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 $\rho&gt;0$ 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.
If $\alpha \sim U$, then how come there is no expectation in your profit function? The $\alpha$ is unknown and which $\alpha$-types the firm gets depends on salary $v$. This should be reflected in the profit function.
Next, your $n(v)$ seems to assume that $\alpha \sim U[0,1]$, but you set $\alpha \sim U[0,2]$. I assume this is a typo and I edited your question. Otherwise, you need to divide your $n(v)$ by two, because your density is $\frac{1}{2}$ instead of 1. This upperbound cancels out anyway.
For any uniformly distributed $\alpha&lt;\widehat v$, you have
$$\mathbb{E}[\alpha \vert \alpha \leq \widehat v] = \int_0^{\widehat v}  \alpha \frac{1}{\widehat v}d\alpha = \left[\frac{1}{2\widehat v}\alpha^2\right]_0^{\widehat v}= \frac{\widehat v}{2} $$
And in your case, it's ${\widehat v}=\sqrt{v/2} \Rightarrow \mathbb{E}[\alpha \vert \alpha \leq \sqrt{v/2}] = \sqrt{v/8}$. If $\alpha \sim U[0,x]$ then the density is $1/x$, but you also account for $\alpha&lt;\widehat v$ by dividing by $\widehat v/x$ such that $x$ cancels out.
Hence, your expected profit function is
$$\mathbb{E}[\pi(v)] = (\rho \sqrt{v/8}-v)\sqrt{v/2}.$$
Acceptance does not skew the distribution. All types below a cutoff accept, all others reject. Therefore, the distribution conditional on acceptance is uniform up to the cutoff.
If you replace your $n(v)$ by $n(v)/2$, taking account for the upperbound 2, you would get the same.