More review of probability:
2.2: Expected Values: *Very Important*
E\left[g\left(X\right)\right] = \int_{-\infty}^{\infty}g\left(x\right)f_{X}\left(x\right)\,\,\mathrm{d}x, when X is a continuous random variable.
E\left[g\left(X\right)\right] = \sum_{x}g\left(x\right)f_{X}\left(x\right), when X is a discrete random variable.
Properties of E\left[\cdot\right]:
- E\left[a\right] = a, a\in\mathbb{R}
- E\left[ag_{1}\left(X_{1}\right) + bg_{2}\left(X_{2}\right) + c\right] = aE\left[g_{1}\left(X_{1}\right)\right] + bE\left[g_{2}\left(X_{2}\right)\right] + c
- If g_{1}\left(x\right)\le g_{2}\left(x\right)\le \cdots \le g_{n}\left(x\right), \forall x, then E\left[g_{1}\left(x\right)\right]\le E\left[g_{2}\left(x\right)\right]\le\cdots\le E\left[g_{n}\left(x\right)\right].
Remark I
In measure theoretic notation the generalized expected value is, E\left[x\right] = \int_{\omega\in\Omega}X\left(\omega\right)\,\,\mathrm{d}P\,\left(\omega\right). I'm not sure exactly what this means; I'll have to come back to it later.
Remark II
What about interchanging sums and integrals?\newcommand{\?}{\stackrel{?}{=}}
\begin{eqnarray*} \int\int f\left(x,y\right)\,\,\mathrm{d}x\,\mathrm{d}y&\?&\int\int f\left(x,y\right)\,\,\mathrm{d}y\,\,\mathrm{d}x\\ \sum_{j}\sum_{k}a_{jk}&\?&\sum_{k}\sum_{j}a_{jk}\\ \sum_{j}\left(\int f_{j}\left(x\right)\,\,\mathrm{d}x\right) &\?& \int\left(\sum_{j}f_{j}\left(x\right)\right)\,\,\mathrm{d}x \end{eqnarray*}
For example, a \ge 0, f \ge 0, f_{j} \ge 0 then it works. Or, take the absolute values inside the sum (integral), and show that one of the sides is finite.
More to follow on classes of distributions.
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