<s>[INST] E(X), var(X) and std dev (X) For some reason, this question didn't show properly in my previous post.
X is given by its Distribution function:
$$
F(x) =\begin{cases}
0 &x \leq 0\\
\frac{x^2}{4} &0 < x \leq 2\\
1 &x > 2
\end{cases}
$$
I'm trying to find $E(X)$, $Var(X)$ and $Std\;dev (X)$
Thanks.
 [/INST] These calculations are under the assumption that $X$ is a continuous variable and that $F(X)$ is the CDF (and not the PDF, at which point $E[X]$ is not defined due to a diverging integral). First we calculate the PDF
$$f(x) = \frac{dF(x)}{dx} = \frac{x}{2}, \quad 0<x<2$$
and $f(x) = 0$ otherwise.
The definition of the expectational value is
\begin{aligned}
E[X] & = \int_{0}^{\infty}xf(x)dx = \\
     & = \int_{0}^{2}x\frac{x}{2}dx = \\
     & = \frac{4}{3}.
\end{aligned}
The variance is defined as
\begin{aligned}
\text{Var}[X] & = E[X^2]-E[X]^2 = \\
 & = \int_{0}^{2}x^2\frac{x^2}{2}dx - \left(\frac{4}{3}\right)^2\\
 & = \frac{16}{5} - \frac{16}{9} = \\
 & = \frac{64}{45}.
\end{aligned}
The variance is the standard deviation squared, which means that
\begin{aligned}
\sigma[X] = \sqrt{\text{Var}[X]} = \sqrt{\frac{64}{45}} = \frac{8}{3\sqrt{5}}.
\end{aligned}
</s>