I have an individual selected for a job. He was competing against 10 candidates. All individuals are independently drawn from uniform distribution U(0,1). Him being chosen means he is better than the other. Normally, expected ability would be -
$$ E(a|chosen) = \frac{\int_{b=0}^1 \int_b^1 a ~ da db}{\int_{b=0}^1 \int_b^1 da db} = \frac{2}{3} $$
Where, a denotes his own ability and b being ability of other Agents.
But I also want to incorporate number of individuals he was competing against because that would boost his confidence further.
How can I incorporate that?
My hunch is tedious and I don't know if it is correct or not -
There is ranking of individuals a>b>c>d>.. >j and expected ability is individual being better than each. Am I right? Is there a simpler way?
$$\frac{\int_{b=0}^1 \int_b^1 a ~ da db}{\int_{b=0}^1 \int_b^1 da db} *... \frac{\int_{i=0}^1 \int_0^i i ~ dj di}{\int_{i=0}^1 \int_0^i dj di} $$
Yes. Rather than multiplying, the method is to nest the integrals.
You have that:
$\qquad\begin{align}\mathsf E(X_{1}\mid X_{1}{\,=\,}\max\{X_1,X_2\}) &=\dfrac{\int_0^1\int_0^{x_1} x_1\,\mathrm d x_2\,\mathrm d x_1}{\int_0^1\int_0^{x_1} \mathrm d x_2\,\mathrm d x_1}\\[1ex]&=\dfrac 23\end{align}$
Similarly, the expected ability of a particular individual given that it is greater than all others among three is:
$\qquad\begin{align}\mathsf E(X_{1}\mid X_{1}{\,=\,}\max\{X_1,X_2,X_3\}) &=\dfrac{\int_0^1\int_0^{x_1}\int_0^{x_1} x_1\,\mathrm d x_3\,\mathrm d x_2\,\mathrm d x_1}{\int_0^1\int_0^{x_1}\int_0^{x_1} \,\mathrm d x_3\,\mathrm d x_2\,\mathrm d x_1}\\[1ex]&=\dfrac {\int_0^1 x^3\,\mathrm d x}{\int_0^1 x^2\,\mathrm d x}\\[1ex]&= \dfrac{3}{4}\end{align}$
And generally, for any integer $n\geqslant 1$, the expected ability of a particular individual given that it is greater than all others among the $n$:
$\qquad\begin{align}\mathsf E(X_{1}\mid X_{1}{\,=\,}\max_{1\leqslant k\leqslant n}\{X_k\}) &=\dfrac{\int_0^1 x^n\,\mathrm d x}{\int_0^1 x^{n-1}\,\mathrm d x}\\[1ex]&= \dfrac{n}{n+1}\end{align}$