# How to reparametrize a probability density function

I have this probability density function (pdf) $$f(t)=\beta \left[\left(\frac{t}{\eta_1}\right)^{\beta} + \left(\frac{t} {\eta_2}\right)^{\beta} \right]t^{-1}e^{- \left(\frac{t}{\eta_1} \right)^{\beta} - \left(\frac{t}{\eta_2}\right)^{\beta}}$$ then using Mathematica 10 (student version), I obtain $$\int_0^\infty f(t) dt= 1.$$ But when I reparametrize $\alpha_i = \left(\frac{t}{\eta_i}\right)^{\beta}$, $i=1,2$, $$g(t) = \beta (\alpha_1 + \alpha_2)t^{-1}e^{- (\alpha_1 + \alpha_2)},$$ I obtain an error warning:

Integrate::idiv: "Integral of 1/t does not converge on {0,∞}.

pdf:

model[t_, β_, η1_, η2_] := \
β  t^-1   ((t/η1)^β + (t/η2)^β) E^(-(t/\
η1)^β - (t/η2)^β)

Integrate[model[t, β, η1, η2], {t, 0, ∞},
Assumptions -> {β > 0, η1 > 0, η2 > 0}]


latter code is equal 1.

pdf reparametrized:

model2[t_, β_, α1_, α2_] := β  t^-1   (α1 + α1) E^(- α1 - α2)


and when

Integrate[model2[t, β, α1, α2], {t, 0, ∞},
Assumptions -> {β > 0, α1 > 0, α2 > 0}]


I obtain this error warning:

Integrate::idiv: "Integral of 1/t does not converge on {0,∞}.

• Please use code, rather than $\LaTeX$. – Feyre Dec 15 '16 at 21:37
• Similar – corey979 Dec 15 '16 at 22:14
• @Feyre, Where am I wrong? I'm novice in to use Mathematica software. – Marco Dec 15 '16 at 22:30
• I do not get the error, in fact your second code evaluates to 1 too, I suggest quitting your kernel and trying again. – Feyre Dec 15 '16 at 22:44
• @Feyre There's an error in the second code: there's model while it should be model2. – corey979 Dec 15 '16 at 23:11

You have to make the substitutions back into the integrand, so Mathematica can know what the definitions of α1 and α2 are.

model2[t_, β_, α1_, α2_] := β(α1 + α2)/t E^(-α1 - α2)
Integrate[
model2[t, β, α1 /. α1 -> (t/n1)^β, α2 /. α2 -> (t/n2)^β], {t, 0, ∞},
Assumptions -> {β > 0, n1 > 0, n2 > 0}]


1

• @Marco, "I can't have $\eta_i$ in my outcomes" - that's kind of muddled; you do realize that $\eta_i$ is constant, while $\alpha_i$ is dependent on $t$, no? Thus, your mean expression cannot involve $\alpha_i$ as you have presented it. – J. M.'s discontentment Dec 16 '16 at 2:23
• @Marco. All these comments should be made as update to your question so they will addressed to the community at large, not just to me. They are not really comments on my answer as far as I can tell. – m_goldberg Dec 16 '16 at 2:23