[proofplan] Subtract the forecast and divide by the conditional standard deviation. The corrected statement assumes the conditional forecast error is normal with mean zero and variance $v_h$, so the standardized error is conditionally standard normal. The displayed interval is exactly the event that this standardized error lies between the symmetric standard-normal quantiles. [/proofplan] [step:Standardize the conditional forecast error] Define \begin{align*} E_{n,h}=X_{n+h}-\widehat{X}_{n+h\mid n}. \end{align*} By assumption, conditional on $\mathcal{F}_n$, \begin{align*} E_{n,h}\sim N(0,v_h). \end{align*} Since $v_h>0$, the standardized error \begin{align*} Z_{n,h}=\frac{E_{n,h}}{\sqrt{v_h}} \end{align*} satisfies \begin{align*} Z_{n,h}\mid\mathcal{F}_n\sim N(0,1). \end{align*} [guided] The zero conditional mean is the centering condition that makes the interval centered at $\widehat{X}_{n+h\mid n}$. With \begin{align*} E_{n,h}=X_{n+h}-\widehat{X}_{n+h\mid n} \end{align*} and \begin{align*} E_{n,h}\mid\mathcal{F}_n\sim N(0,v_h), \end{align*} dividing by the positive conditional standard deviation $\sqrt{v_h}$ gives \begin{align*} Z_{n,h}=\frac{E_{n,h}}{\sqrt{v_h}}\mid\mathcal{F}_n\sim N(0,1). \end{align*} [/guided] [/step] [step:Translate the central normal probability back to the original scale] By the definition of $z_{1-\alpha/2}$ and symmetry of the standard normal distribution, \begin{align*} \mathbb{P}(-z_{1-\alpha/2}\le Z_{n,h}\le z_{1-\alpha/2}\mid\mathcal{F}_n)=1-\alpha. \end{align*} Substituting $Z_{n,h}=E_{n,h}/\sqrt{v_h}$ gives \begin{align*} \mathbb{P}(-z_{1-\alpha/2}\sqrt{v_h}\le E_{n,h}\le z_{1-\alpha/2}\sqrt{v_h}\mid\mathcal{F}_n)=1-\alpha. \end{align*} Finally, since $E_{n,h}=X_{n+h}-\widehat{X}_{n+h\mid n}$, adding $\widehat{X}_{n+h\mid n}$ throughout the event yields \begin{align*} \mathbb{P}\left(\widehat{X}_{n+h\mid n}-z_{1-\alpha/2}\sqrt{v_h}\le X_{n+h}\le \widehat{X}_{n+h\mid n}+z_{1-\alpha/2}\sqrt{v_h}\,\middle|\,\mathcal{F}_n\right)=1-\alpha. \end{align*} This proves the stated conditional coverage. [/step]