbf {A}}~\longrightarrow ~{\mathbf {Y}}^i~\longrightarrow ~{\mathbf {g}}({\mathbf {Y}}^i) = \widehat{{\mathbf {A}}}^i$; (41) by definition; (42) by applying the chain rule (Cover and Thomas 2012) to both terms; (43) by design of the feedback scheme; (44) because conditioning reduces entropy for the first term a,d by virtue of the AWGN model for the second term; (45) because the Gaussian distribution maximizes entropy and ${\mathbf {X}}_i$ and ${\mathbf {Z}}_i$ are independent; (46) by the concavity of the logarithm function. $\square$

Proof of Lemma 1

The proof consists of finding the conditions that make the inequalities in the proof of Theorem 1 equalities. Equality in (37) is equivalent to condition (4); Equality in (38) is equivalent to ${\mathbf {A}} - \widehat{{\mathbf {A}}}^i$ Gaussian; Equality in (40) is equivalent to $H({\mathbf {A}}|{\mathbf {Y}}^i) = H({\mathbf {A}}|{\mathbf {Y}}^i,{\mathbf {g}}({\mathbf {Y}}^i)) = H({\mathbf {A}}|{\mathbf {g}}({\mathbf {Y}}^i))$, so that ${\mathbf {Y}}^i~\longrightarrow ~{\mathbf {g}}({\mathbf {Y}}^i)~\longrightarrow ~{\mathbf {A}}$ form a Markov chain (Cover and Thomas 2012), leading to condition (5); Equality in (44) is equivalent to condition (3); Equality in (45) means the $Y_i$’s are Gaussian; Equality in (46) leads to condition (2) by concavity of the logarithm. Finally, ${\mathbf {X}}_i$ is Gaussian as the result of the sum of two Gaussians ${\mathbf {Y}}_i$ and ${\mathbf {Z}}_i$, and so is $\widehat{{\mathbf {A}}}_i$ as the sum of ${\mathbf {A}}$ and ${\mathbf {A}} - \widehat{{\mathbf {A}}}_i$. This finally yields condition (1). $\square$

Proof of Theorem 3

We start by considering ${\mathbf {X}}_i = {\mathbf {f}}({\mathbf {Y}}^{i-1}, {\mathbf {A}})$, which should be independent of ${\mathbf {Y}}_{i-1},\;\forall i$ by condition (4) of Lemma 1. This implies the decorrelation

$$\begin{aligned} {\mathbb {E}}[{\mathbf {f}}({\mathbf {Y}}^{i-1}, {\mathbf {A}})({\mathbf {Y}}_{i-1})] = 0,\quad \forall i. \end{aligned}$$

(47)

Since ${\mathbf {X}}_i$ is a function of two Gaussians ${\mathbf {A}}$ and ${\mathbf {Y}}^{i-1}$, the conditional expectation ${\mathbf {X}}_i = {\mathbf {E}}[{\mathbf {X}}_i|{\mathbf {A}}, {\mathbf {Y}}^{i-1}]$ is linear, hence ${\mathbf {X}}_i = \alpha _i ({\mathbf {A}} - \tilde{{\mathbf {f}}}({\mathbf {Y}}^{i-1}))$. Plugging this in (47) makes for a direct application of the orthogonality principle, showing that $\tilde{{\mathbf {f}}} = {\mathbb {E}}[{\mathbf {A}}|{\mathbf {Y}}^{i-1}] = {\mathbf {g}}({\mathbf {Y}}^{i-1})$. $\square$

Proof of Theorem 4

The goal of the proof is to evaluate ${\mathbb {E}}[{\mathbf {A}}|{\mathbf {Y}}^{i-1}]$. We first use the operational formula from the orthogonality principle ${\mathbb {E}}[{\mathbf {A}}|{\mathbf {Y}}^{i-1}] = {\mathbb {E}}[{\mathbf {A}}({\mathbf {Y}}^{i-1})^t] {\mathbb {E}}[{\mathbf {Y}}^{i-1}({\mathbf {Y}}^{i-1})^t]^{-1}{\mathbf {Y}}^{i-1}$. Because the channel outputs are independent [conditions (3) and (5) from Lemma 1], and input powers are identical (conditions 2 from Lemma 1), ${\mathbb {E}}[{\mathbf {Y}}^{i-1}({\mathbf {Y}}^{i-1})^t]^{-1} = (P+N)^{-1} {\mathbb {I}}$, where ${\mathbb {I}}$ is the identity matrix of size $i-1$. Then, let ${\mathbf {A}}_i = {\mathbf {X}}_i/\alpha _i$ be the unscaled version of ${\mathbf {X}}_i$ and notice that ${\mathbf {A}} - {\mathbf {A}}_i = {\mathbf {g}}({\mathbf {Y}}^{i-1})$. As the channel outputs are independent, we immediately have that ${\mathbb {E}}[({\mathbf {A}} - {\mathbf {A}}_i)Y_i] = 0$, hence ${\mathbb {E}}[{\mathbf {A}}Y_i] = {\mathbb {E}}[{\mathbf {A}}_iY_i]$.

Combining both results, we get

$$\begin{aligned} {\mathbb {E}}[{\mathbf {A}}|{\mathbf {Y}}^{i-1}]&= (P+N)^{-1}{\mathbb {E}}[{\mathbf {A}}({\mathbf {Y}}^{i-1})^t] {\mathbb {I}}{\mathbf {Y}}^{i-1} \end{aligned}$$

(48)

$$\begin{aligned}&= (P+N)^{-1} \sum _{j=1}^{i-1} {\mathbb {E}}[{\mathbf {A}}{\mathbf {Y}}_j]{\mathbf {Y}}_j \end{aligned}$$

(49)

$$\begin{aligned}&= (P+N)^{-1} \sum _{j=1}^{i-1} {\mathbb {E}}[{\mathbf {A}}_j{\mathbf {Y}}_j]{\mathbf {Y}}_j \end{aligned}$$

(50)

$$\begin{aligned}&= \sum _{j=1}^{i-1} {\mathbb {E}}[{\mathbf {A}}_j|{\mathbf {Y}}_j] \end{aligned}$$

(51)

where

$$\begin{aligned} {\mathbb {E}}[{\mathbf {A}}_i|{\mathbf {Y}}_i]&= (P+N)^{-1}{\mathbb {E}}[{\mathbf {A}}_i{\mathbf {Y}}_i]{\mathbf {Y}}_i \end{aligned}$$

(52)

$$\begin{aligned}&= (P+N)^{-1} {\mathbb {E}}[{\mathbf {X}}_i/\alpha _i \cdot {} ({\mathbf {X}}_i + {\mathbf {Z}}_i)]{\mathbf {Y}}_i \end{aligned}$$

(53)

$$\begin{aligned}&= \frac{1}{\alpha _i}\frac{P}{P+N} {\mathbf {Y}}_i \end{aligned}$$

(54)

$\square$

Proof of Theorem 5

First, notice that we can write $D_{i}$ as

$$\begin{aligned} D_{i}&= {\mathbb {E}}[({\mathbf {A}} - {\mathbb {E}}[{\mathbf {A}}|{\mathbf {Y}}^i])^2] \end{aligned}$$

(55)

$$\begin{aligned}&= {\mathbb {E}}[({\mathbf {A}} - ({\mathbb {E}}[{\mathbf {A}}|{\mathbf {Y}}^{i-1}] + {\mathbb {E}}[{\mathbf {A}}|{\mathbf {Y}}_i]))^2] \end{aligned}$$

(56)

$$\begin{aligned}&= D_{i-1} - {\mathbb {E}}[({\mathbb {E}}[{\mathbf {A}}|{\mathbf {Y}}_i])^2] \end{aligned}$$

(57)

Using (54), one has

$$\begin{aligned} D_{i} = D_{i-1} - \frac{1}{\alpha _i^2} \frac{P^2}{P+N}. \end{aligned}$$

Finally, notice that $D_{i-1} = {\mathbb {E}}[({\mathbf {A}} - {\mathbb {E}}[{\mathbf {A}}|{\mathbf {Y}}^{i-1}])^2] = {\mathbb {E}}[({\mathbf {A}}_i)^2] = P/\alpha _i^2$, to see that

$$\begin{aligned} D_{i} = D_{i-1} \left( 1 - \frac{P}{P+N} \right) = \frac{D_{i-1}}{1 + P/N} \end{aligned}$$

The closed form for the distortion is obtained by applying this equation recursively, starting from $D_0 = {\mathbb {E}}[{\mathbf {A}}^2] = \sigma ^2_0$:

$$\begin{aligned} D_{i} = \frac{\sigma _0^2}{(1 + P/N)^{i}}. \end{aligned}$$

(58)

Finally, we evaluate $\alpha _i$ (with $\alpha _0 = \frac{\sqrt{P}}{\sigma _0}$):

$$\begin{aligned} \alpha _i = \sqrt{\frac{P}{D_i}} = \frac{\sqrt{P}}{\sigma _0} (1 +P/N)^{i/2} = \alpha _0 (1 +P/N)^{i/2}. \end{aligned}$$

(59)

$\square$