I need to implement the following in python: For a given discrete time series Zt (t=0 to T), find smallest t such that:

$c\sum_{s=0}^t e^{[k(Zt-Zs)+m(t-s)]} >= \frac{p*}{1-p*} $

where c,k,m are constants and P* is given by

$\int_0^{0.5} \frac{(1-2y)e^{-d/y}}{(1-y)^{1+d}y^{1-d}}dy=\int_{0.5}^{p*} \frac{(2y-1)e^{-d/y}}{(1-y)^{1+d}y^{1-d}}dy $

where d is another constant(something to be optimized)

I need to implement this for each row of an array Z = input 2d array of shape (N,w) I implemented a loop version:

i implemented the sum in equation 1 above utilizing the fact

$ e^{[k(Zt-Zs)+m(t-s)]} = \frac{e^{kZt+mt}}{e^{kZs+ms}} $


$\sum_{s=0}^t e^{[k(Zt-Zs)+m(t-s)]} =e^{kZt+mt}*\sum_{s=0}^t\frac{1}{e^{kZs+ms}} $ which is reflected in use of cumsum in code below

def f(z):
    return ((1-2*z)*np.exp(-d/z))/(((1-z)**(1+d))*(z**(1-d)))
def rhs(p):
    return integrate.quad(-f,0.5, p)
p_star= fsolve(rhs-lhs,0.75) # will depend on time series only indrectly when d will be optimized
for i in np.arange(N):
    main = np.exp (kz+m*np.arange(w)) 
    final_sum= main*cumsum_t 
    t_solution= # index i where final_sum[i]> p_star/(1-p_star)) # not implemented yet

Is there any way I can vectorize this for Z? In this case N is ~400,000 so vectorization would really help. The function f(z) will be fine with vector inputs but I dont think the function rhs will be fine as it uses integration.

  • $\begingroup$ For the given sum $\sum_{s=0}^{t}$, what does it mean if $t \not \in \mathbb{Z}$ ? $\endgroup$ – tqviet Oct 27 '16 at 5:16
  • $\begingroup$ well, it is a discrete time series. $\endgroup$ – dayum Oct 27 '16 at 5:18
  • $\begingroup$ What are typical values of $p*$ and $d$? How accurate do you need $p*$ to be? If you want to vectorize the calculation of $p*$, I think the only way you can do it is by approximating the solution of the equation defining $p*$... $\endgroup$ – GertVdE Oct 27 '16 at 8:10
  • $\begingroup$ quad is incredibly slow even vectorizing this won't help much. Try a different integration method - I will dig up an old post I had that is way faster $\endgroup$ – Matt Jun 17 '17 at 16:12
  • $\begingroup$ stackoverflow.com/questions/37367688/… $\endgroup$ – Matt Jun 17 '17 at 16:15

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