The largest of $N$ random numbers over a uniform distribution? So I read somewhere than if you have $N$ numbers picked independently from a uniform distribution, say $[0,1]$, the greatest number has an expected value of $\frac{N}{N+1}$. So if you have 2 numbers the greatest has expected value $2/3$. The smallest has expected value of $1/3$. The expected values are uniformly distributed. This makes sense, but is there a clear/intuitive proof of this? Thanks :)
 A: Let $U_1 , ... , U_N$ be the $N$ random variables from the uniform distribution. Let $X$ be their maximum. We will in fact compute the distribution of $X$, that is
$$F(t):= \mathbb P(X < t) .$$
The maximum of $N$ numbers is less than $t$ if and only if all $N$ of them are less than $t$. Therefore
$$\mathbb P(X < t) = \mathbb P(U_1 < t , ... , U_N < t).$$
Since the $U_j$'s are independent, this is the same as
$$\mathbb P(U_1 < t) ... \mathbb P(U_N < t).$$
If $t \in [0,1]$, then since $U_j$'s are uniformly distributed we get
$$F(t) = t...t = t^N.$$
Hence the density of $X$ is $F'(t) = N t^{N-1}$ on $[0,1]$. So, the mean of $X$ is
$$\int_0^1 t\cdot Nt^{n-1} dt = \frac{N}{N+1}.$$
EDIT: The general case of the expectation of the $k$th largest of $N$ random variables follows from a similar argument. Here one uses that if $Y$ is the $k$th largest of $N$ independent random variables, then $Y$ is less than $t$ if and only if $k$ of the random variables are less than $t$ and $N-k$ are greater than or equal to $t$ (and there are $N$ choose $k$ ways for this to happen).
A: There is an easier way to calculate the $E^N_1(U_k)$, where $U_k$ is the $k$th largest number out of $N$ numbers independently and uniformly distributed between $0$ and $1$. Assuming we've already proved that $E^N_1(U_N)=N/(N+1)$ (see user15464's proof). Now let's look at $E^N_1(U_{N-1})$:
Given that the largest number is $t^\prime$, the expectation value of the second largest number out of $N$ uniform random numbers in $(0, 1)$ is then equal to the expectation value of the largest number out of $N-1$ numbers uniformly distributed between $0$ and $t^\prime$:
$$
E^N_1(U_{N-1})\vert_{U_N = t^\prime}= E^{N-1}_{t\prime}(U_{N-1})
$$
And $E^{N-1}_{t\prime}(U_{N-1})$ obviously just scales with $t^\prime$:
$$
E^{N-1}_{t^\prime}(U_{N-1})=t^\prime E^{N-1}_1(U_{N-1})=t^\prime (N-1)/N
$$
Now to remove the condition on $t^\prime$:
$$
E^N_1(U_{N-1}) = E(E^N_1(U_{N-1})\vert_{U_N = t^\prime}) = E(t^\prime (N-1)/N) = E(t^\prime)(N-1)/N
$$
Recall that $t^\prime$ is the largest number out of $N$ numbers in $(0, 1)$, so $E(t^\prime)$ is just $E^N_1(U_N)$. There you have:
$$
E^N_1(U_{N-1}) =  \frac{N-1}{N}E^N_1(U_N)
$$
Now you can deduce the general formula for $E^N_1(U_k)$:
$$
E^N_1(U_k) = \frac{k}{k+1}\frac{k+1}{k+2}\ldots\frac{N}{N+1} = \frac{k}{N+1}
$$
Which also holds for $k=N$.
