Let $X_{1}, X_{2}, ..., X_{n}$ be independent and identically distributed random variables with finite expected value. Find $\mathbb{E}[X_{1}|(X_{1}+X_{2}+ ... + X_{n})=x]$. Assuming that $f_{(X_{1}+X_{2}+ ... + X_{n})}(x)\neq 0$, I must have the expected value is $\frac{x}{n}$. ¿Could you give me a hint to do it, please?
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Use the facts that $$E(X_1+ \ldots + X_n|X_1 + \ldots + X_n =x)=x$$ $$E(X_1|X_1+ \ldots + X_n=x)=E(X_j|X_1+\ldots + X_n=x)$$ for any $j\in {1,\ldots, n}$ – Matthew H. Jan 19 '21 at 03:21
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2The issue I have with the above comment (and the current answer) is that, while correct, they don't explain how some of the assumptions are used. The independence is necessary, but that's not apparent from the argument. – Clement C. Jan 19 '21 at 03:44
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1Does this answer your question? Conditional expectation for a sum of iid random variables: $E(\xi\mid\xi+\eta)=E(\eta\mid\xi+\eta)=\frac{\xi+\eta}{2}$ – StubbornAtom Jan 19 '21 at 14:34
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Also https://stats.stackexchange.com/a/374997/119261. – StubbornAtom Jan 19 '21 at 14:35
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By symmetry, $$\mathbb E[X_1\mid X_1+\dots+X_n]=\mathbb E[X_2\mid X_1+\dots+X_n]=\dots=\mathbb E[X_n\mid X_1+\dots+X_n].$$ But by linearity $$\sum_{i=1}^n\mathbb E\left[X_i\,\middle|\,\sum_{j=1}^n X_j=x\right]=\mathbb E\left[\sum_{i=1}^nX_i\,\middle|\,\sum_{j=1}^nX_j=x\right]=x,$$ so we have $\mathbb E[X_i\mid X_1+\dots+X_n=x]=\frac{x}{n}$.
jlammy
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1Maybe worth pointing out where/if the independence assumption is used ("by symmetry" can be a bit of a magic wand). – Clement C. Jan 19 '21 at 03:32
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1@PedroAmaral It has to be: if not, the first fact stated doesn't hold. For instance, take $X_1,X_2,X_3$ all (marginally) uniform on ${-1,1}$, with $X_3$ independent of $(X_1,X_2)$ but $X_1=-X_2$ always. Then $X_1+X_2+X_3=X_3$ (the first two cancel), so $$\mathbb{E}[X_1 \mid X_1+X_2+X_3] = \mathbb{E}[X_1 \mid X_3] = 0$$ and $$\mathbb{E}[X_2 \mid X_1+X_2+X_3] = \mathbb{E}[X_2 \mid X_3] = 0$$ but $$\mathbb{E}[X_3 \mid X_1+X_2+X_3] = \mathbb{E}[X_3 \mid X_3] = X_3 \in {-1,1}$$ – Clement C. Jan 19 '21 at 04:15
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@ClementC. Ah very tricky! Is there a simple way to prove that the expectations are equal when the $X_i$'s are all independent? I came up with the following but it is somewhat messy:
$$ f_{X_1,\sum X_i}(x_1, x) $$ $$ = \idotsint f_{X_1, \sum X_i,X_2, ...,X_n}(x_1, x, x_2, .., x_n)dx_2...dx_n $$
$$ = \idotsint \delta_{x,\sum x_i} \cdot f_{X_1,X_2, ...,X_n}(x_1, x_2, .., x_n)dx_2...dx_n $$
$$ = \idotsint \delta_{x,\sum x_i} \cdot f_{X_1,X_2, ...,X_n}(x_1, x_2, .., x_n)dx_2...dx_n $$
$$ = \idotsint \delta_{x,\sum x_i} \cdot f(x_1)f(x_2)...f(x_n)dx_2...dx_n $$
– Pedro Amaral Jan 19 '21 at 05:14 -
@PedroAmaral I think the notion needed here is exchangeability, which is implied by i.i.d. Once you have that, the symmetry argument is basically by definition. – Clement C. Jan 19 '21 at 06:35