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Assume $n\ge 2$ and ${\sigma }^2>0$. Up to a constant, the log-likelihood of the normal sample isFor each $s$, the square is minimized at $\mu =\bar{X}$; optimizing the remaining one-variable likelihood gives the maximum-likelihood estimatorsThey are the global maximum when the residual sum is positive, which occurs with probability one. On the exceptional all-equal sample the likelihood increases without bound as $s\downarrow 0$.

To derive the distributions and independence, write $Z_i=(X_i-\mu )/\sigma$ and perform an orthogonal change of coordinates whose first row is $n^{-1/2}(1,\dots ,1)$. The transformed components $Z_1^{\prime },\dots ,Z_n^{\prime }$ remain independent standard normal random variables, since their joint density depends only on the squared Euclidean norm. ThenThe first expression uses only $Z_1^{\prime }$, the second only the other components. ConsequentlyThis also exhibits why the maximum-likelihood variance uses divisor $n$, even though its chi-squared degrees of freedom are $n-1$.

The independent future observation satisfies $X_0-\bar{X}\sim N(0,{\sigma }^2(1+1/n))$. It is independent of the residual sum of squares, because both $X_0$ and $\bar{X}$ are independent of the residual vector. Combining this normal variable with the independent chi-squared variable givesThe event in the proposed prediction interval is equivalent to $|T|<\gamma \sqrt{(n-1)/n}$. By symmetry and continuity of Student's t-distribution, exact joint coverage $1-\alpha$ therefore requiresThis is the normal prediction interval with maximum-likelihood variance. The factor $\sqrt{n/(n-1)}$ cannot be dropped: the interval uses $\hat{\sigma }$, not the unbiased residual standard deviation. Coverage here is over repeated joint sampling of the old and new observations, not a claim of parameter-free conditional coverage for every realized sample.