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a general two-sided 100(1 - a)% confidence interval for the mean /i when o is known: where a1a2 = a. Show that the length of

please help me with probability HW, i will rate the best answer, thank you very much. if you're not sure with your answer please do me a favor and let someone else do it cus i can only post the question once. i appreciate the help !

a general two-sided 100(1 - a)% confidence interval for the mean /i when o is known: where a1a2 = a. Show that the length of the interval, o(za, + za,)//n is minimized when a1 = a2 = a/2.
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Let us denote a-Za,b -Za2

To find which confidence interval is shorter, suppose a- Vn <Xb- νn is a (1 - a) 100% confidence interval. Then, the length of the interval is (b —а) n .

The minimization problem is
minimize L(a,b) (b -a)
Subject to 1 xdx 1 - o /2T h (a, b) = a
Using Lagrange multiplier, the conditions are,

(a, b) h (a, b) 0 да да L (a, b) h (a, b) 0..(1,2) ab ab

Now, using Fundamental Theorem of calculus,

1ea2 1 a h(a, b) a xp x-a- 2T 1 e-b2 2T да aa Ja V2T 1 exdx 2T h(a,b) bь дь a

Substituting in equations (1) and (2),

ea0.. 0.. ..(3) 2T e V2T 1 -b2 0.... ...(4) n

Adding equations (3) and (4), we get-a2 oe eb a tb -b2 e /2T V2元

ab does not make any sense. So a=-b

Thus the interval \bar{x}-z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}<\mu <\bar{x}+z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}} is shorter than

\bar{x}-z_{\frac{2\alpha }{3}}\frac{\sigma }{\sqrt{n}}<\mu <\bar{x}+z_{\frac{\alpha }{3}}\frac{\sigma }{\sqrt{n}}

Here az,b Z a1 a2 a/2
Proof is complete.

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