# Showing $\mathbb{E}[T_n] = \theta \mathbb{E}_1[T_n]$ is scale equivariant?

This is question 5 is from Staudte and Sheather (1990), Robust estimation and testing.

Let $X_1,\ldots , X_n$ be i.i.d with $$F_\theta = F(\frac{x}{\theta}),\quad x>0;\theta>0.$$ Assume that $T_n = T_n(X_1,\ldots ,X_n)$ is scale equivariant. Show that $$\mathbb{E}[T_n] = \theta \mathbb{E}_1[T_n]$$.

Using $\int_a^b f(u) du = \sum_{k=a}^b$ and $X\rightarrow \theta X$, my attempt at the question is as follows. \begin{aligned} \mathbb{E}[T_n] &= \int T_n dF(X_n) \\ &= \sum_x T_n P(X) &= \frac{1}{n}[\theta X_1 +,\ldots ,+ \theta X_n] \\ &= \theta\frac{1}{n}[ X_1 +,\ldots ,+ X_n] \\ &= \theta \sum_x T_n P(X) \\ &=\theta \int T_n dF(X_n) \\ &= \theta \mathbb{E}[T_n] \end{aligned}

Is my working out correct? If not, where and why was I wrong?

I should also point out that this is not an assignment question. I need to study this book to gain some background knowledge in robust statistics.

-
I don't follow some of the equations you've written; some things appear to be missing. The key here is to understand the definitions of scale family and scale equivariant and then to connect the two. Very little is needed in the way of explicit calculations. –  cardinal Aug 20 '12 at 13:41
@cardinal, I do not fully understand the topic and that is probably why the question is unclear. But a very useful comment; I will dig deeper into the scale family –  user7045 Aug 20 '12 at 16:27
No problem. My previous comment was a bit terse. What I meant was more that things like $\int_a^b f(u) \,\mathrm du = \sum_{k=a}^b$ and $[X_1+,\ldots,+X_n]$ are mathematical notation, but with an unclear meaning (to me). (For example, in the first case, there is no actual quantity on the right-hand side.) What you are asking is clear, though, and MansT has indicated the right direction to take. If you have follow-up questions as you try to understand the concepts, please consider posting them. :) –  cardinal Aug 20 '12 at 16:31

Your random variables belong to what is called a scale family. The first step should be to show that if $X\sim F_1$ then $\theta\cdot X\sim F_\theta$.

A statistic $T_n(X_1,\ldots,X_n)$ is usually said to be scale invariant if $$T_n(\theta X_1,\ldots,\theta X_n)=T_n(X_1,\ldots,X_n),$$ i.e. if rescaling the data leaves the statistic unchanged, but sometimes it is taken to mean that $$T_n(\theta X_1,\ldots,\theta X_n)=\theta T_n(X_1,\ldots,X_n),$$ i.e. that the statistic "scales in the right way", which seems to be the case here. If I understand your notation correctly, you wish to show that this is true for its expected value as well.

Let $X\sim F_\theta$ and $Y=\frac{1}{\theta}X\sim F_1$. Then you can show that

$$\mathbb{E}_\theta(T_n)=\mathbb{E}_\theta T_n(X_1,\ldots,X_n)=\mathbb{E}_\theta\theta T_n\Big(\frac{1}{\theta}X_1,\ldots,\frac{1}{\theta}X_n\Big)\\=\mathbb{E}_1 \theta T_n(Y_1,\ldots,Y_n)=\theta\cdot \mathbb{E}_1(T_n).$$

The crucial step is to show the equality in the line break.

-
Did you notice that the title and the question use "equivariant," not "invariant"? –  whuber Aug 20 '12 at 16:19
@manst, this is quit helpful. I will attempt the first step and move on from there. –  user7045 Aug 20 '12 at 16:37
@whuber: yes, but it was tagged with invariance and I've seen the two words used as synonyms. Just wanted to avoid possible misunderstandings :) –  MånsT Aug 20 '12 at 16:47
Good point about the tag. The words aren't synonyms, as you ably discuss. –  whuber Aug 20 '12 at 17:40
\begin{align*} \mathbb{E}_\theta[T_n] &= \int \ldots \int t_n (x_1, \ldots , x_n) f_\theta(x_1), \ldots , f_\theta(x_n) \text{d}x_1, \ldots , \text{d}x_n \\ \end{align*} Since $F_\theta(x) = F(\frac{x}{\theta})$ we can write $f_\theta(x) = \frac{1}{\theta}f(\frac{x}{\theta})$, where $f(u) = F'(u)$ \begin{align*} &= \int \ldots \int t_n (x_1, \ldots , x_n) \frac{1}{\theta}f\left(\frac{x_1}{\theta}\right), \ldots , \frac{1}{\theta}f\left(\frac{x_n}{\theta}\right) \text{d}x_1, \ldots , \text{d}x_n \\ \end{align*} let $\textbf{u} = \textbf{x}/\theta$ or $\theta \textbf{u} = \textbf{x}$ \begin{align*} &= \int \ldots \int t_n (\theta u_1, \ldots , \theta u_n) \frac{1}{\theta} f(u_1), \ldots , \frac{1}{\theta}f(u_n) \theta du_1, \ldots , \theta du_n \\ &= \int \ldots \int t_n (\theta u_1, \ldots , \theta u_n) \ f(u_1), \ldots , f(u_n) du_1, \ldots , du_n \\ &= \theta \int \ldots \int t_n ( u_1, \ldots , u_n) \ f(u_1), \ldots , f(u_n) du_1, \ldots , du_n \\ &= \theta \mathbb{E}_1[T_n] \end{align*}