# Mean of unormalized Gamma distribution

If $$X$$ follows a gamma distribution with parameters $$\alpha$$ and $$\beta$$ then:

$$p(x|\alpha,\beta) = \frac{\beta^{\alpha}x^{\alpha-1}e^{-\beta x}}{\Gamma(\alpha)}$$ with $$\Gamma(\alpha) = (\alpha - 1)!$$

Then, $$mean(x) = \frac{\alpha}{\beta}$$

But, if for $$x$$ we have this probability density function instead:

$$p(x|\alpha,\beta) = K \times \beta^{\alpha}x^{\alpha-1}e^{-\beta x}$$ with no $$\Gamma(\alpha)$$ and $$K$$ a constant value

What would be the $$mean$$ value of $$x$$ in this case ?

Best regards

Aymeric

Simply your density (that is a posterior density) is not exactly a density. The correct one is

$$p(\mathbf{x}|a,b)\propto b^a x^{a-1}e^{-xb}$$

The constant must be calculated in order to let $$p(\mathbf{x}|a,b)$$ be a nice density.

This constant $$K$$ can be calculated using Bayes Theorem (integrating the denominator of Bayes Theorem ) or simply recognizing in your posterior the kernel of a Gamma distribution, thus

$$K=\frac{1}{\Gamma(a)}$$

and your expectation is exactly what you showed, $$\frac{a}{b}$$

• Thank you, so we don't use K in the calculus of the mean ? even if I change K to be of ANY value ? Nov 23, 2020 at 8:38
• @ailauli69 ; as you stated in the title: your posterior is a "unnormalized" distribution...say it is NOT yet a density but only a function that is "similar" to a density...First normalize it than calculate the mean...take a look at the following example: $y=K\cdot e^{-\frac{(1-x)^2}{8}}$. What kind of distibution is this? Which is its expectation? Which is its variance? Nov 23, 2020 at 8:55

Note that $$\int_{0}^{\infty} \frac{\beta^{\alpha}e^{-\beta x}}{\Gamma(\alpha)} x^{\alpha-1} dx=1$$ If you use $$\int_{0}^{\infty} x^{m} e^{-kx}dx=k^{-m-1} \Gamma(m+1).$$ Next the expatation value of $$x$$ is $$E(x)=\int_{0}^{\infty} \frac{\beta^{\alpha}e^{-\beta x}}{\Gamma(\alpha)} x^{\alpha} dx=\frac{\beta^{\alpha}}{\Gamma(\alpha)}\beta^{-\alpha-1} \Gamma(\alpha+1)=\frac{\alpha}{\beta}.$$

You have a misconception about probability densities. If $$f_X(x)$$ is a density for some real-valued random variable $$X$$, then we require $$\int_{x = -\infty}^\infty f_X(x) \, dx = 1.$$ Equivalently, if we wish to avoid using indicator functions, we can write the above as an integral over the support $$\Omega$$ of $$X$$: $$\int_{x \in \Omega} f_X(x) \, dx = 1.$$ In the case of a gamma distribution, the support is $$X \in [0, \infty)$$.

As such, you are not free to choose any value of $$K$$ in the function $$K \beta^\alpha x^{\alpha-1} e^{-\beta x}$$. You are constrained because when integrated on $$[0,\infty)$$, the result must be $$1$$, otherwise it is not a density. This forces $$K = 1/\Gamma(\alpha)$$.

In fact, if we remove all of the factors that are independent of $$x$$, i.e. we consider the kernel of the gamma density $$x^{\alpha-1} e^{-\beta x},$$ the unique constant $$K$$ that makes this kernel a density function is precisely $$K = \frac{\beta^\alpha}{\Gamma(\alpha)}.$$

• So, as long as I have the Kernel of a gamma density in my pdf (proba density function) eventhough it's multiplied by a constant, I can compute the mean(x) as alpha/beta ? Nov 23, 2020 at 10:01