Maximum of uniformly distributed random variables using iterated expectations

Question

I'm working through the problems in Wasserman's 'All of Statistics'. The chapter on expectations and conditional expectations ends with a (seemingly) easy problem:

Let $Y$ be the maximum of $n$ iid uniformly distributed rvs on $(0,1)$. What is $\mathbb{E}(Y)$?

I determined this using conditional expectations but having checked against answers on the site there must be a fault in my reasoning. Can someone please tell me where my fault was?

Suppose $n$ = 2, and the r.v.'s be $X_1, X_2$. For any draw, one will be the larger and one smaller. Let us denote them by $\max(X_1, X_2) = m$ And $\min(X_1, X_2) = l$ The problem should reduce to finding $ \mathbb{E}(m)$.

By the Law of Iterated Expectations, $\mathbb{E}(m) = \mathbb{E}(\mathbb{E}(m|l))$.

Which would be $$ \int_0^1 [\int_l^1 [m.f(m|l)] \text{d}m f(l)] \text{d}l$$

$f(m|l)$ should be $\frac 1 {1 - l}$.

$f(l)$ is simply the uniform pdf so is 1.

Therefore the expression to compute should be

$$ \int_0^1 [\int_l^1 [m.\frac 1 {1 - l}] \text{d}m] \text{d}l $$

$$ = \int_0^1 [\frac {m^2} {2(1-l)}]^1_l \text{d}l $$

$$ = \int_0^1 [\frac {1+l} 2] \text{d}l $$

$$ = [\frac {l} {2}+\frac {l^2} {4} ]_0^1 $$

$$ = \frac 3 4 $$

Furthermore one can generalise to $ 1 - \frac 1 {2^n}$.

This is incorrect, as has been demonstrated by numerous other excellent answers the solution should $ \frac n {n + 1}$.

What have I gotten wrong here?

Answer 1

The probability that the maximum $Z$ of $n$ independent $U(0,1)$ random variables is no larger than $z$, $0 < z < 1$ is simply $\prod P\{X_i \leq z\} = z^n$ and so the density is $nz^{n-1}\mathbf 1_{0 < z < 1}$. From this, it is easy to verify that $$E[Z] = \int_0^1 z\cdot nz^{n-1}\,\mathrm dz = \left.\frac{n}{n+1}z^{n+1}\right|_0^1 = \frac{n}{n+1}.$$

If you want to do it the hard way by first computing $E[Z\mid W]$ first (where $W$ is the minimum of the $n$ random variables), then we need to find the conditional distribution of $Z$ given $W = w$. Let us start by first finding the joint distribution of $(Z,W)$. For $0 < w < z < 1$, we have that $$P\{z< Z < z+\Delta z, w < W < w+\Delta w\} \approx f_{Z,W}(z,w)\cdot\Delta z \Delta w.$$ Any of the $n$ $X_i$ can be $Z$ and any one of the remaining $n-1$ $X_j$ can be $W$, and the remaining $n-2$ random variables must be in the interval $[w,z]$. Thus we have $$\begin{align} f_{Z,W}(z,w) &= n(n-1)(z-w)^{n-2}, 0 < w < z < 1,\\ f_W(w) &= \int_w^1 n(n-1)(z-w)^{n-2}\,\mathrm dz\\ &= n (z-w)^{n-1}\bigr|_w^1\\ &= n\left(1-w\right)^{n-1},\\ f_{Z\mid W}(z\mid W=w) &= \frac{n-1}{(1-w)^{n-1}}(z-w)^{n-2}, ~ w < z < 1.\\ E[(Z -W) \mid W =w] &= \int_w^1 (z-w)\cdot \frac{n-1}{(1-w)^{n-1}}(z-w)^{n-2},\mathrm dz\\ &= \frac{n-1}{n}(1-w),\\ E[Z\mid W = w] &= w + \frac{n-1}{n}(1-w), \end{align}$$ so that we arrive at $E[Z\mid W] = \frac{W + n-1}{n}$. We finally get to use the law of iterated expectation to arrive at $$E[Z] = E[E[Z\mid W]] = E\left[\frac{W + n-1}{n}\right] = \frac{\frac{1}{n+1} + n-1}{n} = \frac{n}{n+1}$$ if you remember that $E[W] = \frac{1}{n+1}$. If not, work it out from the density of $W$ given above (hint: it is a Beta density).

Moral: Don't wrap your left arm twice around your head in order to scratch your right ear, and don't try to find the expected value of the maximum via the law of iterated expectation and conditioning on the value of the minimum