I have come across a representation of a non-standard normal distributed variable square. It is clear for me that assuming
$Z_j \approx N\left ( \theta_j, \frac{\sigma^2}{n} \right )$
we can write $Z_j$ as
$Z_j = \theta_j + \frac{\sigma}{\sqrt n}\epsilon$,
where $\epsilon \sim N(0,1)$.
The point is how do I represent $Z_j^2$ with the use of $\epsilon \sim N(0,1)$?
In All of Nonparametric Statistics p. 185-186 is is stated that for such $Z_j$ we have:
$Z_j^2 = \left ( \theta_j^2 + \frac{\sigma}{\sqrt n}\epsilon \right )^2.$
It is not clear for me how to reach this result and I kindly ask for some theoretical explanation or references to them.
It's a misprint. Instead of
$$Z_j^2 = \left( \theta_j^{\color{red}{2}} + \frac{\sigma}{\sqrt{n}} \epsilon \right)^2 \tag{1}$$
it should read
$$Z_j^2 = \left( \theta_j + \frac{\sigma}{\sqrt{n}} \epsilon \right)^2.$$
One way to see that $(1)$ cannot hold true goes as follows: It is well-known that for any random variable $X$ we have
$$\text{var} (X) = \mathbb{E}(X^2)- (\mathbb{E}X)^2.$$
For $Z_j \sim N(\theta_j, \frac{\sigma^2}{n})$ this implies
$$\mathbb{E}(Z_j^2) = \text{var}(Z_j) + (\mathbb{E}Z_j)^2 = \frac{\sigma^2}{n}+ \theta_j^2. \tag{2}$$
On the other hand, by expanding the square, we find
$$\mathbb{E} \left( \left[ \theta_j^2 + \frac{\sigma}{\sqrt{n}} \epsilon \right]^2 \right) = \theta_j^4 + \frac{\sigma^2}{n}. \tag{3}$$
(Note that the mixed terms vanish as $\mathbb{E}\epsilon=0$.) Since the expressions at the right-hand side of $(2)$ and $(3)$ are not equal, the identity $(1)$ cannot hold true.