How do we show that minimal expected length for confidence interval for normal with unknown mean is the following?

Question

How do we show that the $(1-\alpha)100\%$ confidence interval $(\bar{Y}-z_{\alpha-\alpha_1} \frac{\sigma}{\sqrt{n}}, \bar{Y}+z_{\alpha_1} \frac{\sigma}{\sqrt{n}})$ for the unknown normal mean $\mu$ has a minimal expected length when it is symmetric, that is, $\alpha_1 =\frac{\alpha}{2}$

Intuitively the middle has the largest percentages. But how do we formally prove it?

Answer 1

I think you have a typo: the $\sigma/\sqrt{n}$ should multiply the quantile, and not be in the subscripts.

You basically want to choose $\alpha_1$ to minimize $z_{\alpha - \alpha_1} + z_{\alpha_1}$. So now our problem is purely about the standard normal distribution.

Without loss of generality suppose $\alpha_1 < \alpha/2$. (The other case $\alpha_1 < \alpha/2$ can be handled symmetrically.) It suffices to show $z_{\alpha_1} - z_{\alpha/2} \ge z_{\alpha/2} - z_{\alpha-\alpha_1}$.

By definition, \begin{align} \int_{z_{\alpha/2}}^{z_{\alpha_1}} \phi(t) \mathop{dt} = \frac{\alpha}{2} - \alpha_1 =\int_{z_{\alpha-\alpha_1}}^{z_{\alpha/2}} \phi(t) \mathop{dt}. \end{align}

Since $0 < z_{\alpha-\alpha_1} < z_{\alpha/2} < z_{\alpha_1}$, we know $\phi$ is strictly decreasing in the range of these integrals. Specifically, we know $\phi(t) < \phi(z_{\alpha/2})$ for $t$ in the range of the first integral, and $\phi(t) > \phi(z_{\alpha/2})$ in the range of the second integral. Thus $$(z_{\alpha_1} - z_{\alpha/2})\cdot \phi(z_{\alpha/2}) > \int_{z_{\alpha/2}}^{z_{\alpha_1}} \phi(t) \mathop{dt} =\int_{z_{\alpha-\alpha_1}}^{z_{\alpha/2}} \phi(t) \mathop{dt} > (z_{\alpha/2} - z_{\alpha- \alpha_1}) \cdot \phi(z_{\alpha/2}),$$ yielding the desired inequality $z_{\alpha_1} - z_{\alpha/2} > z_{\alpha/2} - z_{\alpha-\alpha_1}$.

Answer 2

Assuming $\sigma$ is known, the width of your confidence interval is deterministic and equals $$ z_{\alpha_1}\frac\sigma{\sqrt n}+z_{\alpha-\alpha_1}\frac\sigma{\sqrt n}=\frac\sigma{\sqrt n}\left(z_{\alpha_1}+z_{\alpha-\alpha_1}\right). $$ To minimize this over all $\alpha_1$ between $0$ and $\alpha$ is the same as minimizing $$ H(t):= z_t + z_{\alpha -t}\tag1 $$ over all $t$ between $0$ and $\alpha$, where $z_t$ is a function of $t$ that satisfies $$ t=\int_{z_t}^\infty\phi(z)\,dz\tag2 $$ and $\phi$ is the standard normal density. To get the derivative of $z_t$ with respect to $t$, apply implicit differentation to (2), using the fundamental theorem of calculus and the chain rule to obtain $$ 1= -\phi(z_t)\frac{d}{dt}z_t. $$ Therefore the derivative of (1) is $$ H'(t) = -\frac1{\phi(z_t)}+\frac1{\phi(z_{\alpha-t})} $$ and equals zero where $$\phi(z_{\alpha-t})=\phi(z_t)\tag1.$$ Since $\phi$ is a decreasing function of $t$ for $t>0$, the only positive value of $t$ where (1) is true is such that $z_{\alpha-t}=z_t$. This in turn implies $\alpha-t=t$ which means $t=\alpha/2$. Argue that $H$ is minimized there by checking the second derivative.