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I have a black box called $F(t)$ ($~$($P~(X\le t)~$, $X$ is random variable) with me where I don't have any information on the exact expression of $F(t)$. But if I supply a $t\ge 0$ I will get a value of $F(t)$ from the black box as output. I want to calculate $E[X]$. I want to fit $F(t)$ values against a polynomial of the form $\sum a_{i}t^{i}$ and then integrate over a suitable range to get the expectation. Is this a good approach to calculate $E[X]$? Also, how should I choose $t$ values. Suppose I want to choose hundred $t$ values. Should I choose them as equally-spaced.

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    Presumably $F(0) = 0$ because if not, getting a good estimate of $E[X]$ might be difficult. Also, you say "Suppose I want to choose hundred $t$ values. Should I choose them as equally-spaced?" You might want to use an _adaptive_ strategy. No point in deciding on the $100$ values ahead of time and asking for, say, $F(0)$, $F(1), \ldots, F(99)$ and getting response $0$ in all cases because $X$ takes on values only in $[200, 800]$, e.g. a GRE score. So, use the first few calls to the black box to learn a little about $F(t)$, and then decide on the strategy.2011-11-18

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It seems to me that this is just a standard quadrature problem and there's no need to invoke any specific approximations. You have

$\int_{-\infty}^\infty xp(x)\mathrm dx=\lim_{L\to\infty}\left(\left[xF(x)\right]_{-L}^L-\int_{-L}^L F(x)\mathrm dx\right)\;,$

and you want to approximate that last integral, given the ability to sample values of $F$. You can apply any quadrature methods, e.g. Gaussian quadrature, that seem suitable for the problem; this will implicity approximate $F$ by polynomials, but in a particularly efficient way. You may be able to use any knowledge about the structure of $F$ that you have in choosing the quadrature method or perhaps transforming the integral before evaluating it.

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    joriki: Since $LF(-L)\to0$ and $L(1-F(L))\to0$ (see [here](http://math.stackexchange.com/q/82820/6179)), one may replace $[xF(x)]_{-L}^L$ by $L$ in the RHS of the displayed equation. This might be what @Dilip had in mind.2011-11-27