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

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?

• $X_i\mid \overline X$ has the same normal distribution for each $i$. – StubbornAtom May 6 at 8:35

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 $$E(T|\overset {-} {X})=2\overset {-} {X}$$.
• 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). – Kavi Rama Murthy May 6 at 8:55
• $$n\overline{X}=\mathbb{E}\left(n\overline{X}\mid\overline{X}\right)=\mathbb{E}\left(\sum_{i=1}^{n}X_{i}\mid\overline{X}\right)=\sum_{i=1}^{n}\mathbb{E}\left(X_{i}\mid\overline{X}\right)$$.
• $$\mathbb{E}\left(X_{i}\mid\overline{X}\right)$$ does not depend on $$i$$.
• $$\mathbb{E}\left(X_{i}^2\mid\overline{X}\right)$$ does not depend on $$i$$.