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I have the following exponential function

$$f(t)= e^{-(\theta/t)^{\beta}-c\frac{\Gamma\Big(\alpha,\big(\frac{\theta}{t}\big)^\beta\Big)}{\Gamma(\alpha)}} $$

where $c,\ \alpha,\ \theta,$ and $\beta$ are constants. $\Gamma(,)$ and $\Gamma()$ are the upper incomplete gamma function and the gamma function, respectively. They are given by

$$ \Gamma\bigg(\alpha,\Big(\frac{\theta}{t}\Big)^\beta\bigg)=\int_{(\theta/t)^\beta}^\infty t^{\alpha-1}e^{-t}dt\\ \Gamma(\alpha)=\int_0^\infty t^{\alpha-1}e^{-t}dt $$

I need to take the logarithm of $f(t)$. EDIT: But I have the following questions:

1. How does one perform the logarithm on the fractional part?

2. If $f(t)$ is a distribution function such that $t_1,t_2,...,t_n$ be an independent random sample from the distribution, how can we perform the logarithm on $f(t)$.

Taking these questions into account, should the logarithm of $f(t)$ be

$$ \ln\Big(f(t)\Big)=-\sum_i^n(\theta/t_i)^\beta-nc\Bigg[\frac{\sum_i^n\Gamma\Big(\alpha,\big(\theta/t_i\big)^\beta\Big)}{\sum_i^n\Gamma_i(\alpha)}\Bigg] $$

Or should it be $$ \ln\Big(f(t)\Big)=-\sum_i^n(\theta/t_i)^\beta-nc-\sum_i^n\Gamma\Big(\alpha,\big(\theta/t_i\big)^\beta\Big)+\sum_i^n\Gamma_i(\alpha) $$

Also, should $\Gamma(,)$ and $\Gamma()$ also be transformed to the logarithm. If so, applying logarithm yields $$ \ln\Bigg(\Gamma\bigg(\alpha,\Big(\frac{\theta}{t}\Big)^\beta\bigg)\Bigg)=\int_{(\theta/t)^\beta}^\infty -(\alpha-1)t^2dt\\ \Gamma(\alpha)=\int_0^\infty -(\alpha-1)t^2dt $$

Are these correct? Should $\int$ sign be transformed to $\prod$ sign?

Any help would be appreciated.

Thanks in advance.

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    $\begingroup$ Why is this any more complicated than $\ln(f(t))=-(\theta/t)^{\beta}-c\frac{\Gamma\Big(\alpha,\big(\frac{\theta}{t}\big)^\beta\Big)}{\Gamma(\alpha)}$ ? $\endgroup$ – Wouter Jun 20 at 5:18
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    $\begingroup$ I also came up with the formula from the previous comment, but I notice you say "where $t = 1,2,\ldots,n$. What is that about? And where does $t_i$ come from--if $t=1,2,\ldots,n$ shouldn't that just be $i$? $\endgroup$ – David K Jun 20 at 5:22
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    $\begingroup$ I don't know what you are doing there, but if your approach transforms Euler's $\Gamma$ function into $\Gamma(\alpha) = \int _0 ^\infty -(\alpha-1) t^2 \ \mathrm dt$ then it's definitely wrong. $\endgroup$ – Alex M. Jun 20 at 5:48
  • $\begingroup$ @Wouter I was wondering if $\ln(f(t))=-(\theta/t)^{\beta}-c\frac{\Gamma\Big(\alpha,\big(\frac{\theta}{t}\big)^\beta\Big)}{\Gamma(\alpha)}$ was the right one. Anyway, now it makes it clearer. Thanks for the tip. $\endgroup$ – nashynash Jun 20 at 7:47
  • $\begingroup$ @DavidK Thank you for pointing out the error. I have revised the notation as per your suggestion. $\endgroup$ – nashynash Jun 20 at 7:50
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It is always true that $$\ln(\exp(g))=g$$ no matter how complex an expression $g$ may be. There is no need to take the logarithm of $g$, or transform products into sums, or any other such thing: the logarithm undoes the exponential, no more, no less.

So $$f(t)= e^{-(\theta/t)^{\beta}-c\frac{\Gamma\Big(\alpha,\big(\frac{\theta}{t}\big)^\beta\Big)}{\Gamma(\alpha)}} $$

$$\ln(f(t))= -(\theta/t)^{\beta}-c\frac{\Gamma\Big(\alpha,\big(\frac{\theta}{t}\big)^\beta\Big)}{\Gamma(\alpha)} $$

Aside: in a complex context, it is true that $$\exp(x)=y$$ has solutions other than $x=\ln(y)$, those being $x=\ln(y)+ n i 2 \pi$, where $n$ is an integer and $i$ the imaginary unit, so some care is occasionally needed.

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  • $\begingroup$ How about if $t_1,t_2,....,t_n$ is an independent random sample from the distribution? (I have edited the main post for more clarity.) Then would we sum the terms in the exponents? Is that correct? $\endgroup$ – nashynash Jun 20 at 8:18
  • $\begingroup$ @Wouter, there should be a limit to how low one should go for a bounty. This question is not even a question; if I had been you, I wouldn't have answered it. Plus, you have 6000+ reputation, so you're an experienced user by now. $\endgroup$ – Alex M. Jun 20 at 9:32
  • $\begingroup$ @AlexM. The post presents the confusion of the postmaster: 1. Is the logarithm transformation to the exponent term correctly done? 2. Should the logarithm transformation be applied to the $\Gamma()$ and $\Gamma(,)$ terms? The postmaster asked for clarifications and the correct solutions if his approach was incorrect, to which Wouter's answer satisfies 1 and your earlier comment seems to clear the doubt in 2. If there are improvements that can be made to the post, kindly suggests so. There will always be someone that benefits from a post, besides the postmaster. $\endgroup$ – nashynash Jun 20 at 10:04
  • $\begingroup$ @AlexM. The postmaster has revised the title of the post and some of the texts in the post. Hopefully, the revisions improve the meaning of the problem that faces the postmaster. $\endgroup$ – nashynash Jun 20 at 10:42
  • $\begingroup$ @nashynash You have defined $f(t)$ as a function of one (presumably real) number. You now tell us that it is a distribution. Now you are asking about multiple samples from that distribution, but you still want $\ln(f(t))$. Do you define $f(t)=f({t_1,t_2,\cdots,t_n})=f(t_1)f(t_2)\cdots f(t_n)$, or something? $\endgroup$ – Wouter Jun 20 at 11:10

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