Statistics, confidence intervalI have a sample of x i where x i = ξ + η, ξ ∼ N ( 0 , σ 2 ) and η ∼ N ( a , 1 ) i.i.d. So I need to construct confidence interval with confidence level γ for σ 2 with unknown a. My attempt is, I used the following statistic: S 2 = ∑ ( x i − x ¯ ) 2 n − 1 but I get an interval which could be with negative endpoints.

Question

Statistics, confidence intervalI have a sample of       x    i   where       x    i    =  ξ  +  η,   ξ  ∼  N  (  0  ,      σ    2    ) and   η  ∼  N  (  a  ,  1  ) i.i.d. So I need to construct confidence interval with confidence level   γ for       σ    2   with unknown a. My attempt is, I used the following statistic:       S    2    =            ∑      (              x        i            −              x        ¯                    )        2                    n      −      1       but I get an interval which could be with negative endpoints.

Marcel Mccullough · Accepted Answer

Step 1At this point I am inclined to address the problem as follows, and maybe tomorrow I&#039;ll know a more explicit argument in favor of this approach.You have                              x          i                ∣        η                            ∼        N        ⁡        (        0        ,                  σ          2                )         for         i        =        1        ,        …        ,        n         and                                             x          1                ,        …        ,                  x          n                 are conditionally independent given         η        ,         and                            η                            ∼        N        ⁡        (        a        ,        1        )        .            Therefore  (      x    i    −  η  )  ∣  η  ∼  N  ⁡  (  0  ,      σ    2    )  .This expression “  “  N  ⁡  (  0  ,      σ    2    )  &#039;&#039;&#039;&#039; does not depend on   η  .. Since the conditional distribution of       x    i    −  η does not depend on   η  ,, it follows that       x    i    −  η and   η are independent, and that the marginal (or &quot;unconditional&quot;, if you like) distribution of       x    i    −  η is the same as its conditional distribution given   η  .. And       x    i    −  η  ,    i  =  1  ,  …  ,  n are not only conditionally independent given   η  ,, but are marginally independent. (Note that       x    i    ,    i  =  1  ,  …  ,  n are not marginally independent.)Step 2A well-known result then tells us that      A          σ      2        =      1          σ      2            ∑          i      =      1        n    (      x    i    −      x    ¯          )    2    ∼      χ          n      −      1        2    .(where A is defined by the equality above). (The reason why A/σ2 has this distribution has been posted here as a question and has been answered. There are several ways to do it.)So one can find numbers C,D for which the following works:                                  Pr        (        C        &amp;lt;                  χ                      n            −            1                    2                &amp;lt;        D        )        =        γ        .                                          Pr                  (          C          &amp;lt;                      A                          σ              2                                &amp;lt;          D          )                =        γ        .                                          Pr                  (                      A            D                    &amp;lt;                      σ            2                    &amp;lt;                      A            C                    )                =        γ        .

I have a sample of x_i where x_i=epsilon+η , epsilon∼N(0,sigma^2) and η∼N(a,1) i.i.d.

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