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I know that the expected value can be computed as :

$\mathbb{E}(X) = \int_{-\infty}^{\infty}xf(x)dx$

What if we do not do the integral over the whole range but only up to some value? Would there be any meaning? For instance, what if we have ?

$\int_{0}^{a}xf(x)dx$

where $f(x)$ is a pdf specifed over $[0,b]$ and $b > a$

Thank you in advance for your explanations.

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  • $\begingroup$ It depends, if most of the mass is in [0,a] then this could be a good enough approximation. $\endgroup$ May 27, 2022 at 9:43
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    $\begingroup$ This is a partial moment. Search this siete (I will try to edite an answer later) $\endgroup$ May 27, 2022 at 10:40
  • $\begingroup$ In the meantime, look at some of our answers that use partial expectations for examples and applications. $\endgroup$
    – whuber
    May 27, 2022 at 13:53

1 Answer 1

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Start by noting the conditional density:

$$f(x|0 \leqslant X \leqslant a) = \frac{f(x) \cdot \mathbb{I}(0 \leqslant x \leqslant a)}{\mathbb{P}(0 \leqslant X \leqslant a)}.$$

Using this conditional density, you can rewrite the latter integral as:

$$\begin{align} \int \limits_0^a x f(x) \ dx &= \int \limits_{-\infty}^{\infty} x f(x) \cdot \mathbb{I}(0 \leqslant x \leqslant a) \ dx \\[6pt] &= \mathbb{P}(0 \leqslant X \leqslant a) \times \int \limits_{-\infty}^{\infty} x f(x|0 \leqslant X \leqslant a) \ dx \\[6pt] &= \mathbb{P}(0 \leqslant X \leqslant a) \times \mathbb{E}(X|0 \leqslant X \leqslant a). \\[6pt] \end{align}$$

So, this kind of integral (taken over a strict subset of the support of the random variable) can be interpreted as the product of the conditional expectation of the random variable conditional on an occurrence in that subset, multiplied by the probability of the random variable falling in that subset. As noted in the comments, this type of quantity is sometimes called a partial moment (or more specifically in this case, a partial expectation).

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