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Could you help me to understand whether and where I need assumption (2) to assert the following result:

Consider the real-valued random variables $X_1,\ldots,X_n$ all defined on the same probability space such that:

(1) $X_1,\ldots,X_n$ are independently distributed

(2) $X_1,\ldots,X_n$ are identically distributed

(3) $T:=T(X_1,\ldots,X_n)$ is a permutation-symmetric measurable function

Then

$E(T\mid X_i=a)=E(T(X_1,\ldots,X_{i-1},X_i,X_{i+1},\ldots,X_n)\mid X_i=a) \underbrace{=}_{(1)}E(T(X_1,\ldots,X_{i-1}, a, X_{i+1}, \ldots, X_n)) \underbrace{=}_{(3)} E(T(a,X_2,\ldots,X_n))$

In case you conclude that (2) is not required, could you clarify why it seems necessary in van der vaart "Asymptotic Statistics" p. 157 in the last paragraph before section 11.4 (here "Consider the special case..." )

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Without assumption (2) your last step is wrong. With (1) and (3) you can only get to the equation $$E[T \mid X_i = a] = E[T(a, X_2, \ldots, X_{i - 1}, X_1, X_{i + 1}, \ldots, X_n)].$$

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  • $\begingroup$ Could you give more details? I don't understand why $X_i \sim X_j$ helps to show above $\endgroup$ – STF Jan 8 '16 at 19:34
  • $\begingroup$ Assumptions (1) and (2) let you replace $X_1$ with $X_i$ in my last term. More precisely, these assumptions guarantee $(X_2, \ldots, X_{i - 1}, X_1, X_{i + 1}, \ldots, X_n) \stackrel{d}{=} (X_2, \ldots, X_n)$, which makes the replacement possible. $\endgroup$ – Dominik Jan 8 '16 at 19:37
  • $\begingroup$ Ok, perfect, thanks $\endgroup$ – STF Jan 8 '16 at 19:38

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