# Average ability conditioning on having accepted an offer

There is a continuum of workers between 0 and 1. These have ability $$\alpha\sim U[0,2]$$. A firm offers them a salary $$v$$ and has profits

$$\pi = (\rho \alpha-v) n(v)$$

where $$n(v)$$ is the fraction of workers accepting the job at $$v$$ and $$\rho>0$$ is a productivity parameter.

Workers accept the offer if the salary is higher then the outside option $$2\alpha^2$$. Compute the expected profits of the firm.

I think share of people accepting the offer is (by uniformity of $$\alpha$$):

$$n(v) = \mathbb{P}[2\alpha^2 \leq v] = \mathbb{P}[\alpha \leq \sqrt{v/2}] = \sqrt{v/2}.$$

Expected profits are conditioned on acceptance and so is expected ability, hence:

$$\mathbb{E}[\alpha \vert \alpha \leq \sqrt{v/2}] = \int_0^\sqrt{v/2} \frac{\alpha f(\alpha)}{\mathbb{P}[\alpha \leq \sqrt{v/2}]}d\alpha = \frac{1}{4n(v)}\left[\alpha^2\right]_0^\sqrt{v/2}$$

by definition of conditional probability.

I would like to double check that I am correct here and that average ability given acceptance is not simply the midpoint between $$\sqrt{v/2}$$ and 0 (i.e. ability is uniformly distributed between the proposed salary and 0). I think that this is incorrect, since acceptance is not uniformly distributed, conditioning on acceptance skews the distribution of ability observed after acceptance.