Is the squared iid process iid?

Question

I wanted to know whether squared iid process is iid or not. I mean, e(t)~iid => e(t)^2~iid?? I think, the squared process is iid also by my intuition but I'm not sure.

help me!!

Answer 1

The distribution of $e^2$ is entirely determined by the distribution of $e,$ thus: $$ \Pr( e(t)^2 \le x) = \Pr(-\sqrt x \le e(t)\le \sqrt x\,) \text{ if } x\ge0 \tag 1 $$ and of course $\Pr(e(t)^2 \le x)=0$ if $x<0.$ If the right side of line $(1)$ above is the same for every value of $t,$ then the left side of line $(1)$ is the same for every value of $t,$ i.e. the squares are identically distributed.

For sets $A_1,\ldots,A_n$ of possible values of $e(t),$ and times $t_1,\ldots,t_n,$ \begin{align} & \Pr( e(t_1)^2 \in A_1\ \& \cdots \ \&\ e(t_n)^2\in A_n) \\[8pt] = {} & \Pr\big( e(t_i) \in \{ \pm\sqrt a : a\in A_i \} \text{ for } i=1,\ldots,n \big) \\[8pt] = {} & \prod_{i=1}^n \Pr\big( e(t_i) \in \{ \pm\sqrt a : a\in A_i \} \big) \\ & \quad \text{because $e(t_1),\ldots, e(t_n)$ are independent} \\[8pt] = {} & \Pr\big( e(t_1)^2 \in A_1\big) \cdots \cdots \Pr\big( e(t_n)^2 \in A_n\big). \end{align} Therefore $e(t)^2$ are independent.