Ratio between highest number among $n$ and $n+1$ samples When $n$ numbers are drawn independently and uniformly from $[0,1]$, the expected value of the highest number is $A=n/(n+1)$. When $n+1$ numbers are drawn under the same condition, the expected value of the highest number is $B=(n+1)/(n+2)$. The ratio is $\frac{A}{B}=\frac{n^2+2n}{n^2+2n+1}$.
For fixed $n$ and any distribution $F$ with support $[0,1]$, let $A(F)$ denote the expected value of the highest of $n$ numbers drawn independently from this distribution, and $B(F)$ denote the expected value of the highest of $n+1$ numbers drawn independently from this distribution. What is the infimum of $\frac{A(F)}{B(F)}$ over all distributions $F$ with support $[0,1]$? 
 A: Fix the distribution $F$ on $[0,1]$ other than the constant $0$ distribution. Let $E_n$ denote the expectation of the maximum of $n$ samples of $F$.
Let $\cal X = X_1,\ldots,X_{n+1}$ be $n+1$ samples from $F$. We claim that
$$
n\max(X_1,\ldots,X_{n+1}) \leq \sum_{i=1}^{n+1} \max(\cal X \setminus X_i).
$$
Indeed, suppose that the maximum is $X_j$, so that the left-hand side is $nX_j$. On the right-hand side we have $nX_j + \max(\cal X \setminus X_j) \geq nX_j$.
Taking expectations, we obtain
$$
n E_{n+1} \leq (n+1) E_n.
$$
In other words,
$$
\frac{E_n}{E_{n+1}} \geq \frac{n}{n+1}.
$$
Another way of looking at this is
$$
\frac{E_n}{n} \geq \frac{E_{n+1}}{n+1}.
$$

It is easy to construct examples in which these inequalities are almost tight. Let $F$ be the distribution which is $0$ with probability $1-\epsilon$ and $1$ with probability $\epsilon$. Then
$$ E_n = 1-(1-\epsilon)^n = n\epsilon + O(\epsilon^2). $$
Therefore
$$ \frac{E_n}{E_{n+1}} = \frac{n\epsilon + O(\epsilon^2)}{(n+1)\epsilon + O(\epsilon^2)} = \frac{n}{n+1} + O(\epsilon). $$
(Here the hidden constant depends on $n$.)
Hence $E_n/E_{n+1}$ can get as close to $n/(n+1)$ as we wish. Can we get all the way to $E_n/E_{n+1}$? If we could, then the first inequality above will have to be tight with probability $1$, which only happens if $\cal X \setminus \max \cal X$ is all $0$ with probability $1$. This, in turn, can only happen if $F$ is the constant $0$ distribution, which we have forbidden. We conclude that our inequalities are all strict.
A: Let us suppose $X$ follows some continuous probability distribution on $[0,1]$.  Then $$F_{X_{(n)}}(x) = F_X(x)^n,$$ and we note $$\operatorname{E}[X_{(n)}] = \int_{x=0}^1 S_{X_{(n)}}(x) \, dx = 1 - \int_{x=0}^1 F_X(x)^n \, dx,$$ where we require that $F_X$ is a monotone function such that $0 = F_X(0) \le F_X(x) \le F_X(1) = 1$ for all $x \in [0,1]$.  If we consider the class of functions of the form $F_X(x) = x^a$ for $a \in (0,\infty)$, we find that the above expectation is $$\operatorname{E}[X_{(n)}] = 1 - (1+a n)^{-1}$$ and the ratio of expectations is $$g_n(a) = \frac{\operatorname{E}[X_{(n)}]}{\operatorname{E}[X_{(n+1)}]} = \frac{1 - (1+a n)^{-1}}{1 - (1 + a(n+1))^{-1}} = \frac{n(an + a + 1)}{(n+1)(an + 1)}.$$  The derivative with respect to $a$ is $$g'_n(a) = \frac{n}{(n+1)(an+1)^2} > 0,$$ thus the function is strictly increasing for all $a > 0$ and is bounded below by the value at $a = 0$, which is $n/(1+n)$ for a fixed $n$.
Although we have not investigated the class of all possible distribution functions for $X$, the above result suggests that we might find it challenging to do better than this.  
