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Suppose $X_1,\ldots,X_n$ are a random sample of $U(-\theta,\theta)$ with $\theta>0$. How can find $$ E\left(|X_1|\ \big|\max |X_i| \right) $$ that $1\leq i\leq n$?

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A nearly identical question has been asked on stats.SE and tagged as homework there. The stats.SE question is more restrictive in asking that Basu's theorem be used in the computation. –  Dilip Sarwate Dec 19 '12 at 13:40

2 Answers 2

Idea: Find a $\sigma(\max |X_i|)$-measurable random variable $Y$ such that

$$\int_F |X_1| \, d\mathbb{P} = \int_F Y \, d\mathbb{P} \qquad (1)$$

for all $F:=[\max|X_i| \leq c]$, $c \in [0,\theta]$. Then $Y=\mathbb{E}(|X_1| \, |\max |X_i|)$ holds.

Let $c \in [0,\theta]$ We have

$$\int_F |X_1| \, d\mathbb{P} = \prod_{j=2}^n \underbrace{\int 1_{[0,c]}(|X_j|) \, d\mathbb{P}}_{=\frac{2c}{2\theta}} \cdot \underbrace{\int |X_1| \cdot 1_{[0,c]}(|X_1|) \, d\mathbb{P}}_{\frac{1}{2\theta} \int_{|x| \leq c} |x| \, dx = \frac{c^2}{2} \cdot \frac{1}{\theta}} = \frac{c^{n+1}}{\theta^n} \cdot \frac{1}{2} \qquad (2)$$

where we used $|X_j| \sim U(0,\theta)$ and the independence of the random variables. On the other hand we know that

$$\mathbb{P} \left[ \max |X_i| \leq c \right] = \prod_{j=1}^n \mathbb{P}[|X_j| \leq c] = \left( \frac{c}{\theta}\right)^n = \frac{n}{\theta^n} \cdot \int_0^c x^{n-1} \, dx \\ \Rightarrow \max |X_i| \sim \frac{n}{\theta^n} \cdot x^{n-1} \cdot 1_{[0,\theta]}(x) \, dx$$

Thus

$$\int_F \max(|X_i|) \, d\mathbb{P} = \int \max(|X_i|) \cdot 1_{[0,c]}(\max |X_i|) \, d\mathbb{P} \\ = \frac{n}{\theta^n} \int_{0}^c x \cdot x^{n-1} \, dx = \frac{n}{\theta^n} \cdot \frac{c^{n+1}}{n+1} \qquad (3)$$

If we compare (2) and (3) we see that

$$Y:= \frac{1}{2} \frac{n+1}{n} \cdot \max(|X_i|)$$

fulfills (1).

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First, we simplify a little by defining $Y_i=|X_i|$, with $Y_i \sim U(0,\theta)$ and we get the equivalent problem of finding $E\left[Y_1| \max(Y_i)\right]$

Lets call $Z=\max(Y_i)$ and let $k$ be the index that attains the maximum ($Y_k = \max(Y_i)$). By the symmetry of the problem, $k$ is a discrete random variable, uniform in $1, 2 \cdots n$

Then

$$E[Y_1 | Z, k ] = \left\{ \begin{array}{ll} Z, & \hbox{if }k = 1, \\ Z/2, & \hbox{if }k\ne 1. \end{array} \right. $$

where the first is rather obvious, not so much the second: one should see (or prove) that conditioning on the event that the maximum value was $Z$ and that this was not attained by $Y_1$, only tells us, regarding $Y_1$, that its value is below $Z$; i.e. that the conditional density of $Y_1$ is uniform between $0$ and $Z$, and hence the expectation is $Z/2$.

But, by tower property, $$E[Y_1 | Z ] = E [E [Y_1 | Z,k]] = \frac{Z}{n} + \frac{Z}{2}\frac{n-1}{n}= \frac{Z}{2} \frac{n+1}{n}$$

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