$\mathbb{E}({T|T_1)}$ over i.i.d random variables

Question

The problem is the following:

Let $(T_n)_{n\leq1}$ a succession of continuous integrable, i.i.d. random variables and let $T=T_1+T_2+...T_n$. Calculate

a)$\mathbb{E}({T_1|T})$

b)$\mathbb{E}({T|T_1)}$

Which for a) i got that $\mathbb{E}({T_1|T})$=$T\over n$

But b) i got the following result which is pretty simple.

$\mathbb{E}({T|T_1})$=$\mathbb{E}(T_1|T_1)$+ $\mathbb{E}(T_2|T_1)$ +$\mathbb{E}(T_3|T_1)$...

$T_1$+$\mathbb{E}(T_2|T_1)$ +$\mathbb{E}(T_3|T_1)+...+\mathbb{E}({T_n|T_1})$

But because of all the are iid, they're the same

$\mathbb{E}({T|T_1})= n*T_1$

Being this easy lead me to think i was wrong, but in some way the result makes sense. Any advice or confirm that this is ok

Answer 1

But because of all the are iid, they're the same

No; i.i.d. means independent and identically distributed.

Independence entails $\forall k\in\{2..n\}:\mathsf E(T_k\mid T_1)=\mathsf E(T_k)$.

• The realised value of $T_1$ does not affect the expected value for any of the other terms in the sequence.

Identical distribution entails $\forall k\in\{1..n\}~.\forall j\in\{1..n\}~:\mathsf E(T_k)=\mathsf E(T_j)$

• All the terms in the sequence do have the same expected value.

However $\mathsf E(T_1\mid T_1)=T_1$ but it is not generally true that $T_1=\mathsf E(T_1)$.

• The realised value of a term may not be the expected value, and quite often is not.

So all that may be said is that: $$\mathsf E(T\mid T_1)=T_1+(n-1)\mathsf E(T_2)$$

Answer 2

Your answer for a) is correct but the answer for b) is wrong. By independence $E(T_i|T_1)=ET_i$ for $i >1$. So the answer is $T_1+ET_2+...+ET_n$.