= Solution
{c}
Put $Q=\sum_iX_i^2$. For $\theta>0$, the <log-likelihood> is $\ell(\theta)=\text{constant}-(n/2)\log\theta-Q/(2\theta)$. Its derivative is $-n/(2\theta)+Q/(2\theta^2)$, changing from positive to negative at $Q/n$. Thus
$$
\boxed{\widehat\theta=\frac1n\sum_iX_i^2.}
$$
The known mean is zero, so no centring at the sample mean is needed. Taking the expected negative second derivative gives $I_n(\theta)=n/(2\theta^2)$. The large-sample variance is consequently $2\theta^2/n$. With $z_{1-\alpha/2}$ the indicated standard-normal <quantile>, a <Wald confidence interval> is
$$
\boxed{\widehat\theta\ \pm\ z_{1-\alpha/2}\widehat\theta\sqrt{2/n}.}
$$
The interval may be restricted to the parameter space $\theta>0$. The parameter is the variance itself, so the standard error is proportional to $\widehat\theta$, not to its square root.
Back to article page
