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How to create Non-Central Student’s T distribution and what priors to use with the distribution?

I have been working with the following link,
Fitting empirical distribution to theoretical ones with Scipy (Python)?
I have been using my data to the code from the link and found out that the common distribution for my data is the Non-Central Student’s T distribution. I couldn’t find the distribution in the pymc3 package, so, I decided to have a look with scipy to understand how the distribution is formed. I created a custom distribution and I have few questions:
I would like to know if my approach to creating the distribution is right?
How can I implement the custom distribution into models?
Regarding the prior distribution, do I use same steps in normal distribution priors (mu and sigma) combined with halfnormed for degree of freedom and noncentral value?
My custom distribution:
import numpy as np
import theano.tensor as tt
from scipy import stats
from scipy.special import hyp1f1, nctdtr
import warnings
from pymc3.theanof import floatX
from pymc3.distributions.dist_math import bound, gammaln
from pymc3.distributions.continuous import assert_negative_support, get_tau_sigma
from pymc3.distributions.distribution import Continuous, draw_values, generate_samples
class NonCentralStudentT(Continuous):
"""
Parameters
----------
nu: float
Degrees of freedom, also known as normality parameter (nu > 0).
mu: float
Location parameter.
sigma: float
Scale parameter (sigma > 0). Converges to the standard deviation as nu increases. (only required if lam is not specified)
lam: float
Scale parameter (lam > 0). Converges to the precision as nu increases. (only required if sigma is not specified)
"""
def __init__(self, nu, nc, mu=0, lam=None, sigma=None, sd=None, *args, **kwargs):
super().__init__(*args, **kwargs)
super(NonCentralStudentT, self).__init__(*args, **kwargs)
if sd is not None:
sigma = sd
warnings.warn("sd is deprecated, use sigma instead", DeprecationWarning)
self.nu = nu = tt.as_tensor_variable(floatX(nu))
self.nc = nc = tt.as_tensor_variable(floatX(nc))
lam, sigma = get_tau_sigma(tau=lam, sigma=sigma)
self.lam = lam = tt.as_tensor_variable(lam)
self.sigma = self.sd = sigma = tt.as_tensor_variable(sigma)
self.mean = self.median = self.mode = self.mu = mu = tt.as_tensor_variable(mu)
self.variance = tt.switch((nu > 2) * 1, (1 / self.lam) * (nu / (nu - 2)), np.inf)
assert_negative_support(lam, 'lam (sigma)', 'NonCentralStudentT')
assert_negative_support(nu, 'nu', 'NonCentralStudentT')
assert_negative_support(nc, 'nc', 'NonCentralStudentT')
def random(self, point=None, size=None):
"""
Draw random values from Non-Central Student's T distribution.
Parameters
----------
point: dict, optional
Dict of variable values on which random values are to be
conditioned (uses default point if not specified).
size: int, optional
Desired size of random sample (returns one sample if not
specified).
Returns
-------
array
"""
nu, nc, mu, lam = draw_values([self.nu, self.nc, self.mu, self.lam], point=point, size=size)
return generate_samples(stats.nct.rvs, nu, nc, loc=mu, scale=lam ** -0.5, dist_shape=self.shape, size=size)
def logp(self, value):
"""
Calculate log-probability of Non-Central Student's T distribution at specified value.
Parameters
----------
value: numeric
Value(s) for which log-probability is calculated. If the log probabilities for multiple
values are desired the values must be provided in a numpy array or theano tensor
Returns
-------
TensorVariable
"""
nu = self.nu
nc = self.nc
mu = self.mu
lam = self.lam
n = nu * 1.0
nc = nc * 1.0
x2 = value * value
ncx2 = nc * nc * x2
fac1 = n + x2
trm1 = n / 2. * tt.log(n) + gammaln(n + 1)
trm1 -= n * tt.log(2) + nc * nc / 2. + (n / 2.) * tt.log(fac1) + gammaln(n / 2.)
Px = tt.exp(trm1)
valF = ncx2 / (2 * fac1)
trm1 = tt.sqrt(2) * nc * value * hyp1f1(n / 2 + 1, 1.5, valF)
trm1 /= np.asarray(fac1 * tt.gamma((n + 1) / 2))
trm2 = hyp1f1((n + 1) / 2, 0.5, valF)
trm2 /= np.asarray(np.sqrt(fac1) * tt.gamma(n / 2 + 1))
Px *= trm1 + trm2
return bound(Px, lam > 0, nu > 0, nc > 0)
def logcdf(self, value):
"""
Compute the log of the cumulative distribution function for Non-Central Student's T distribution
at the specified value.
Parameters
----------
value: numeric
Value(s) for which log CDF is calculated. If the log CDF for multiple
values are desired the values must be provided in a numpy array or theano tensor.
Returns
-------
TensorVariable
"""
nu = self.nu
nc = self.nc
return nctdtr(nu, nc, value)
My Custom model:
with pm.Model() as model:
# Prior Distributions for unknown model parameters:
mu = pm.Normal('sigma', 0, 10)
sigma = pm.Normal('sigma', 0, 10)
nc= pm.HalfNormal('nc', sigma=10)
nu= pm.HalfNormal('nu', sigma=1)
# Observed data is from a Likelihood distributions (Likelihood (sampling distribution) of observations):
=> (input custom distribution) observed_data = pm.Beta('observed_data', alpha=alpha, beta=beta, observed=data)
# draw 5000 posterior samples
trace = pm.sample(draws=5000, tune=2000, chains=3, cores=1)
# Obtaining Posterior Predictive Sampling:
post_pred = pm.sample_posterior_predictive(trace, samples=3000)
print(post_pred['observed_data'].shape)
print('\nSummary: ')
print(pm.stats.summary(data=trace))
print(pm.stats.summary(data=post_pred))
Edit 1:
I redesigned the custom model to include the custom distribution, however, I keep on getting error based on the equations used to get the likelihood distribution or sometimes tensor locks down and the code just freeze. Find my code below,
with pm.Model() as model:
# Prior Distributions for unknown model parameters:
mu = pm.Normal('mu', mu=0, sigma=1)
sd = pm.HalfNormal('sd', sigma=1)
nc = pm.HalfNormal('nc', sigma=10)
nu = pm.HalfNormal('nu', sigma=1)
# Custom distribution:
# observed_data = pm.DensityDist('observed_data', NonCentralStudentT, observed=data_list)
# Observed data is from a Likelihood distributions (Likelihood (sampling distribution) of observations):
observed_data = NonCentralStudentT('observed_data', mu=mu, sd=sd, nc=nc, nu=nu, observed=data_list)
# draw 5000 posterior samples
trace_S = pm.sample(draws=5000, tune=2000, chains=3, cores=1)
# Obtaining Posterior Predictive Sampling:
post_pred_S = pm.sample_posterior_predictive(trace_S, samples=3000)
print(post_pred_S['observed_data'].shape)
print('\nSummary: ')
print(pm.stats.summary(data=trace_S))
print(pm.stats.summary(data=post_pred_S))
Edit 2:
I am looking online in order to convert the function to theano, the only thing that I found to define the function is from the following GitHub link hyp1f1 function GitHub
Will this be enough to use in order to convert the function into theano?
In addition, I have a question, it is okay to use NumPy arrays with theano?
Also, I thought of another way but I am not sure if this can be implemented, I looked into the nct function in scipy and they wrote the following,
If Y is a standard normal random variable and V is an independent
chi-square random variable ( chi2 ) with k degrees of freedom, then
X=(Y+c) / sqrt(V/k)
has a non-central Student’s t distribution on the real line. The
degrees of freedom parameter k (denoted df in the implementation)
satisfies k>0 and the noncentrality parameter c (denoted nc in the
implementation) is a real number.
The probability density above is defined in the “standardized” form.
To shift and/or scale the distribution use the loc and scale
parameters. Specifically, nct.pdf(x, df, nc, loc, scale) is
identically equivalent to nct.pdf(y, df, nc) / scale with y = (x -
loc) / scale .
So, I thought of only using the priors as normal and chi2 random variables code part in their distributions and use the degree of freedom variable as mentioned before in the code into the equation mentioned in SciPy, will it be enough to get the distribution?
Edit 3:
I managed to run the code in the link about fitting empirical distribution and found out the second best was the student t distribution, so, I will be using this. Thank you for your help. I just have a side question, I ran my model with student t distribution but I got these warnings:
There were 52 divergences after tuning. Increase target_accept or
reparameterize. The acceptance probability does not match the target.
It is 0.7037574708196309, but should be close to 0.8. Try to increase
the number of tuning steps. The number of effective samples is smaller
than 10% for some parameters.
I am just confused about these warnings, Do you have any idea what it means? I know that this won't affect my code, but, I can reduce the divergences? and regarding the effective samples, Do I need to increase the number of samples in the trace code?

Using python built-in functions for coupled ODEs

THIS PART IS JUST BACKGROUND IF YOU NEED IT
I am developing a numerical solver for the Second-Order Kuramoto Model. The functions I use to find the derivatives of theta and omega are given below.
# n-dimensional change in omega
def d_theta(omega):
return omega
# n-dimensional change in omega
def d_omega(K,A,P,alpha,mask,n):
def layer1(theta,omega):
T = theta[:,None] - theta
A[mask] = K[mask] * np.sin(T[mask])
return - alpha*omega + P - A.sum(1)
return layer1
These equations return vectors.
QUESTION 1
I know how to use odeint for two dimensions, (y,t). for my research I want to use a built-in Python function that works for higher dimensions.
QUESTION 2
I do not necessarily want to stop after a predetermined amount of time. I have other stopping conditions in the code below that will indicate whether the system of equations converges to the steady state. How do I incorporate these into a built-in Python solver?
WHAT I CURRENTLY HAVE
This is the code I am currently using to solve the system. I just implemented RK4 with constant time stepping in a loop.
# This function randomly samples initial values in the domain and returns whether the solution converged
# Inputs:
# f change in theta (d_theta)
# g change in omega (d_omega)
# tol when step size is lower than tolerance, the solution is said to converge
# h size of the time step
# max_iter maximum number of steps Runge-Kutta will perform before giving up
# max_laps maximum number of laps the solution can do before giving up
# fixed_t vector of fixed points of theta
# fixed_o vector of fixed points of omega
# n number of dimensions
# theta initial theta vector
# omega initial omega vector
# Outputs:
# converges true if it nodes restabilizes, false otherwise
def kuramoto_rk4_wss(f,g,tol_ss,tol_step,h,max_iter,max_laps,fixed_o,fixed_t,n):
def layer1(theta,omega):
lap = np.zeros(n, dtype = int)
converges = False
i = 0
tau = 2 * np.pi
while(i < max_iter): # perform RK4 with constant time step
p_omega = omega
p_theta = theta
T1 = h*f(omega)
O1 = h*g(theta,omega)
T2 = h*f(omega + O1/2)
O2 = h*g(theta + T1/2,omega + O1/2)
T3 = h*f(omega + O2/2)
O3 = h*g(theta + T2/2,omega + O2/2)
T4 = h*f(omega + O3)
O4 = h*g(theta + T3,omega + O3)
theta = theta + (T1 + 2*T2 + 2*T3 + T4)/6 # take theta time step
mask2 = np.array(np.where(np.logical_or(theta > tau, theta < 0))) # find which nodes left [0, 2pi]
lap[mask2] = lap[mask2] + 1 # increment the mask
theta[mask2] = np.mod(theta[mask2], tau) # take the modulus
omega = omega + (O1 + 2*O2 + 2*O3 + O4)/6
if(max_laps in lap): # if any generator rotates this many times it probably won't converge
break
elif(np.any(omega > 12)): # if any of the generators is rotating this fast, it probably won't converge
break
elif(np.linalg.norm(omega) < tol_ss and # assert the nodes are sufficiently close to the equilibrium
np.linalg.norm(omega - p_omega) < tol_step and # assert change in omega is small
np.linalg.norm(theta - p_theta) < tol_step): # assert change in theta is small
converges = True
break
i = i + 1
return converges
return layer1
Thanks for your help!

You can wrap your existing functions into a function accepted by odeint (option tfirst=True) and solve_ivp as
def odesys(t,u):
theta,omega = u[:n],u[n:]; # or = u.reshape(2,-1);
return [ *f(omega), *g(theta,omega) ]; # or np.concatenate([f(omega), g(theta,omega)])
u0 = [*theta0, *omega0]
t = linspan(t0, tf, timesteps+1);
u = odeint(odesys, u0, t, tfirst=True);
#or
res = solve_ivp(odesys, [t0,tf], u0, t_eval=t)
The scipy methods pass numpy arrays and convert the return value into same, so that you do not have to care in the ODE function. The variant in comments is using explicit numpy functions.
While solve_ivp does have event handling, using it for a systematic collection of events is rather cumbersome. It would be easier to advance some fixed step, do the normalization and termination detection, and then repeat this.
If you want to later increase efficiency somewhat, use directly the stepper classes behind solve_ivp.

How to do a gradient descent problem (machine learning)?

could somebody please explain how to do a gradient descent problem WITHOUT the context of the cost function? I have seen countless tutorials that explain gradient descent using the cost function, but I really don't understand how it works in a more general sense.
I am given a 3D function:
z = 3*((1-xx)2) * np.exp(-(xx2) - (yy+1)2) \
- 10*(xx/5 - xx3 - yy5) * np.exp(-xx2 - yy2)- (1/3)* np.exp(-(xx+1)**2 - yy2)
And I am asked to:
Code a simple gradient algorithm. Set the parameters as follows:
learning rate = step size: 0.1
Max number of iterations: 20
Stopping criterion: 0.0001 (Your iterations should stop when your gradient is smaller than the threshold)
Then start your algorithm at
(x0 = 0.5, y0 = -0.5)
(x0 = -0.3, y0 = -0.3)
I have seen this piece of code floating around wherever gradient descent is talked about:
def update_weights(m, b, X, Y, learning_rate):
m_deriv = 0
b_deriv = 0
N = len(X)
for i in range(N):
# Calculate partial derivatives
# -2x(y - (mx + b))
m_deriv += -2*X[i] * (Y[i] - (m*X[i] + b))
# -2(y - (mx + b))
b_deriv += -2*(Y[i] - (m*X[i] + b))
# We subtract because the derivatives point in direction of steepest ascent
m -= (m_deriv / float(N)) * learning_rate
b -= (b_deriv / float(N)) * learning_rate
return m, b
enter code here
But I don't understand how to use it for my problem. How does my function fit in there? What do I adjust instead of m and b? I'm very very confused.
Thank you.

Gradient Descent is optimization algorithm for finding the minimum of a function.
Very simplified view
Lets start with a 1D function y = f(x)
Lets start at an arbitrary value of x and find the gradient (slope) of f(x).
If the slope is decreasing at x then it means we have to go further toward (right of number line) x (for reaching the minimum)
If the slope is increasing at x then it means we have to go away from (left of number line) x
We can get the slope by taking the derivative of the function. The derivative is -ve if the slop is decreasing and +ve if the slope is increasing
So we can start at some arbitrary value of x and slowly move toward the minimum using the derivatives at that value of x. How slowly we are moving is determined by the learning rate or step size. so we have the update rule
x = x - df_dx*lr
We can see that if the slope is decreasing the derivative (df_dx) is -ve and x is increasing and so x is moving to further right. On the other hand if slope is increasing the df_dx is +ve which decreases x and so we are moving toward left.
We continue this either for some large number of times or until the derivative is very small
Multivariate function z = f(x,y)
The same logic as above applies except now we take the partial derivatives instead of derivative.
Update rule is
x = x - dpf_dx*lr
y = y - dpf_dy*lr
Where dpf_dx is the partial derivative of f with respect to x
The above algorithm is called the gradient decent algorithm. In Machine learning the f(x,y) is a cost/loss function whose minimum we are interested in.
Example
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d.axes3d import Axes3D
from pylab import meshgrid
from scipy.optimize import fmin
import math
def z_func(a):
x, y = a
return ((x-1)**2+(y-2)**2)
x = np.arange(-3.0,3.0,0.1)
y = np.arange(-3.0,3.0,0.1)
X,Y = meshgrid(x, y) # grid of point
Z = z_func((X, Y)) # evaluation of the function on the grid
fig = plt.figure()
ax = fig.gca(projection='3d')
surf = ax.plot_surface(X, Y, Z, rstride=1, cstride=1,linewidth=0, antialiased=False)
plt.show()
The min of z_func is at (1,2). This can be verified using the fmin function of scipy
fmin(z_func,np.array([10,10]))
Now lets write our own gradient decent algorithm to find the min of z_func
def gradient_decent(x,y,lr):
while True:
d_x = 2*(x-1)
d_y = 2*(y-2)
x -= d_x*lr
y -= d_y*lr
if d_x < 0.0001 and d_y < 0.0001:
break
return x,y
print (gradient_decent(10,10,0.1)
We are starting at some arbitrary value x=10 and y=10 and a learning rate of 0.1. The above code prints 1.000033672997724 2.0000299315535326 which is correct.
So if you have a continuous differentiable convex function, to find its optimal (which is minimal for a convex) all you have to do is find the partial derivatives of the function with respect to each variable and use the update rule mentioned above. Repeat the steps until the gradients are small which mean we have reached the minima for a convex function.
If the function is not convex, we might get stuck in a local optima.

Is there Implementation of Hawkes Process in PyMC?

I want to use Hawkes process to model some data. I could not find whether PyMC supports Hawkes process. More specifically I want an observed variable with Hawkes Process and learn a posterior on its params.
If it is not there, then could I define it in PyMC in some way e.g. #deterministic etc.??

It's been quite a long time since your question, but I've worked it out on PyMC today so I'd thought I'd share the gist of my implementation for the other people who might get across the same problem. We're going to infer the parameters λ and α of a Hawkes process. I'm not going to cover the temporal scale parameter β, I'll leave that as an exercise for the readers.
First let's generate some data :
def hawkes_intensity(mu, alpha, points, t):
p = np.array(points)
p = p[p <= t]
p = np.exp(p - t)
return mu + alpha * np.sum(p)
def simulate_hawkes(mu, alpha, window):
t = 0
points = []
lambdas = []
while t < window:
m = hawkes_intensity(mu, alpha, points, t)
s = np.random.exponential(scale=1/m)
ratio = hawkes_intensity(mu, alpha, points, t + s)
t = t + s
if t < window:
points.append(t)
lambdas.append(ratio)
else:
break
points = np.sort(np.array(points, dtype=np.float32))
lambdas = np.array(lambdas, dtype=np.float32)
return points, lambdas
# parameters
window = 1000
mu = 8
alpha = 0.25
points, lambdas = simulate_hawkes(mu, alpha, window)
num_points = len(points)
We just generated some temporal points using some functions that I adapted from there : https://nbviewer.jupyter.org/github/MatthewDaws/PointProcesses/blob/master/Temporal%20points%20processes.ipynb
Now, the trick is to create a matrix of size (num_points, num_points) that contains the temporal distance of the ith point from all the other points. So the (i, j) point of the matrix is the temporal interval separating the ith point to the jth. This matrix will be used to compute the sum of the exponentials of the Hawkes process, ie. the self-exciting part. The way to create this matrix as well as the sum of the exponentials is a bit tricky. I'd recommend to check every line yourself so you can see what they do.
tile = np.tile(points, num_points).reshape(num_points, num_points)
tile = np.clip(points[:, None] - tile, 0, np.inf)
tile = np.tril(np.exp(-tile), k=-1)
Σ = np.sum(tile, axis=1)[:-1] # this is our self-exciting sum term
We have points and we have a matrix containg the sum of the excitations term.
The duration between two consecutive events of a Hawkes process follow an exponential distribution of parameter λ = λ0 + ∑ excitation. This is what we are going to model, but first we have to compute the duration between two consecutive points of our generated data.
interval = points[1:] - points[:-1]
We're now ready for inference:
with pm.Model() as model:
λ = pm.Exponential("λ", 1)
α = pm.Uniform("α", 0, 1)
lam = pm.Deterministic("lam", λ + α * Σ)
interarrival = pm.Exponential(
"interarrival", lam, observed=interval)
trace = pm.sample(2000, tune=4000)
pm.plot_posterior(trace, var_names=["λ", "α"])
plt.show()
print(np.mean(trace["λ"]))
print(np.mean(trace["α"]))
7.829
0.284
Note: the tile matrix can become quite large if you have many data points.

Artefacts from Riemann sum in scipy.signal.convolve

Short summary: How do I quickly calculate the finite convolution of two arrays?
Problem description
I am trying to obtain the finite convolution of two functions f(x), g(x) defined by
To achieve this, I have taken discrete samples of the functions and turned them into arrays of length steps:
xarray = [x * i / steps for i in range(steps)]
farray = [f(x) for x in xarray]
garray = [g(x) for x in xarray]
I then tried to calculate the convolution using the scipy.signal.convolve function. This function gives the same results as the algorithm conv suggested here. However, the results differ considerably from analytical solutions. Modifying the algorithm conv to use the trapezoidal rule gives the desired results.
To illustrate this, I let
f(x) = exp(-x)
g(x) = 2 * exp(-2 * x)
the results are:
Here Riemann represents a simple Riemann sum, trapezoidal is a modified version of the Riemann algorithm to use the trapezoidal rule, scipy.signal.convolve is the scipy function and analytical is the analytical convolution.
Now let g(x) = x^2 * exp(-x) and the results become:
Here 'ratio' is the ratio of the values obtained from scipy to the analytical values. The above demonstrates that the problem cannot be solved by renormalising the integral.
The question
Is it possible to use the speed of scipy but retain the better results of a trapezoidal rule or do I have to write a C extension to achieve the desired results?
An example
Just copy and paste the code below to see the problem I am encountering. The two results can be brought to closer agreement by increasing the steps variable. I believe that the problem is due to artefacts from right hand Riemann sums because the integral is overestimated when it is increasing and approaches the analytical solution again as it is decreasing.
EDIT: I have now included the original algorithm 2 as a comparison which gives the same results as the scipy.signal.convolve function.
import numpy as np
import scipy.signal as signal
import matplotlib.pyplot as plt
import math
def convolveoriginal(x, y):
'''
The original algorithm from http://www.physics.rutgers.edu/~masud/computing/WPark_recipes_in_python.html.
'''
P, Q, N = len(x), len(y), len(x) + len(y) - 1
z = []
for k in range(N):
t, lower, upper = 0, max(0, k - (Q - 1)), min(P - 1, k)
for i in range(lower, upper + 1):
t = t + x[i] * y[k - i]
z.append(t)
return np.array(z) #Modified to include conversion to numpy array
def convolve(y1, y2, dx = None):
'''
Compute the finite convolution of two signals of equal length.
#param y1: First signal.
#param y2: Second signal.
#param dx: [optional] Integration step width.
#note: Based on the algorithm at http://www.physics.rutgers.edu/~masud/computing/WPark_recipes_in_python.html.
'''
P = len(y1) #Determine the length of the signal
z = [] #Create a list of convolution values
for k in range(P):
t = 0
lower = max(0, k - (P - 1))
upper = min(P - 1, k)
for i in range(lower, upper):
t += (y1[i] * y2[k - i] + y1[i + 1] * y2[k - (i + 1)]) / 2
z.append(t)
z = np.array(z) #Convert to a numpy array
if dx != None: #Is a step width specified?
z *= dx
return z
steps = 50 #Number of integration steps
maxtime = 5 #Maximum time
dt = float(maxtime) / steps #Obtain the width of a time step
time = [dt * i for i in range (steps)] #Create an array of times
exp1 = [math.exp(-t) for t in time] #Create an array of function values
exp2 = [2 * math.exp(-2 * t) for t in time]
#Calculate the analytical expression
analytical = [2 * math.exp(-2 * t) * (-1 + math.exp(t)) for t in time]
#Calculate the trapezoidal convolution
trapezoidal = convolve(exp1, exp2, dt)
#Calculate the scipy convolution
sci = signal.convolve(exp1, exp2, mode = 'full')
#Slice the first half to obtain the causal convolution and multiply by dt
#to account for the step width
sci = sci[0:steps] * dt
#Calculate the convolution using the original Riemann sum algorithm
riemann = convolveoriginal(exp1, exp2)
riemann = riemann[0:steps] * dt
#Plot
plt.plot(time, analytical, label = 'analytical')
plt.plot(time, trapezoidal, 'o', label = 'trapezoidal')
plt.plot(time, riemann, 'o', label = 'Riemann')
plt.plot(time, sci, '.', label = 'scipy.signal.convolve')
plt.legend()
plt.show()
Thank you for your time!

or, for those who prefer numpy to C. It will be slower than the C implementation, but it's just a few lines.
>>> t = np.linspace(0, maxtime-dt, 50)
>>> fx = np.exp(-np.array(t))
>>> gx = 2*np.exp(-2*np.array(t))
>>> analytical = 2 * np.exp(-2 * t) * (-1 + np.exp(t))
this looks like trapezoidal in this case (but I didn't check the math)
>>> s2a = signal.convolve(fx[1:], gx, 'full')*dt
>>> s2b = signal.convolve(fx, gx[1:], 'full')*dt
>>> s = (s2a+s2b)/2
>>> s[:10]
array([ 0.17235682, 0.29706872, 0.38433313, 0.44235042, 0.47770012,
0.49564748, 0.50039326, 0.49527721, 0.48294359, 0.46547582])
>>> analytical[:10]
array([ 0. , 0.17221333, 0.29682141, 0.38401317, 0.44198216,
0.47730244, 0.49523485, 0.49997668, 0.49486489, 0.48254154])
largest absolute error:
>>> np.max(np.abs(s[:len(analytical)-1] - analytical[1:]))
0.00041657780840698155
>>> np.argmax(np.abs(s[:len(analytical)-1] - analytical[1:]))
6

Short answer: Write it in C!
Long answer
Using the cookbook about numpy arrays I rewrote the trapezoidal convolution method in C. In order to use the C code one requires three files (https://gist.github.com/1626919)
The C code (performancemodule.c).
The setup file to build the code and make it callable from python (performancemodulesetup.py).
The python file that makes use of the C extension (performancetest.py)
The code should run upon downloading by doing the following
Adjust the include path in performancemodule.c.
Run the following
python performancemodulesetup.py build
python performancetest.py
You may have to copy the library file performancemodule.so or performancemodule.dll into the same directory as performancetest.py.
Results and performance
The results agree neatly with one another as shown below:
The performance of the C method is even better than scipy's convolve method. Running 10k convolutions with array length 50 requires
convolve (seconds, microseconds) 81 349969
scipy.signal.convolve (seconds, microseconds) 1 962599
convolve in C (seconds, microseconds) 0 87024
Thus, the C implementation is about 1000 times faster than the python implementation and a bit more than 20 times as fast as the scipy implementation (admittedly, the scipy implementation is more versatile).
EDIT: This does not solve the original question exactly but is sufficient for my purposes.