Let $X_1,\dots,X_n \sim U([0,1])$ be $n$ i.i.d. random variables uniformly distributed over $[0,1]$.

Define for all $i = 1,\dots,n$, $Y_i = X_i/\sum_{i = 1}^n X_i$. Does this correspond to some well-known family of distributions? What is the expected value of $\max Y_i$ or of $\min Y_i$?


Since the $Y_i$ all have the same normalization, $$ \dfrac{\max Y_i}{\min Y_i} = \dfrac{\max X_i}{\min X_i}$$ Note that for $0 < t < s < 1$, $$ \mathbb P(t < \min X_i, \max X_i \le s) = \mathbb P({\text all }\ X_i \in (t,s]) = (s-t)^n$$ Thus the joint density of $(\min X_i, \max X_i)$ is $f(x,y) = n (n-1) (y-x)^{n-2}$ for $0 < x < y < 1$. However, this has nonzero limits as $x \to 0$, and $\mathbb E[(\max X_i)/(\min X_i)] = \infty$ (the integral diverges).

EDIT: For the revised question, condition on $\max X_i$.

The density of $\max X_i$ is $n x^{n-1}$. Given the maximum is $X_j = t$, the other $X_i$ are independent uniform on $[0,t]$. Thus the conditional expectation of $\max Y_i$ given $\max X_i = t$ is the expected value of $1/(1+S)$ where $S$ is the sum of $n-1$ independent $U[0,1]$ random variables. Now $$\mathbb E \left[ \dfrac{1}{1+S} \right] = \int_0^\infty \exp(-t) M_S(-t)\; dt$$ where $M_S(t) = \mathbb E[\exp(St)]$ is the moment generating function of $S$. Now $$M_S(t) = \left(\dfrac{e^t-1}{t}\right)^{n-1}$$so that $$ \mathbb E \left[ \dfrac{1}{1+S} \right] = \int_0^\infty \exp(-t) \left(\dfrac{1-e^{-t}}{t}\right)^{n-1}\; dt $$ I don't know if there is a closed form in general.

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  • $\begingroup$ Sorry! I was not very clear in my writing. I meant what is the expected value of max Y_i OR min Y_i $\endgroup$ – Spark Jul 16 '15 at 19:21
  • $\begingroup$ In any case, I'm more interested whether this is a canonical family of vector distributions, since it is essentially sampling a probability distribution over n elements at random from the space of all distributions. $\endgroup$ – Spark Jul 16 '15 at 19:23
  • $\begingroup$ I see, thank you very much! This has been a lot of help (though it mostly tells me that this distribution family is not very pleasant...). $\endgroup$ – Spark Jul 16 '15 at 20:03

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