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

Question

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.

Answer 1

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.

Answer 2

For completeness here is the solution (adding in the missing steps that @manst wanted me to think about).

\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*}