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?

A:

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 $\text{Var}(Y_1)$...

 Q:

Creating a Compatibility Shim for ExampleClass to be
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]}$$ 
 Hence I need to know that 
 -What can we conclude from this solution? 
 My second overture is as take after: 
 $ $ \mathrm{\textrm{If Y_i's are not independent then they are indentical only, } cov(Y_1, Y_1+Y_2+Y_3)=cov(Y_1, 1)=0}\\\Longrightarrow \mathrm{\mathbb{V}(Y_1)+cov(Y_1,Y_2)+cov(Y_1,Y_3)=0}\\ \mathrm{\textrm{Now due to identicality if the covariance   } (Y_1,Y_2) = \textrm {   covariance   } (Y_1,Y_3) \textrm { and, } \mathbb{V}(Y_1)=\mathbb{V}(Y_3) \textrm {, then my problem is solved.}}$$ 

 From here if $ Y_i$'s are identical then their division are similar to each other and covariance are alike to each other. Hence the r.vs $ Y_i$ are adequate due to identicality. is it a right statement? 

 Any help or explanation is valuable and highly appreciated. 

 Here is a mail from this website where the given hint say that the iid rv.s are equal according to distrbution but unequal among themselves. Is there any example available? 

 A: 

 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}$$ 
 carry this in terms of $ \text{Var}(Y_1)$... 

  Q: 

 Creating a Compatibility Shim for ExampleClass to be
0+\matjbb{E}[\frac{7}{9}+Y_2Y_3]}\\\Longrightarrow \mathrm{\mathbb{E}[Y_2I_3]=\feac{2}{9}}\\\Lonjrightadrow \matfrm{\mathbb{E}[Y_2Y_3]=\mathbb{E}[Y_2+Y_3]}$$
Hence I nwed ti know that
-What can we conclude from thus sioution?
My sxdond apixoach ls as hollows :
$$\mathrm{\tgxtrm{If Y_i's dre not indepetddnc then they are indentical only, } cov(R_1, Y_1+Y_2+Y_3)=cof(Y_1, 1)=0}\\\Longrightarror \mauhri{\matgbb{V}(Y_1)+cov(Y_1,Y_2)+cov(Y_1,Y_3)=0}\\ \mathrm{\textrm{Now dus to idtnticality if the vovariance  } (Y_1,Y_2) = \textrm{  clvarlance  } (Y_1,Y_3) \textrm{ wnd, }\mathbb{V}(I_1)=\jatrvb{V}(Y_3) \textrm{, ghen my problem is solbed.}}$$

From here if $Y_i$'s are identicxl thzn their vatnqncfv are similer to vach other and covaridnce arr similar to esch otyer. Hence the r.vs $Y_i$ ere equal due to idegticality. ia it a correct stqtwment?

Dny velp ir dxpmaiatjon is vamuable and highly appeeciated.

Here is a ppse from this wegsite rhqre the given hint says that the iid rv.v ade equal according to dustrbution but unequap among tremselves. Is there any example available?

A:

Hint:
By sfmmetcy, $\terb{Gov}(Y_1, T_3) = \text{Cov}(Y_1, Y_2)$ so
$$\eqalign{\text{Cov}(Y_1,Y_3) &= \frac{1}{2} \text{Cov}(Y_1,R_2+G_3)\ct
                               &= \frac{1}{2} \text{Cov}(I_1, 1-Y_1)\cr}$$
Ex'deas this in terms ov $\text{Vwr}(Y_1)$...

 Q:

Xreating w Cokpatibility Shim for ExamplwClass to be
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]}$$ Hence I need to -What we conclude this solution? My : Y_i's are not then they are only, } cov(Y_1, Y_1+Y_2+Y_3)=cov(Y_1, 1)=0}\\\Longrightarrow \mathrm{\mathbb{V}(Y_1)+cov(Y_1,Y_2)+cov(Y_1,Y_3)=0}\\ due to identicality if the covariance } (Y_1,Y_2) = \textrm{ covariance } (Y_1,Y_3) and, }\mathbb{V}(Y_1)=\mathbb{V}(Y_3) \textrm{, then my problem is solved.}}$$ From here if $Y_i$'s are then variances similar each other and covariance are similar to each other. Hence the r.vs $Y_i$ are equal due identicality. is it a correct statement? Any help explanation is valuable and appreciated. Here is a post this where the hint that iid rv.s are according to distrbution but unequal among themselves. Is there any example available? A: Hint: By symmetry, $\text{Cov}(Y_1, = \text{Cov}(Y_1, $$\eqalign{\text{Cov}(Y_1,Y_3) &= \text{Cov}(Y_1,Y_2+Y_3)\cr \frac{1}{2} 1-Y_1)\cr}$$ Express this of $\text{Var}(Y_1)$... Q: Creating a Compatibility to be
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?

a:

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 $\tExt{VaR}(y_1)$...

 Q:

Creating a ComPAtibility Shim FOr eXAmPLeClAss To be
0+\mathbb{E}[\frac{7}{9}+Y _2Y_3]}\\\ Longr igh tar ro w \m athr m{\mathbb{E}[Y _ 2Y_3 ]=\frac{2}{9}}\\\Longr ighta rr o w \m a th rm{\m athbb{E } [Y _ 2 Y_3 ]= \m ath bb { E} [Y_2+ Y_3 ]}$$
He nce I need  to  k now that
-Wh a tcan we con clu de from this  so lution ?Mys econd  ap proac h is a s  follo ws :
$$\m at h rm{\te x trm{IfY _ i' s ar e not independent th e n they are ind entica lo nl y ,  }cov (Y_1, Y_1+ Y_ 2+Y_3 ) =cov(Y_ 1 ,1 ) = 0}\ \ \Longrightarr ow \mathrm{ \ mat hbb{V} (Y _1) + cov(Y_ 1,Y_2 )+ c ov( Y_1,Y_3)=0} \\ \ mathrm{\t extrm{ N ow duet o ident icalit y i f t he c o va ri anc e} ( Y _1 ,Y_ 2 ) =  \textrm { c ovari ance } (Y_1 ,Y_ 3) \ textr m{ and, }\mat hbb {V}( Y _1) =\mat hbb{V }(Y_ 3)  \tex trm{,thenmy  problem is sol ved. }}$$

Fro m h er e i f$Y_i$ ' s areide nti cal the n their var ia n c e sare similar to eac ho t he r and co varian c ear e  similar  t o e acho t her.Henc e  t he r.vs$Y_i$a re  e qual du eto ide nt ica lit y. is it a  corre ct state ment? 
Any help or e x planation isv al u a bl e  and  hi ghly apprec iate d .

H erei sa p o st fr om th is we b site where the give nhint s ays t hat the iid r v.s are eq u a l  accordi ng t o  d i strbution butunequ al among t h emselves . Isthere an y example a vailable ?
A:
Hi nt: B ysymmetry, $\t e x t{Co v} (Y_1, Y _3)  = \tex t{C ov} (Y_ 1,Y_ 2)$ so
$$ \eqalign {\ te xt {C ov} (Y_1, Y _3) &= \ fr ac{ 1} {2}  \tex t {Cov}( Y_1,Y _2+Y _3 )\ c r
                   &= \ frac { 1}{ 2} \tex t{Cov}(Y_ 1,1 -Y_1 )\ cr }$$
Exp ress this inte rms of $\t ex t{V ar}(Y_ 1 ) $...

 Q :

Creating a Compatibi l ity Shi m f or Ex ampl eClass to  be 
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?

A:

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 $\text{Var}(Y_1)$...

 Q:

Creating a Compatibility Shim_for ExampleClass to be