Conditional expectation of second moment given sum of iid variables.

Question

We have $\xi_i \geq 0$, $\forall i = \overline{1,n}$ (i.i.d. variables). Assume that $S_n = \xi_1 +...+ \xi_n$.

It is easy to show that $\mathrm{E} (\xi_1\vert S_n = 1) = \frac{1}{n}$.

Now we want to look at the second moment: $\mathrm{E}(\xi_1^2\vert S_n = 1)$.

1) Since $0\leq\xi_i\leq 1$ then $\xi_i\geq \xi_i^2,\: \forall i=\overline{1,n},$ finally, $\mathrm{E}(\xi_1^2\vert S_n = 1) \leq \mathrm{E}(\xi_1\vert S_n = 1) = \frac{1}{n}$.

2) On the other hand, by Jensen's inequality we have: $(\mathrm{E}(\xi_1\vert S_n = 1))^2\leq \mathrm{E}(\xi_1^2\vert S_n = 1),$ where$(\mathrm{E}(\xi_1\vert S_n = 1))^2 = \frac{1}{n^2}$.

The question is: how can we prove that the second moment has order of $\frac{1}{n^2}$; namely how we can get upper bound as $\frac{c}{n^2}$, where $c$ is a constant?

Answer 1

You can't prove it, because it might not be true.

Suppose $\xi_i$ take nonnegative integer values. Then the only way to have $S_n = 1$ is that one $\xi_i = 1$ while all the rest are $0$. So for such a distribution, $E(\xi_i^2 | S_n = 1) = 1/n$, and indeed $E(\xi_i^p | S_n = 1) = 1/n$ for all positive $p$.

More generally, if $P(0 < \xi_i < \epsilon) = 0$ for some $\epsilon > 0$, then $\xi_i^2 \ge \epsilon \xi_i$ a.s. so $E(\xi_1^2 | S_n) \ge \epsilon E(\xi_1 | S_n)$.