<p>If <span class="math-container">$\alpha \sim U$</span>, then how come there is no expectation in your profit function? The <span class="math-container">$\alpha$</span> is unknown and which <span class="math-container">$\alpha$</span>-types the firm gets depends on salary <span class="math-container">$v$</span>. This should be reflected in the profit function.</p>
<p>Next, your <span class="math-container">$n(v)$</span> seems to assume that <span class="math-container">$\alpha \sim U[0,1]$</span>, but you set <span class="math-container">$\alpha \sim U[0,2]$</span>. I assume this is a typo and I edited your question. Otherwise, you need to divide your <span class="math-container">$n(v)$</span> by two, because your density is <span class="math-container">$\frac{1}{2}$</span> instead of 1. This upperbound cancels out anyway.</p>
<p>For any uniformly distributed <span class="math-container">$\alpha&lt;\widehat v$</span>, you have
<span class="math-container">$$\mathbb{E}[\alpha \vert \alpha \leq \widehat v] = \int_0^{\widehat v}  \alpha \frac{1}{\widehat v}d\alpha = \left[\frac{1}{2\widehat v}\alpha^2\right]_0^{\widehat v}= \frac{\widehat v}{2} $$</span>
And in your case, it's <span class="math-container">${\widehat v}=\sqrt{v/2} \Rightarrow \mathbb{E}[\alpha \vert \alpha \leq \sqrt{v/2}] = \sqrt{v/8}$</span>. If <span class="math-container">$\alpha \sim U[0,x]$</span> then the density is <span class="math-container">$1/x$</span>, but you also account for <span class="math-container">$\alpha&lt;\widehat v$</span> by dividing by <span class="math-container">$\widehat v/x$</span> such that <span class="math-container">$x$</span> cancels out.</p>
<p>Hence, your expected profit function is
<span class="math-container">$$\mathbb{E}[\pi(v)] = (\rho \sqrt{v/8}-v)\sqrt{v/2}.$$</span></p>
<p>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.</p>
<p>If you replace your <span class="math-container">$n(v)$</span> by <span class="math-container">$n(v)/2$</span>, taking account for the upperbound 2, you would get the same.</p>

CC BY-SA 4.0
2020-11-16T13:19:07.967Z
2020-11-16T17:43:18.897Z
2020-11-16T17:43:18.897Z
Bayesian
<p>There is a continuum of workers between 0 and 1. These have ability <span class="math-container">$\alpha\sim U[0,2]$</span>. A firm offers them a salary <span class="math-container">$v$</span> and has profits</p>
<p><span class="math-container">$$
\pi = (\rho \alpha-v) n(v)
$$</span></p>
<p>where <span class="math-container">$n(v)$</span> is the fraction of workers accepting the job at <span class="math-container">$v$</span> and <span class="math-container">$\rho&gt;0$</span> is a productivity parameter.</p>
<p>Workers accept the offer if the salary is higher then the outside option <span class="math-container">$2\alpha^2$</span>. Compute the expected profits of the firm.</p>
<p>I think share of people accepting the offer is (by uniformity of <span class="math-container">$\alpha$</span>):</p>
<p><span class="math-container">$$n(v) = \mathbb{P}[2\alpha^2 \leq v] = \mathbb{P}[\alpha \leq \sqrt{v/2}] = \sqrt{v/2}.$$</span></p>
<p>Expected profits are conditioned on acceptance and so is expected ability, hence:</p>
<p><span class="math-container">$$
\mathbb{E}[\alpha \vert \alpha \leq \sqrt{v/2}] = \int_0^\sqrt{v/2}  \frac{\alpha f(\alpha)}{\mathbb{P}[\alpha \leq \sqrt{v/2}]}d\alpha = \frac{1}{4n(v)}\left[\alpha^2\right]_0^\sqrt{v/2} 
$$</span></p>
<p>by definition of conditional probability.</p>
<p>I would like to double check that I am correct here and that average ability given acceptance is not simply the midpoint between <span class="math-container">$\sqrt{v/2}$</span> 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.</p>

<labor-economics><probability><profits>
Average ability conditioning on having accepted an offer
Three Diag
# Average ability conditioning on having accepted an offer

(Asked by: username_0 on 2020-06-05)

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.

**Tags:** labor-economics, probability, profits


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## 1 answers

### Answer by username_1 on 2020-11-16. Score: 3 | Accepted answer

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.


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