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If $Z_1,\dots ,Z_{\nu }$ are independent standard normal random variables and $\lambda ={\sum }_j{\mu }_j^2$, then ${\sum }_j(Z_j+{\mu }_j{)}^2$ has a noncentral chi-squared distribution with $\nu$ degrees of freedom and noncentrality $\lambda$. For $\lambda =0$ it reduces to the chi-squared distribution. Squaring a shifted signed normal Wald statistic gives the case $\nu =1$.

For a one-sided Wald test use the signed normal Wald statistic, rather than its square. The maximum-likelihood estimate is the sample mean ${\bar{Y}}_n$, with exact variance ${\sigma }^2/n$, soA sum of independent normal random variables is normal, giving ${\bar{Y}}_n\sim N(\delta ,{\sigma }^2/n)$ and hence $W_n\sim N(\sqrt{n}\delta /\sigma ,1)$. Under the null hypothesis its mean is zero:Thus its null distribution is exactly the standard normal distribution, without an asymptotic approximation. If the squared Wald statistic convention is used, $W_n^2$ has a chi-squared distribution with one degree of freedom; the signed form is needed to distinguish the two directions.