conditional expectation with uniform distribution

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If $r$ follows a uniform distribution on $[-10,20]$, what is the conditional expectation of $E[r|r<0]$? I feel very confused dealing with the $r<0$ condition. This comes from a question where I am asked to compute $E[r|r<0]*P(r<0)$.

Thanks in advance!

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If $r$ follows a uniform distribution on $[−10;20]$, what is the conditional expectation of $E[r|r<0]$?

It is just the expected value of $r$ where it is less than $0$ , divided by the probability that it is less than $0$.

$$\begin{align}\mathsf E(r\mid r<0) &= \dfrac{\mathsf E(r~\mathbf 1_{r<0})}{\mathsf P(r<0)} \\[1ex]&= \dfrac{\int_{-10}^0 r~f(r)\operatorname d r}{\int_{-10}^0 f(r)\operatorname d x} \\[1ex]&= \int_{-10}^{0}r~f(r\mid r< 0)\operatorname d r\end{align}$$

Now, since $r$ is uniformly distributed over $[-10;20]$, then ...


PS:

This comes from a question where I am asked to compute $E[r|r<0]∗P(r<0) $

Of course, that is just $\mathsf E(r~\mathbf 1_{r<0}) = \int_{-10}^0 r~f(r)\operatorname d r$

PPS: $\mathbf 1_{r<0}$ is an Indicator function; a piecewise function that is $1$ only where indicated, and $0$ elsewhere. $$\mathbf 1_{r<0}~=~\begin{cases}1&:& r<0\\ 0& :&\text{elsewhere}\end{cases}$$