A random variable is uniformly distributed over[0,1].What is its skew?
[plagarism]
The skew of a RV X is:
$\displaystyle E\left( \frac{(x-\mu)^3}{\sigma^3}\right)=\frac{E( (x-\mu)^3)}{\sigma^3}$
where $\displaystyle \mu$ and $\displaystyle \sigma$ are the RV's mean and standard deviation.
Now for the uniform distribution on $\displaystyle [0,1]$
$\displaystyle p(x)=\begin{cases}1, & x \in [0,1]\\ 0, & \mbox{otherwise} \end{cases}$
and $\displaystyle \mu=1/2$
$\displaystyle \sigma=\sqrt{\int_0^1 (x-1/2)^2\ dx}$
(you may well know the value of the above, it's in most statistics books somewhere and engraved on the heart of everyone who has ever used Monte-Carlo methods in anger) and:
$\displaystyle E((x-1/2)^3)=\int_0^1 (x-1/2)^3\ dx$
[/plagarism]
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