Transformation of a discret random variable, with different support size.












0














Let X a random variable with probability mass function:



$$f_X(x)= frac{1}{4} I_{{-2,-1,1,2}} (x)$$



and let $Y:= X^2$, Proof that the Corr(X,Y)=0 and yet $X$ and $Y$ are not independent.



What I have:



$y=g(x)=x^2 Rightarrow x=g^{-1}(y)=sqrt{y}$



$A_x={-2,-1,1,2}, B_y={1,4}$



And I think, $f_Y(y)=frac{1}{4} I_{{1,4}} (y)$



But that doesn't make sense, also when I try to calculate the Expected value, I can't because of the support of the function.










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  • Sorry but none of this is necessary since the correlation you are after only involves $E(XY)=E(X^3)$, $E(X)$ and $E(Y)=E(X^2)$, and since, by definition, for every function $g$, $$E(g(X))=frac14(g(-2)+g(-1)+g(1)+g(2))$$
    – Did
    Dec 2 '18 at 23:00












  • thanks a lot, that was super simple.
    – pin_r
    Dec 2 '18 at 23:33










  • You have earned a marginal density function of $Y$ from that of $X$. But in general, to evaluate $E[XY]$, you need the joint distribution of $(X,Y)$.
    – Song
    Dec 2 '18 at 23:45
















0














Let X a random variable with probability mass function:



$$f_X(x)= frac{1}{4} I_{{-2,-1,1,2}} (x)$$



and let $Y:= X^2$, Proof that the Corr(X,Y)=0 and yet $X$ and $Y$ are not independent.



What I have:



$y=g(x)=x^2 Rightarrow x=g^{-1}(y)=sqrt{y}$



$A_x={-2,-1,1,2}, B_y={1,4}$



And I think, $f_Y(y)=frac{1}{4} I_{{1,4}} (y)$



But that doesn't make sense, also when I try to calculate the Expected value, I can't because of the support of the function.










share|cite|improve this question






















  • Sorry but none of this is necessary since the correlation you are after only involves $E(XY)=E(X^3)$, $E(X)$ and $E(Y)=E(X^2)$, and since, by definition, for every function $g$, $$E(g(X))=frac14(g(-2)+g(-1)+g(1)+g(2))$$
    – Did
    Dec 2 '18 at 23:00












  • thanks a lot, that was super simple.
    – pin_r
    Dec 2 '18 at 23:33










  • You have earned a marginal density function of $Y$ from that of $X$. But in general, to evaluate $E[XY]$, you need the joint distribution of $(X,Y)$.
    – Song
    Dec 2 '18 at 23:45














0












0








0







Let X a random variable with probability mass function:



$$f_X(x)= frac{1}{4} I_{{-2,-1,1,2}} (x)$$



and let $Y:= X^2$, Proof that the Corr(X,Y)=0 and yet $X$ and $Y$ are not independent.



What I have:



$y=g(x)=x^2 Rightarrow x=g^{-1}(y)=sqrt{y}$



$A_x={-2,-1,1,2}, B_y={1,4}$



And I think, $f_Y(y)=frac{1}{4} I_{{1,4}} (y)$



But that doesn't make sense, also when I try to calculate the Expected value, I can't because of the support of the function.










share|cite|improve this question













Let X a random variable with probability mass function:



$$f_X(x)= frac{1}{4} I_{{-2,-1,1,2}} (x)$$



and let $Y:= X^2$, Proof that the Corr(X,Y)=0 and yet $X$ and $Y$ are not independent.



What I have:



$y=g(x)=x^2 Rightarrow x=g^{-1}(y)=sqrt{y}$



$A_x={-2,-1,1,2}, B_y={1,4}$



And I think, $f_Y(y)=frac{1}{4} I_{{1,4}} (y)$



But that doesn't make sense, also when I try to calculate the Expected value, I can't because of the support of the function.







probability statistics






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asked Dec 2 '18 at 22:54









pin_r

124




124












  • Sorry but none of this is necessary since the correlation you are after only involves $E(XY)=E(X^3)$, $E(X)$ and $E(Y)=E(X^2)$, and since, by definition, for every function $g$, $$E(g(X))=frac14(g(-2)+g(-1)+g(1)+g(2))$$
    – Did
    Dec 2 '18 at 23:00












  • thanks a lot, that was super simple.
    – pin_r
    Dec 2 '18 at 23:33










  • You have earned a marginal density function of $Y$ from that of $X$. But in general, to evaluate $E[XY]$, you need the joint distribution of $(X,Y)$.
    – Song
    Dec 2 '18 at 23:45


















  • Sorry but none of this is necessary since the correlation you are after only involves $E(XY)=E(X^3)$, $E(X)$ and $E(Y)=E(X^2)$, and since, by definition, for every function $g$, $$E(g(X))=frac14(g(-2)+g(-1)+g(1)+g(2))$$
    – Did
    Dec 2 '18 at 23:00












  • thanks a lot, that was super simple.
    – pin_r
    Dec 2 '18 at 23:33










  • You have earned a marginal density function of $Y$ from that of $X$. But in general, to evaluate $E[XY]$, you need the joint distribution of $(X,Y)$.
    – Song
    Dec 2 '18 at 23:45
















Sorry but none of this is necessary since the correlation you are after only involves $E(XY)=E(X^3)$, $E(X)$ and $E(Y)=E(X^2)$, and since, by definition, for every function $g$, $$E(g(X))=frac14(g(-2)+g(-1)+g(1)+g(2))$$
– Did
Dec 2 '18 at 23:00






Sorry but none of this is necessary since the correlation you are after only involves $E(XY)=E(X^3)$, $E(X)$ and $E(Y)=E(X^2)$, and since, by definition, for every function $g$, $$E(g(X))=frac14(g(-2)+g(-1)+g(1)+g(2))$$
– Did
Dec 2 '18 at 23:00














thanks a lot, that was super simple.
– pin_r
Dec 2 '18 at 23:33




thanks a lot, that was super simple.
– pin_r
Dec 2 '18 at 23:33












You have earned a marginal density function of $Y$ from that of $X$. But in general, to evaluate $E[XY]$, you need the joint distribution of $(X,Y)$.
– Song
Dec 2 '18 at 23:45




You have earned a marginal density function of $Y$ from that of $X$. But in general, to evaluate $E[XY]$, you need the joint distribution of $(X,Y)$.
– Song
Dec 2 '18 at 23:45










1 Answer
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Use the idea in Did's comment to get the following: $EX=0, EY=2.5,EXY=0$ so $EXY-EXEY=0$. Also, $P{Y=1,X=2}=P{X=1}=frac 1 4, P{Y=1}P{X=1}=frac 1 2 frac 1 4$ which shows that $X$ and $Y$ are not independent.






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    1 Answer
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    1 Answer
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    active

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    Use the idea in Did's comment to get the following: $EX=0, EY=2.5,EXY=0$ so $EXY-EXEY=0$. Also, $P{Y=1,X=2}=P{X=1}=frac 1 4, P{Y=1}P{X=1}=frac 1 2 frac 1 4$ which shows that $X$ and $Y$ are not independent.






    share|cite|improve this answer


























      0














      Use the idea in Did's comment to get the following: $EX=0, EY=2.5,EXY=0$ so $EXY-EXEY=0$. Also, $P{Y=1,X=2}=P{X=1}=frac 1 4, P{Y=1}P{X=1}=frac 1 2 frac 1 4$ which shows that $X$ and $Y$ are not independent.






      share|cite|improve this answer
























        0












        0








        0






        Use the idea in Did's comment to get the following: $EX=0, EY=2.5,EXY=0$ so $EXY-EXEY=0$. Also, $P{Y=1,X=2}=P{X=1}=frac 1 4, P{Y=1}P{X=1}=frac 1 2 frac 1 4$ which shows that $X$ and $Y$ are not independent.






        share|cite|improve this answer












        Use the idea in Did's comment to get the following: $EX=0, EY=2.5,EXY=0$ so $EXY-EXEY=0$. Also, $P{Y=1,X=2}=P{X=1}=frac 1 4, P{Y=1}P{X=1}=frac 1 2 frac 1 4$ which shows that $X$ and $Y$ are not independent.







        share|cite|improve this answer












        share|cite|improve this answer



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        answered Dec 2 '18 at 23:21









        Kavi Rama Murthy

        50.1k31854




        50.1k31854






























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