Starting off an IIR with plain averages to accelerate slow rise time
You are omitting the initial condition of the filter:

$$y(t)=\alpha x(t-1)+(1-\alpha)y(t-1)$$

which is 

$$y(0)=x(0)=x_0$$

which is implemented by correcting this line before the cycle:

if t == 1:
    y=a*x;


You dont need to switch anything. Only provide that value and an initial evaluation with that index, and you will not have a slow rise problem.

2016-11-22
The following image shows the result of putting the proper IC to the filter. In red there is the standard "zero" IC, while the green shows the correct IC. The $x(t)$ is a step periodic signal. Note that the IC is perfectly matched with the expected result. 



2016-11-23

Another depict for a biased Laplacian Noise, and different parameters for a linear filter, with IC settled as indicated. Note that regardless of the IC setting, all the signals are around the bias, which is the correct result.