3 parameter Weibull distribution integration issues

Questions
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1

Hi all,

I was trying to solve this following Integration:

p[x_, a_, b_, c_] := PDF[WeibullDistribution[a, b, c], x]
Integrate[p[x,a1,b1,c1]*Log[x-c1], {x,0,Infinity}]

But it did not solve the integral. Therefore, I simply tried to integrate over the 3 parameter Weibull distribution:

Integrate[PDF[WeibullDistribution[a, b, c], x], {x, 0, Infinity}]

To my surprise, it did not integrate to 1…instad, its answer is the integral form of the 3-param Weibull.
When I tried the sanity check to integrate over the more common 2-param Weibull, it does integrates to 1.

Therefore, I was wondering if anybody has any idea how to at least make the integration of 3-parameter Weibull distribution to 1 in mathematica? I admit I am very new to this software…any help is appreciated, thanks!!

2

@Yu,
Usually this is best handled by CDF function if you are looking for such result:

In[20]:= Limit[Simplify[
                        CDF[WeibullDistribution[a,b,u],x],
               Assumptions-> x>u],
         x->Infinity,Assumptions-> a>0 &&b>0]
Out[20]= 1
3

oh i see.. thanks!!

In fact, I was trying to solve this integration relating to 3-parameter Weibull distribution:

Integrate[PDF[WeibullDistribution[a, b, c], x]*Log[x - c], {x, 0, Infinity}]

Because mathemateca didn’t simply anything from the above integration, therefore I tried to just simply integrate the 3-parameter Weibull to see what happens. Can you shed some lights on how to incorporate your solution to this integration? Thanks!

4

Hi,

The integrand

PDF[WeibullDistribution[a, b, c], x]

is a Piecewise statement and is

(a E^-((-c + x)/b)^a)/b ((-c + x)/b)^(-1 + a)

for x>c and 0 otherwise. c is assumed to be positive. Integrate does not work with the Piecewise as is in this case.

I tried to directly integrate the above expression over the range c<x<Infinity, but it failed. Instead, using the variable transformation

t == ((-c + x)/b)^a

the integral can be transformed to

Integrate[E^-t, {t, 0, \[Infinity]}]

which is equal to 1.

Your original integral can be evaluated in a similar manner. The result is

Integrate[E^-t Log[b t^(1/a)], {t, 0, \[Infinity]}]

(* Out: -(EulerGamma/a) + Log[b] *)

Youngjoo Chung

5

Hi Youngjoo,

Thank you so much for your suggestion. However, I wasn’t able to use your variable transformation and arrive at E^-t, from the 3-parameter Weibull distribution…using your transformation of

t == ((-c + x)/b)^a

I arrived at:

(a/b) ((x - c)/b)^(a - 1) E^-t

I suppose it is of 3-param Weibull form, but not the exact transformation right? Or am I mistaken somewhere?

Futhermore, I used your method to transform my original integral and tried it:

Integrate[E^-t Log[b t^(1/a)], {t, 0, ?}]

However, it did not result in the answer that you showed (with the EulerGamma), it was still an integral with -(real+x/b)^a in it, although I did set the t transformation before hand…is there some other setting that I neeed to do..? Thanks!

6

Hi,

All x in the integral needs to be transformed, so that thre is no x left:

{(a E^-((-c + x)/b)^a)/b ((-c + x)/b)^(-1 + a) -> (a E^-t t^((-1 + a)/a))/b,
\[DifferentialD]x -> (b t^(-1 + 1/a) \[DifferentialD]t)/a}

This gives

(a E^-((-c + x)/b)^a)/b ((-c + x)/b)^(-1 + a) \[DifferentialD]x -> E^-t \[DifferentialD]t

By the way, the result

Integrate[E^-t Log[b t^(1/a)], {t, 0, ?}]

is intermediate, and the Log function should be expanded before evaluation like

Integrate[E^-t Log[t], {t, 0, \[Infinity]}]/a + Integrate[E^-t, {t, 0, \[Infinity]}] Log[b]

This gives the result shown above.

Youngjoo Chung