I need to understand some underlying concepts and facts regarding i.i.d random variables.
The problem is $$\textrm{Suppose } \mathrm{X_1, X_2, X_3 } \textrm{ are i.i.d positive valued r.v.s.}\\ \textrm{Define } \mathrm{Y_i=\frac{X_i}{X_1+X_2+X_3} \textrm{, i=1,2,3. Find the correlation between } Y_1 \textrm{ and } Y_3.}$$
Here is the first approach to solve, $$\mathrm{\sum_{i=1}^{3} Y_i=1 \textrm{ and since X_i's are iid, hence each of } Y_i=\frac{1}{3}. Therefore, \mathbb{E}(Y_i)=\frac{1}{3} \textrm{ for i=1,2,3.}}\\ \mathrm{\textrm{To find } \mathbb{E}(Y_{1}^{2}) \textrm{ and } \mathbb{E}(Y_{3}^{2}) \textrm{ we have }}\\ \mathrm{1=\mathbb{E}\bigg[\frac{(X_1+X_2+X_3)^2}{(X_1+X_2+X_3)^2}\bigg]}\\=\mathrm{\mathbb{E} \bigg[\frac{3X_{1}^{2}}{(X_1+X_2+X_3)^2}+\frac{2(X_1X_2+X_2X_3+X_3X_1)}{(X_1+X_2+X_3)^2}\bigg]}\\=\mathrm{\mathbb{E} \bigg[3Y_{1}^{2}+2(Y_1Y_2+Y_2Y_3+Y_3Y_1)\bigg]}\\=\mathrm{\mathbb{E} \bigg[3Y_{1}^{2}+2[Y_1(1-Y_1)+Y_2Y_3]\bigg]}\\=\mathrm{\mathbb{E} \bigg[Y_{1}^{2}+2Y_1+Y_2Y_3\bigg]}\\=\mathrm{\mathbb{E} \bigg[Y_{1}^{2}+2Y_1+Y_2Y_3\bigg]}\\=\mathrm{\mathbb{E} \bigg[Y_{1}^{2}-\frac{2Y_1}{3}+\frac{1}{9}+\frac{8}{9}+\frac{8Y_1}{3}+Y_2Y_3-1\bigg]}\\=\mathrm{\mathbb{V}(Y_1)+\mathbb{E} \bigg[\frac{8}{9}+\frac{8Y_1}{3}+Y_2Y_3-1\bigg]}\\=\mathrm{0+\mathbb{E}[\frac{7}{9}+Y_2Y_3]}\\\Longrightarrow \mathrm{\mathbb{E}[Y_2Y_3]=\frac{2}{9}}\\\Longrightarrow \mathrm{\mathbb{E}[Y_2Y_3]=\mathbb{E}[Y_2+Y_3]}$$
Hence I need to know that
-What can we conclude from this solution?
My second approach is as follows :
$$\mathrm{\textrm{If Y_i's are not independent then they are indentical only, } cov(Y_1, Y_1+Y_2+Y_3)=cov(Y_1, 1)=0}\\\Longrightarrow \mathrm{\mathbb{V}(Y_1)+cov(Y_1,Y_2)+cov(Y_1,Y_3)=0}\\ \mathrm{\textrm{Now due to identicality if the covariance } (Y_1,Y_2) = \textrm{ covariance } (Y_1,Y_3) \textrm{ and , }\mathbb{V}(Y_1)=\mathbb{V}(Y_3) \textrm{, then my problem is solved.}}$$
- From here if $Y_i$'s are identical then their variances are similar to each other and covariance are similar to each other. Hence the r.vs $Y_i$ are equal due to identicality. is it a correct statement?
Any help or explanation is valuable and highly appreciated.
Here is a post from this website where the given hint says that the iid rv.s are equal according to distrbution but unequal among themselves. Is there any example available?
Hint: By symmetry, $\text{Cov}(Y_1, Y_3) = \text{Cov}(Y_1, Y_2)$ so $$\eqalign{\text{Cov}(Y_1,Y_3) &= \frac{1}{2} \text{Cov}(Y_1,Y_2+Y_3)\cr &= \frac{1}{2} \text{Cov}(Y_1, 1-Y_1)\cr}$$ Express this in terms of $\text{Var}(Y_1)$...