Estimating latency from inhibitory input

Abstract

Stimulus response latency is the time period between the presentation of a stimulus and the occurrence of a change in the neural firing evoked by the stimulation. The response latency has been explored and estimation methods proposed mostly for excitatory stimuli, which means that the neuron reacts to the stimulus by an increase in the firing rate. We focus on the estimation of the response latency in the case of inhibitory stimuli. Models used in this paper represent two different descriptions of response latency. We consider either the latency to be constant across trials or to be a random variable. In the case of random latency, special attention is given to models with selective interaction. The aim is to propose methods for estimation of the latency or the parameters of its distribution. Parameters are estimated by four different methods: method of moments, maximum-likelihood method, a method comparing an empirical and a theoretical cumulative distribution function and a method based on the Laplace transform of a probability density function. All four methods are applied on simulated data and compared.

Appendices

Appendix 1: Conditional and unconditional distribution of $T$ in Model A and B

General expressions for the cdf, Laplace transform of the pdf, mean and variance of $T_\mathrm{A}$, see (4) and (7), and $T_\mathrm{B}$, see (5) and (8), are given below, first conditionally on $\theta$, then the unconditional expressions.

Model A From the definition of $T_\mathrm{A}$ in (2) and using that $W$ and $U$ are independent, or by integrating (4), the cdf of $T_\mathrm{A}$ is

$$\begin{aligned} F_{T_\mathrm{A}|\,\theta }(t) = {\left\{ \begin{array}{ll} 1-\mathrm{e}^{-\lambda t} &{} t \in [0,\theta ] \\ 1-\mathrm{e}^{-\lambda \theta } + \mathrm{e}^{-\lambda t}F_U(t-\theta ) &{} t \in (\theta , \infty ). \end{array}\right. } \end{aligned}$$

(27)

To determine the mean and variance of $T_\mathrm{A}$, we derive the Laplace transform of the pdf of $T_\mathrm{A}$ given by (4),

$$\begin{aligned} \widehat{f_{T_\mathrm{A}|\,\theta }}(s) = \frac{\lambda }{s+\lambda }\left( 1-\mathrm{e}^{-(s+\lambda )\theta }\right) + \mathrm{e}^{-(s+\lambda )\theta } \widehat{f_U}(s), \end{aligned}$$

(28)

where $\widehat{f_U}(s)$ is the Laplace transform of $f_U(t)$. Thus, it holds

$$\begin{aligned} \mathbb {E}(T_\mathrm{A}\,|\,\theta ) = \frac{1}{\lambda }\left[ 1-\mathrm{e}^{-\lambda \theta } \right] + \mathbb {E}(U)\mathrm{e}^{-\lambda \theta } \end{aligned}$$

(29)

$$\begin{aligned} \mathrm{Var}(T_\mathrm{A}\,|\,\theta )&= \mathrm{e}^{-\lambda \theta }\left[ \mathbb {E}(U^2) + 2\mathbb {E}(U)\left( \theta - \frac{1}{\lambda } \right) - \frac{2\theta }{\lambda }\right] \nonumber \\&\quad -\, \mathrm{e}^{-2\lambda \theta }\left[ \mathbb {E}(U)-\frac{1}{\lambda } \right] ^2 + \frac{1}{\lambda ^2}. \end{aligned}$$

(30)

The unconditional expresions are

$$\begin{aligned}&\widehat{f_{T_\mathrm{A}}}(s) = \frac{1}{s+\lambda +1/\theta ^*}\left( \lambda + \frac{\widehat{f_U}(s)}{\theta ^*} \right) \end{aligned}$$

(31)

$$\begin{aligned}&\mathbb {E}(T_\mathrm{A}) = \frac{\mathbb {E}(U)+\theta ^*}{\lambda \theta ^*+1} \end{aligned}$$

(32)

$$\begin{aligned}&\mathrm{Var}(T_\mathrm{A}) = \frac{{\theta ^*}^2+(\lambda \theta ^*+1)\mathbb {E}(U^2)-\left[ \mathbb {E}(U)\right] ^2}{(\lambda \theta ^*+1)^2} \end{aligned}$$

(33)

where we have used the relations $\mathbb {E}(T) = \mathbb {E}\left[ \mathbb {E}(T\,|\,\varTheta ) \right]$ and $\mathrm{Var}(T) = \mathbb {E}\left[ \mathrm{Var}(T\,|\,\varTheta ) \right] + \mathrm{Var}\left[ \mathbb {E}(T\,|\,\varTheta ) \right]$.

Model B From the independence of $W$ and $U$ and based on (3) we get the cdf of $T_\mathrm{B}$,

$$\begin{aligned} F_{T_\mathrm{B}|\,\theta }(t) = {\left\{ \begin{array}{ll} 1-\mathrm{e}^{-\lambda t} &{} t \in [0,\theta ] \\ 1-\mathrm{e}^{-\lambda \theta }+\int _\theta ^t \lambda \mathrm{e}^{-\lambda y}F_U(t-y)\,\mathrm{d}y &{} t \in (\theta , \infty ). \end{array}\right. } \end{aligned}$$

(34)

The Laplace transform of the pdf of $T_\mathrm{B}$ is

$$\begin{aligned} \widehat{f_{T_\mathrm{B}|\,\theta }}(s) = \frac{\lambda }{\lambda +s} \left[ 1-\mathrm{e}^{-(\lambda +s)\theta } \left( 1-\widehat{f_U}(s) \right) \right] . \end{aligned}$$

(35)

Using the Laplace transform $\widehat{f_{T_\mathrm{B}|\,\theta }}(s)$ yields the mean and variance of $T_\mathrm{B}$,

$$\begin{aligned} \mathbb {E}(T_\mathrm{B}\,|\,\theta ) = \frac{1}{\lambda } + \mathrm{e}^{-\lambda \theta }\mathbb {E}(U) \end{aligned}$$

(36)

$$\begin{aligned} \mathrm{Var}(T_\mathrm{B}\,|\,\theta )&= \mathrm{e}^{-\lambda \theta }\left[ \mathbb {E}(U^2) + 2\theta \mathbb {E}(U) \right] \nonumber \\&\quad - \mathrm{e}^{-2\lambda \theta } \left[ \mathbb {E}(U) \right] ^2 + \frac{1}{\lambda ^2}. \end{aligned}$$

(37)

The unconditional expressions are

$$\begin{aligned}&\widehat{f_{T_\mathrm{B}}}(s) = \frac{\lambda }{s+\lambda +1/\theta ^*} \left[ 1 + \frac{\widehat{f_U}(s)}{\theta ^*(s+\lambda )} \right] \end{aligned}$$

(38)

$$\begin{aligned}&\mathbb {E}(T_\mathrm{B}) = \frac{1}{\lambda } + \frac{\mathbb {E}(U)}{\lambda \theta ^*+1} \end{aligned}$$

(39)

$$\begin{aligned}&\mathrm{Var}(T_\mathrm{B}) = \frac{\mathbb {E}(U^2)}{\lambda \theta ^*+1} + \frac{\mathbb {E}(U)[2\theta ^*-\mathbb {E}(U)]}{(\lambda \theta ^*+1)^2} + \frac{1}{\lambda ^2}. \end{aligned}$$

(40)

Appendix 2: Laplace transform and moments in Model 1–5

Model 1 The Laplace transform of (10) is

$$\begin{aligned} \widehat{f_{T_{M1}}}(s) = \frac{\lambda }{s+\lambda }\left( 1-\mathrm{e}^{-(\lambda +s)\theta ^*} \right) + \frac{\kappa }{s+\kappa } \mathrm{e}^{-(\lambda +s)\theta ^*}. \end{aligned}$$

(41)

From (29) and (30) with $E(U) = 1/\kappa$ and $E(U^2)=2/\kappa ^2$ we get

$$\begin{aligned} \mathrm{E}(T_{M1}) = \frac{1-\mathrm{e}^{-\lambda \theta ^*}}{\lambda } + \frac{\mathrm{e}^{-\lambda \theta ^*}}{\kappa } \end{aligned}$$

(42)

$$\begin{aligned} \mathrm{Var}(T_{M1})&= \frac{1}{\lambda ^2} + \frac{2 (\lambda - \kappa ) (1 + \kappa \theta ^*)}{\lambda \kappa ^2} \mathrm{e}^{-\lambda \theta ^*}\nonumber \\&\quad - \left( \frac{\lambda - \kappa }{\lambda \kappa } \mathrm{e}^{-\lambda \theta ^*} \right) ^2. \end{aligned}$$

(43)

Model 2 The Laplace transform of (11) is

$$\begin{aligned} \widehat{f_{T_{M2}}}(s) = \frac{\lambda }{s+\lambda } + \frac{\lambda \mathrm{e}^{-(\lambda +s)\theta ^*}}{s+\lambda }\left[ \left( \frac{\lambda }{s+\lambda } \right) ^k-1 \right] . \end{aligned}$$

(44)

From (36) and (37) follows (with $E(U)=k/\lambda$ and $E(U^2)=k(k+1)/\lambda ^2$)

$$\begin{aligned}&\mathbb {E}(T_{M2}) = \frac{1}{\lambda } + \frac{k}{\lambda }\mathrm{e}^{-\lambda \theta ^*} \end{aligned}$$

(45)

$$\begin{aligned}&\mathrm{Var}(T_{M2}) = \frac{1}{\lambda ^2} + \frac{k(2\lambda \theta ^*\!+k+\!1)}{\lambda ^2}\mathrm{e}^{-\lambda \theta ^*} \! - \frac{k^2}{\lambda ^2}\mathrm{e}^{-2\lambda \theta ^*}. \end{aligned}$$

(46)

Model 3 The Laplace transform, mean and variance of (12) are

$$\begin{aligned} \widehat{f_{T_{M3}}}(s)&= \frac{\lambda }{s+\lambda + 1/\theta ^*} \nonumber \\&\quad +\frac{\kappa }{(\lambda -\kappa )\theta ^* +1}\left[ \frac{1}{s+\kappa } - \frac{1}{s+\lambda +1/\theta ^*} \right] \end{aligned}$$

(47)

$$\begin{aligned}&\mathbb {E}(T_{M3}) = \frac{\kappa \theta ^*+1}{\kappa (\lambda \theta ^*+1)} \end{aligned}$$

(48)

$$\begin{aligned}&\mathrm{Var}(T_{M3}) = \frac{(\kappa \theta ^*)^2 + 2\lambda \theta ^* + 1}{\kappa ^2(\lambda \theta ^*+1)^2}. \end{aligned}$$

(49)

Model 4 From (38)–(40), the Laplace transform, mean and variance are

$$\begin{aligned}&\widehat{f_{T_{M4}}}(s) = \frac{\lambda }{s+\lambda +1/\theta ^*}\left[ 1 + \frac{1}{\theta ^*(s+\lambda )}\left( \frac{\lambda }{s+\lambda }\right) ^k\right] \end{aligned}$$

(50)

$$\begin{aligned}&\mathbb {E}(T_{M4}) = \frac{\lambda \theta ^* + k+1}{\lambda (\lambda \theta ^* + 1)} \end{aligned}$$

(51)

$$\begin{aligned}&\mathrm{Var}(T_{M4}) = \frac{1}{\lambda ^2}\left\{ 1+\frac{k\left[ (k+3)\lambda \theta ^* + 1 \right] }{(\lambda \theta ^*+1)^2} \right\} \end{aligned}$$

(52)

Model 5 Using (13) with $\widehat{f_E}(s)=\left( \lambda /(s+\lambda )\right) ^k$, the Laplace transform of $f_X(t)$ is

$$\begin{aligned} \widehat{f_{X_{M5}}}(s) = \frac{\left[ \lambda (s+\lambda )\right] ^{k}}{\left[ \left( s+\lambda )(s +\lambda +\frac{1}{\theta ^*}\right) \right] ^{k} + \left[ \lambda (s+\lambda )\right] ^{k} - \left[ \lambda \!\left( s+\lambda +\frac{1}{\theta ^*}\right) \right] ^{k}}. \end{aligned}$$

(53)

The mean of $X_{M5}$, calculated from (14), is

$$\begin{aligned} \mathbb {E}(X_{M5}) = \frac{k}{\lambda }\left( \frac{\lambda \theta ^* + 1}{\lambda \theta ^*} \right) ^{k} . \end{aligned}$$

(54)

The Laplace transform calculated from (17) is

$$\begin{aligned}&\widehat{f_{T_{M5}}}(s)\nonumber \\&\quad = \frac{-\lambda }{k(s+1/\theta ^*)}\left\{ \left( \frac{\lambda }{s+\lambda +1/\theta ^*}\right) ^k - 1 + \left( \frac{\lambda }{s+\lambda +1/\theta ^*}\right) ^k \right. \nonumber \\&\qquad \left. \times \frac{\left[ s \left( \left( \frac{\lambda }{s+\lambda }\right) ^k - \left( \frac{\lambda }{s+\lambda +1/\theta ^*}\right) ^k\right) + \frac{1}{\theta ^*}\left( \left( \frac{\lambda }{s+\lambda }\right) ^k-1\right) \right] }{s \left[ 1 - \left( \frac{\lambda }{s+\lambda }\right) ^k + \left( \frac{\lambda }{s+\lambda +1/\theta ^*}\right) ^k\right] } \right\} . \end{aligned}$$

(55)

Again, from $\widehat{f_{T_{M5}}}(s)$ we can determine the mean of $T_{M5}$,

$$\begin{aligned} \mathbb {E}(T_{M5}) = \left( \frac{k}{\lambda } - \theta ^* \right) \left( \frac{\lambda \theta ^* + 1}{\lambda \theta ^*} \right) ^k + \frac{k+1}{2\lambda } + \theta ^*. \end{aligned}$$

(56)

In general, it is difficult to find the inverse transform of $\widehat{f_{X_{M5}}}(s)$ and $\widehat{f_{T_{M5}}}(s)$. However, for $k=1$, such that the excitatory pulses form a Poisson process, it is possible. It holds for a Poisson process that $f^+_E(t)$ is equal to $f_E(t)$ and therefore the distribution of $T_{M5}$ is the same as the distribution of $X_{M5}$, i.e. $f_{X_{M5}}(t) = f_{T_{M5}}(t)$. The pdf is therefore

$$\begin{aligned} f_{T_{M5}}(t)&= \frac{1}{2}\frac{\lambda }{4\lambda \theta ^* + 1} \mathrm{e}^{-\frac{t}{2}(2\lambda +1/\theta ^*)} \nonumber \\&\quad \times \left[ \left( 1+4\lambda \theta ^*+ \sqrt{1+4\lambda \theta ^*} \right) \mathrm{e}^{-\frac{t}{2\theta ^*} \sqrt{1+4\lambda \theta ^*}} \right. \nonumber \\&\quad + \left. \left( 1+4\lambda \theta ^*- \sqrt{1+4\lambda \theta ^*} \right) \mathrm{e}^{\frac{t}{2\theta ^*}\sqrt{1+4\lambda \theta ^*}} \right] . \end{aligned}$$

(57)

Appendix 3: Moment estimators

Models with constant latency (Model 1 and 2) By plugging (42), (43), (45) and (46) into (21), we get moment equations for the models. In both cases, the obtained system of equations has no analytical solution with respect to $\theta$. However, parameters $\kappa$ or $k$ can be isolated in the first equation and inserted into the second equation. The estimates of $\theta ^*$ are then calculated numerically as solutions to these equations:

$$\begin{aligned} \text{ Model } \text{1: } s^2&= \frac{1}{\lambda ^2}- \left( \bar{t} - \frac{1}{\lambda }\right) ^2 \nonumber \\&\quad +\,2\left( \bar{t} - \frac{1}{\lambda }\right) \left[ \left( \bar{t} - \frac{1}{\lambda }\right) \mathrm{e}^{\lambda \theta ^*} + \frac{1}{\lambda } + \theta ^*\right] \end{aligned}$$

(58)

$$\begin{aligned} \text{ Model } \text{2: } s^2&= \frac{1}{\lambda ^2}\Big [ (2\lambda \theta ^* + 1)(\lambda \bar{t} - 1) \nonumber \\&\quad \left. +\, (\lambda \bar{t} - 1)^2\mathrm{e}^{\lambda \theta ^*} - \lambda \bar{t}\left( \lambda \bar{t} - 2\right) \right] . \end{aligned}$$

(59)

Models with random latency (Model 3 and 4) The moment equations (21) for Model 3 and 4 have analytical solutions. The estimators of $\theta ^*$ are

$$\begin{aligned} \hat{\theta }^*_{M,M3}&= \left( \frac{2\bar{t}(\lambda \bar{t}-1)}{2(s^2-\bar{t}^2)} - \frac{3}{2}\lambda \right. \nonumber \\&\quad \left. +\frac{\sqrt{\lambda ^2(\bar{t}^2+s^2)^2+4\lambda \bar{t}(\bar{t}-3s^2)-4(\bar{t}^2-2s^2)}}{2(s^2-\bar{t}^2)}\right) ^{-1} \end{aligned}$$

(60)

$$\begin{aligned} \hat{\theta }^*_{M,M4}&= \left( \frac{\lambda ^2\bar{t}^2 - 1}{2(\lambda s^2 - \bar{t})} -\frac{\lambda }{2} \right. \nonumber \\&\quad \left. +\frac{\sqrt{\lambda ^4 (s^2 + \bar{t}^2)^2 - 2\lambda ^3\bar{t}(5s^2 + \bar{t}) + \lambda ^2(6s^2 + 7\bar{t}^2) - 6\lambda \bar{t} + 1}}{2(\lambda s^2 - \bar{t})}\right) ^{-1} \end{aligned}$$

(61)

Model with selective interaction (Model 5) Since $\theta ^*$ is the only unknown parameter in Model 5, we only use the first moment in (21). The estimate is denoted by $\hat{\theta }^*_{M,T}$. Because the distribution of $X$ is known, we can also use observations of $X$ for the estimation. The moment equation $\mathbb {E}(X) = \bar{x}$ is employed, where $\bar{x}$ is the average of observations $x_1,\ldots ,x_n$ of $X$. This moment estimator is denoted by $\hat{\theta }^*_{M,X}$. Another alternative is to use observations $t_i$ and $x_i$ together. Let the time from the stimulus onset to the second observed spike after it be a random variable denoted by $Z$, i.e. $Z = T + X$. The mean of $Z$ is $\mathbb {E}(Z) = \mathbb {E}(T) + \mathbb {E}(X)$ and the moment equation is thus $\mathbb {E}(T) + \mathbb {E}(X) = \bar{z}$, where $\bar{z}$ is the average of $t_i+x_i,\,i=1,\ldots ,n$. The corresponding moment estimator is denoted by $\hat{\theta }^*_{M,Z}$.

The solution to $\mathbb {E}(X)=\bar{x}$ with $\mathbb {E}X$ given by (54) yields

$$\begin{aligned} \hat{\theta }^*_{M,X} = \frac{1}{\lambda } \left[ \left( \frac{\lambda }{k}\bar{x} \right) ^{1/k} - 1 \right] ^{-1}. \end{aligned}$$

(62)

No closed forms exist for $\hat{\theta }^*_{M,T}$ and $\hat{\theta }^*_{M,Z}$, they are found as numerical solutions to the equations

$$\begin{aligned}&\left( \frac{k}{\lambda } - \hat{\theta }^*_{M,T} \right) \left( \frac{\lambda \hat{\theta }^*_{M,T} + 1}{\lambda \hat{\theta }^*_{M,T}} \right) ^k + \frac{k+1}{2\lambda } + \hat{\theta }^*_{M,T} = \bar{t} \\&\left( \frac{2k}{\lambda } - \hat{\theta }^*_{M,Z} \right) \left( \frac{\lambda \hat{\theta }^*_{M,Z} + 1}{\lambda \hat{\theta }^*_{M,Z}} \right) ^k + \frac{k+1}{2\lambda } + \hat{\theta }^*_{M,Z} = \bar{z}. \end{aligned}$$

If excitatory pulses form a Poisson process, i.e. $k=1$, all three moment equations have analytical solutions, namely

$$\begin{aligned} \hat{\theta }^*_{M,T}&= \left[ \lambda \left( \lambda \bar{t} - 1 \right) \right] ^{-1} \\ \hat{\theta }^*_{M,X}&= \left[ \lambda \left( \lambda \bar{x} - 1 \right) \right] ^{-1} \\ \hat{\theta }^*_{M,Z}&= \left[ \lambda \left( \frac{1}{2}\lambda \bar{z}-1\right) \right] ^{-1}. \end{aligned}$$

Appendix 4: Log-likelihood functions

The pdfs of $T_\mathrm{A}$ and $T_\mathrm{B}$ are given in (4) and (5) and the log-likelihood functions of $\theta ^*$ are

$$\begin{aligned} l^A(\theta ^* )&= n_{\theta ^*}\ln \lambda -\lambda \Big ( (n-n_{\theta ^*})\theta ^* + \sum _{t_i\le \theta ^*} t_i \Big )\nonumber \\&\quad + \sum _{t_i > \theta ^*} \ln f_U(t_i-\theta ^*) \end{aligned}$$

(63)

and

$$\begin{aligned} l^B(\theta ^* ) \!=\! n\ln \lambda \!-\! \lambda \sum _{t_i\le \theta ^*} t_i + \sum _{t_i>\theta ^*} \ln \int \limits _{\theta ^*}^{t_i} \mathrm{e}^{-\lambda y} f_U(t_i-y)\,\mathrm{d}y, \end{aligned}$$

(64)

where $n_{\theta ^*}$ is the number of observations $t_i \le \theta ^*$.

For Model 1, 2 and 3, the log-likelihood functions are

$$\begin{aligned} l^{M1}(\theta ^* ,\kappa )&= \sum _{t_i\le \theta ^*} (\ln \lambda - \lambda t_i)\nonumber \\&\quad + \sum _{t_i> \theta ^*} \left[ \ln \kappa - \lambda \theta ^* - \kappa (t_i-\theta ^*)\right] \end{aligned}$$

(65)

$$\begin{aligned} l^{M2}(\theta ^* ,k)&= n \ln \lambda - \lambda \sum _{i=1}^n t_i \nonumber \\&\quad + \sum _{t_i>\theta ^*} \left( k\ln [\lambda (t_i-\theta ^*)]-\ln \Gamma (k+1) \right) \end{aligned}$$

(66)

$$\begin{aligned} l^{M3}(\theta ^*,\kappa )&= -\frac{\lambda \theta ^*+1}{\theta ^*}\sum _{i=1}^n t_i \nonumber \\&+ \sum _{i=1}^n \ln \left[ \lambda + \frac{\kappa }{(\lambda -\kappa )\theta ^*+1}\left( \mathrm{e}^{(\lambda -\kappa +1/\theta ^*)t_i} - 1 \right) \right] . \end{aligned}$$

(67)

The log-likelihood function for Model 5 when $k=1$ is

$$\begin{aligned} l^{M5}(\theta ^*)&= -n\ln 2 + n\ln \lambda - n\ln \theta ^*\nonumber \\&\quad -\, n\ln \left[ 4\lambda +\frac{1}{\theta ^*}\right] - \frac{2\lambda \theta ^*+1}{2\theta ^*}\sum _{i=1}^n t_i \nonumber \\&\quad + \sum _{i=1}^n\ln \left[ \left( 1+4\lambda \theta ^* +\sqrt{1+4\lambda \theta ^*}\right) \mathrm{e}^{-\frac{t_i}{2\theta ^*} \sqrt{1+4\lambda \theta ^*}} \right. \nonumber \\&\quad \left. + \left( 1+4\lambda \theta ^* -\sqrt{1+4\lambda \theta ^*}\right) \mathrm{e}^{\frac{t_i}{2\theta ^*} \sqrt{1+4\lambda \theta ^*}}\right] . \end{aligned}$$

(68)