Finding the mean of an unknown distribution.

Question

I solved the integral and it's theta, but i'm not sure how to get the mean of the distribution because i don't know the kind of distribution i'm dealing with.

Any ideas? i'd really appreciate your help. =)

Answer 1

Suppose $X$ is a random variable with density $x\mapsto\frac1\theta x^{(1-\theta)/\theta}\mathbf{1}_{0<x<1}$ for some $\theta>0$. Then the mean of $X$ (or of the distribution) is $$ {\rm E}[X]=\int_{\mathbb{R}}x\frac1\theta x^{(1-\theta)/\theta}\mathbf{1}_{0<x<1}\,\mathrm dx=\frac1\theta\int_0^1 x^{1/\theta}\,\mathrm dx. $$

Answer 2

The expectation of a random variable, $X$, is defined as $$\int xd\mu$$ where $\mu$ is the probability measure that $X$ follows. In other words the expectation of this distribution will be simply $$\int x\left\{\frac{1}{\theta}x^{(1-\theta)/\theta}\right\}dx$$