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I know that the equation $$\Bbb{E}[\frac {X_1} {X_1+...+X_n}]=\frac{1}{n}$$ is true if $X_1,...,X_n$ are independent identically distributed non-negative integer-valued random variables.

Now there is a follow-up question:

Is that still true if $X_1,...,X_n$ are not independent?

I tried to find a counterexample, but I couldn't find any that works:

$(1)$ Let $\Omega=\{1,...,6\}$, $X_1,X_2$ Laplace-distributed random variables with $$X_1: \Omega \to \Omega, X_1(\omega)=\omega$$ $$X_2: \Omega \to \Omega, X_2(\omega)=7-\omega$$

Then $X_1$,$X_2$ are identically distributed but not independent and

$$\Bbb{E}[\frac{X_1}{X_1+X_2}] = \Bbb{E}[\frac{X_1}{7}] = \frac{1}{2}$$

$(2)$ Let

$\Omega = \{1,...,6\}^3$,

$X_k: \Omega \to \{1,...,6\}$ a Laplace-distributed random variable with $X_k(\omega_1,\omega_2,\omega_3)=\omega_k, k \in \{1,2,3\}$.

Define $S_1 := X_1+X_2, S_2:=X_2+X_3$. Then $S_1$,$S_2$ are identically distributed but not independent and

$$\Bbb{E}\left[\frac{S_1}{S_1 + S_2} \right] = \Bbb{E}\left[ \frac{X_1 + X_2}{X_1 + 2X_2 + X_3}\right] = \sum_{\omega\in\Omega}\frac{\omega_1 + \omega_2}{\omega_1 + 2 \omega_2 + \omega_3} \underbrace{\Bbb{P}(X_1= \omega_1, X_2 = \omega_2, X_3 = \omega_3)}_{= \frac{1}{6^3}}\\ = \frac{1}{6^3}\sum_{i=1}^6\sum_{j=1}^6\sum_{k=1}^6 \frac{i + j}{i + 2 j + k} = \frac{1}{2}$$

So unfortunately non of the examples worked out to prove that $X_1,...,X_n$ need to be independent. Can anyone think of a proper Counterexample that works?

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    $\begingroup$ Under what conditions? If $n=2$, if $P(X_1=1)=P(X_2=2)=1$, then $E(X_1/(X_1+X_2))=1/3$, but I don't think that's what you have in mind. $\endgroup$ – kimchi lover Dec 10 '17 at 20:19
  • $\begingroup$ Let $X_2 = 2X_1$. Then $X_1$ and $X_2$ are dependent and $\mathbb{E}(X_1/(X_1+X_2)) = 1/3$. $\endgroup$ – induction601 Dec 10 '17 at 20:44
  • $\begingroup$ Btw, could you provide a reference for the proof on $\mathbb{E}(X_1/\sum_{i=1}^nX_i) = 1/n$ when $X_i$'s are iid with non-negative integer values? $\endgroup$ – induction601 Dec 10 '17 at 20:46
  • $\begingroup$ $X_1,...,X_n$ need to be identically distributed which as far as I know is not the case in either of your examples. I will provide the proof asap. $\endgroup$ – newbie Dec 10 '17 at 20:53
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Let $\Omega=\{1,2,3\}$, $\mathcal F=2^\Omega$, $X_1(\omega)=\omega$ and set equal probabilities for elementary events. Then $$ \mathbb P(X_1=1)=\mathbb P(X_1=2)=\mathbb P(X_1=3)=\frac13 $$ Define $X_2(1)=2$, $X_2(2)=3$, $X_2(3)=1$. Random variables $X_1$ and $X_2$ are dependent but identically distributed.

Consider r.v. $Y=\frac{X_1}{X_1+X_2}$: $$ Y(1)=\frac{1}{1+2}=\frac13; \, Y(2)=\frac{2}{2+3}=\frac25; \, Y(3)=\frac{3}{3+1}=\frac34 $$ So $$ \mathbb E[Y]=\frac13\cdot\left(\frac13+\frac25+\frac34\right)=\frac{89}{180}\neq \frac12. $$ Consider r.v. $Z=\frac{X_2}{X_1+X_2}$: $$ Z(1)=\frac{2}{1+2}=\frac23; \, Z(2)=\frac{3}{2+3}=\frac35; \, Z(3)=\frac{1}{3+1}=\frac14 $$ You can see that the distributions of $\frac{X_1}{X_1+X_2}$ and $\frac{X_2}{X_1+X_2}$ are not the same. And $$ \mathbb E[Z]=\frac13\cdot\left(\frac23+\frac35+\frac14\right)=\frac{91}{180}\neq \frac12. $$

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