Average ability conditioning on having accepted an offer

Question

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.

Answer 1

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 $\alpha<\widehat v$, 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 $\alpha<\widehat v$ 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.