# Is an average of power-law curves a power-law curve?

If I look at an exponential function, $p(t) = e^{-\mu t}$ where the parameter $\mu$ varies over a gamma distribution given by the density function

$f(\mu) = \frac{1}{\Gamma(a)b^a}\mu^{a-1}e^{-\mu/b}$ (parameters $a, b$ and $\Gamma$ the gamma function)

Then I can talk about the "average" of all exponentials with respect to $\mu$ as the integral

$\int_{0}^{\infty}p(t)f(\mu)d\mu$

I'm reading a paper where they present, without proof, the claim that this has closed form

$(1+bt)^{-a}$

That is, the "average" of an exponential curve is a power-law curve. (I'm slightly simplifying the actual claim, so I'm hoping I didn't make a mistake)

On one hand, I'm interested to know what results they used in probability calculus to get to this closed form (it is probably pretty basic, but I just never learned this stuff). More importantly, I'm interested in whether the same technique applies to a family of power-law curves. That is if I instead write $p(t) = t^{-\mu}$, can I easily derive a closed form, for this, and if so, is it a power-law?

As a side note, if the distribution is instead uniform, the answer is (sort of) yes: the integral is trivial and the closed form quickly converges to a power-law curve, as it does in the case of an exponential.

• edited my typo of "gamma distribution" to "gaussian distribution" – JeremyKun Oct 13 '12 at 4:29
• Your distribution is a gamma distribution, not a Gaussian distribution. You should roll back your edit. – Dilip Sarwate Oct 13 '12 at 14:08

## 2 Answers

Hey man so here's one approach for the first part. Suppose you believe that the Gamma distribution is indeed a probability distribution, i.e. it integrates to 1. This tells you that $$\int_0^\infty x^{a-1}e^{-\frac{x}{b}}dx =\frac{1}{\Gamma(a) b^a}$$ Now when we try to evaluate your integral, up to fiddling with constants we run into the same integral: $$\int_0^\infty \frac{1}{\Gamma(a)b^a} x^{a-1} e^{\frac{-x}{b}} e^{-xt} dx$$$$= \frac{1}{\Gamma(a)b^a} \int_0^\infty x^{a-1} e^{-x(\frac{1}{b} + t)} dx$$Now just note that $\frac{1}{b} + t = (\frac{b}{tb + 1})^{-1}$, so plugging into the first formula, we our integral equals $$\frac{1}{\Gamma(a)b^a} \cdot (\Gamma(a) (\frac{b}{(tb+1)})^a ) = (tb + 1)^{-a}$$as desired.

You can attack the second part with the same method. First suppose $t > 0$, then we note first that $t^{-x} = (e^{\log t})^{-x} = e^{-x \log t}$, so when we evaluate $$\frac{1}{\Gamma(a)b^a} \int_0^\infty t^{-x} x^{a-1} e^{-\frac{x}{b}} dx = \frac{1}{\Gamma(a)b^a} \int_0^\infty x^{a-1} e^{-x(\frac{1}{b} + \log t)} dx$$So it's the same problem as before except with $\log t$ instead of $t$, and thus we get this equals $$(b \log t + 1)^{-a} \quad \quad t > 0$$(At the corner case $t = 0$, we just get 1)

This follows directly from the definition of the Gamma function as an integral. \begin{align} \int_0^{\infty}\frac{1}{\Gamma(a)b^a}\mu^{a-1}\exp(-\mu/b)\exp(-\mu t) \,\mathrm d\mu &= \int_0^{\infty}\frac{1}{\Gamma(a)b^a}\mu^{a-1}\exp(-\mu(b^{-1}+ t)) \,\mathrm d\mu\\ &= \int_0^{\infty}\frac{1}{\Gamma(a)b^a}\frac{x^{a-1}}{(b^{-1} + t)^{a}}\exp(-x) \,\mathrm dx\\ &= \frac{1}{b^a(b^{-1} + t)^a}\int_0^{\infty}\frac{x^{a-1}\exp(-x)}{\Gamma(a)} \,\mathrm dx\\ &= \frac{1}{(1+bt)^a} \end{align}

More generally, for any probability density function $f(x)$, the value of the integral $\int e^{-\lambda x}f(x)\,\mathrm dx$ is the moment-generating function of the random variable evaluated at $-\lambda$.