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Let $X_1,X_2,...,X_n$ be iid observations from a normal distribution with mean $\mu$ and variance $\sigma^2$, $\sigma^2>0$ is known and $\mu$ is an unknown real number. Let $g(\mu)=2\mu$ be the parameter of interest and

$$T(X_1,X_2,...,X_n)=X_1^2+2X_3-X_4^2$$

How d I compute $E(T|\bar{X})=2\bar{X}$?

My approach:

$$E(T|\bar{X})=E(X_1^2+2X_3-X_4^2|\bar{X})=E(X_1^2|\bar{X})+2E(X_3|\bar{X})-E(X_4^2|\bar{X})$$

What do I do from here?

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  • $\begingroup$ $X_i\mid \overline X$ has the same normal distribution for each $i$. $\endgroup$ – StubbornAtom May 6 at 8:35
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The formula is true for all i.i.d. random variables with finite variance!. $E(X_1^{2}|\overset {-} {X})=E(X_4^{2}|\overset {-} {X})$ because $X_i$'s are i.i.d.. Hence we are left with $2E(X_3|\overset {-} {X})$. Now $E(X_i|\overset {-} {X})=E(X_j|\overset {-} {X})$ for all $i,j$ (again because $X_i$'s are i.i.d.). Adding over $i$ and dividing by $n$ we get $E(X_j|\overset {-} {X})=E(\overset {-} {X}|\overset {-} {X})=\overset {-} {X}$. Hence $E(T|\overset {-} {X})=2\overset {-} {X}$.

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  • $\begingroup$ Why are we adding over i and dividing by n? $\endgroup$ – Lady May 6 at 8:53
  • $\begingroup$ By doing that LHS becomes $E(\overset{-} {X} |\overset{-} {X} )$ which is just $\overset{-} {X} $. (This is like saying that if $n$ numbers are equal then each of them is also equal to their average). $\endgroup$ – Kavi Rama Murthy May 6 at 8:55
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Hints:

  • $n\overline{X}=\mathbb{E}\left(n\overline{X}\mid\overline{X}\right)=\mathbb{E}\left(\sum_{i=1}^{n}X_{i}\mid\overline{X}\right)=\sum_{i=1}^{n}\mathbb{E}\left(X_{i}\mid\overline{X}\right)$.

  • $\mathbb{E}\left(X_{i}\mid\overline{X}\right)$ does not depend on $i$.

  • $\mathbb{E}\left(X_{i}^2\mid\overline{X}\right)$ does not depend on $i$.

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