Faculty Selection Cut-Off

Relevant wiki: Continuous Probability Distributions - Uniform Distribution

For ease of typing, we will rescale the range of score to $[0, 1]$ by dividing the score of each candidate by $100$ .

Let the cut-off score of the first candidate be $t$ . Also, let the random variable $Z_n$ denote the score of the selected candidate (from $n$ candidates) and $X_1$ denote the score of the first candidate. Then we can write down $Z_n$ in terms of $Z_{n-1}$ as follows: $Z_n= 1(X_1 \geq t)X_1 + 1(X_1 < t)Z_{n-1}.$ Taking expectations of both sides and recalling that $X_1 \perp Z_{n-1}$ , we have $\mathbb{E}(Z_n)=\mathbb{E}(X_1 1(X_1\geq t))+ \mathbb{P}(X_1<t)\mathbb{E}(Z_{n-1}) = \int_{t}^{1} x dx + t\mathbb{E}(Z_{n-1})=\frac{1}{2}(1-t^2)+t\mathbb{E}(Z_{n-1}).$ The optimal threshold $t=M^*_n$ is obtained by setting the derivative of $\mathbb{E}(Z_n)$ w.r.t. $t$ to $0$ , yielding $M^*_n= \mathbb{E}(Z_{n-1}).$ Thus, we have the following recursion for $M^*_n$ : $M^*_n= \frac{1}{2}\big(1+(M^*_{n-1})^2\big),$ with $M^*_1=0$ . Define $\epsilon_n=(1-M^*_n), n\geq 1$ . This yields the following recursion in $\epsilon_n$ : $\epsilon_{n}=\epsilon_{n-1}-\frac{1}{2}\epsilon_{n-1}^2, \hspace{20pt}(1)$ with $\epsilon_1=1$ . It is easy to show that $\epsilon_n \searrow 0$ . Hence, we have $\frac{\epsilon_n}{\epsilon_{n-1}}=1-\frac{1}{2}\epsilon_{n-1} \stackrel{n\to \infty}{\longrightarrow} 1. \hspace{20pt}(2)$ From Eqn. (1), we have $\forall k \geq 2$ : $\frac{\epsilon_k-\epsilon_{k-1}}{\epsilon_k \epsilon_{k-1}}=-\frac{1}{2}\frac{\epsilon_{k-1}}{\epsilon_k}.$ Summing up the above for $k=2$ to $n$ and dividing both sides by $n$ , we have $\frac{1}{n}-\frac{1}{n\epsilon_n}=-\frac{1}{2}\bigg(\frac{1}{n}\sum_{k=2}^{n}\frac{\epsilon_{k-1}}{\epsilon_k}\bigg).$ Taking limit as $n\to \infty$ and using the Cesaro mean theorem in conjunction with (2), we have

$\lim_{n \to \infty}n(1-M^*_n)=2.$

The answer finally follows by scaling back the limit, i.e., multiplying both sides by $100$ .