Hi Mamali
what is one of the most useful techniques of probability?
if you don't know, assume uniform distribution
Which is it all about probability, how to walk through a dark room without knowing where the furniture is, if you understand what I mean.
So, you brought a CDF and pdf without any further information: again, let if be uniform.
pdf for uniform is as flat as it gets, and CDF of a flat constant is? yes, t.
had a first look to this monster function:
t=[-100:.01:100]
k1=sqrt(.85);k2=sqrt(1-.85);k3=3;x=((k3+k1*t)/k2).^7
plot(t,x)
grid on
grid minor
both sides diverge but it looks more or less symmetric, so it may be integrable. replace pdf(t)=1 and let me get to the CDF a bit later.
What would happen if you integrate the function without the CDF (big PHI)?
y=@(t) ((k3+k1*t)/k2).^7
y =
@(t)((k3+k1*t)/k2).^7
>> integral(y,-Inf,Inf)
Warning: Minimum step size reached near x = -Inf. There may be
a singularity, or the tolerances may be too tight for this
problem.
> In integralCalc/checkSpacing (line 456)
In integralCalc/iterateScalarValued (line 319)
In integralCalc/vadapt (line 132)
In integralCalc (line 103)
In integral (line 88)
ans =
13226991826182229000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.00
to get more precision, need the Symbolic tool box, and use function vpa, variable precision arithmetic.
However, MATLAB already warning that there may be a singularity, and if not knowing anything else of that odd spots where integral may walk through, or may stuck before being able to carry on integrating towards Inf, there isn't much more that can be done.
Now the CDF of a uniform distribution, that has the pdf multiplying the function to integrate. uniform[0 1] or [-A A] makes sense, because how do you apply the concept of probability to an infinite set? you need to define a subset to then define the probabilities.
uniform pdf is a pulse, from any 2 fixed values[a b]. The value of the uniform pdf is 1/abs(a-b) throughout the interval where the pdf is defined, AND ZERO ELSEWHERE. For convenience I define the interval [0 1] pdf is 1 between 0 and 1.
The CDF of a uniform pdf [a b] is a steady ramp from zero on a to 1 on b. Here the integral interval has been reduced to [0 1] therefore I would rid the CDF (big PHI) by replacing it with a simple multiplying t.
Through the above reasonable simplifications, your reference vector t, your function x, and the integral, now are:
t=[0:.001:1]
x=@(t) t.*((k3+k1*t)/k2).^7
integral(x,0,1)
3307946.04
good to go?
if they keep throwing at you these kind of functions, it would be greatly appreciated the distributions to be defined in advance. Hope it helps John jgb2012@sky.com 
