# Is there a meaning to the integral of $x \times f(x)$ over a range that is not infinite?

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 $$b > a$$

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

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