Transformation of a discret random variable, with different support size.
$begingroup$
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
asked Dec 2 '18 at 22:54
pin_rpin_r
124
$endgroup$
add a comment |
$begingroup$
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
asked Dec 2 '18 at 22:54
pin_rpin_r
124
$endgroup$
$begingroup$
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 $g$, $$E(g(X))=frac14(g(-2)+g(-1)+g(1)+g(2))$$
$endgroup$
– Did
Dec 2 '18 at 23:00
$begingroup$
thanks a lot, that was super simple.
$endgroup$
– pin_r
Dec 2 '18 at 23:33
$begingroup$
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 $(X,Y)$.
$endgroup$
– Song
Dec 2 '18 at 23:45
add a comment |
$begingroup$
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
asked Dec 2 '18 at 22:54
pin_rpin_r
124
$endgroup$
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
probability statistics
asked Dec 2 '18 at 22:54
pin_rpin_r
124
asked Dec 2 '18 at 22:54
pin_rpin_r
124
asked Dec 2 '18 at 22:54
pin_rpin_r
124
asked Dec 2 '18 at 22:54
pin_rpin_r
124
asked Dec 2 '18 at 22:54
pin_rpin_r
124
124
$begingroup$
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 $g$, $$E(g(X))=frac14(g(-2)+g(-1)+g(1)+g(2))$$
$endgroup$
– Did
Dec 2 '18 at 23:00
$begingroup$
thanks a lot, that was super simple.
$endgroup$
– pin_r
Dec 2 '18 at 23:33
$begingroup$
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 $(X,Y)$.
$endgroup$
– Song
Dec 2 '18 at 23:45
add a comment |
$begingroup$
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 $g$, $$E(g(X))=frac14(g(-2)+g(-1)+g(1)+g(2))$$
$endgroup$
– Did
Dec 2 '18 at 23:00
$begingroup$
thanks a lot, that was super simple.
$endgroup$
– pin_r
Dec 2 '18 at 23:33
$begingroup$
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 $(X,Y)$.
$endgroup$
– Song
Dec 2 '18 at 23:45
$begingroup$
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 $g$, $$E(g(X))=frac14(g(-2)+g(-1)+g(1)+g(2))$$
$endgroup$
– Did
Dec 2 '18 at 23:00
$begingroup$
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 $g$, $$E(g(X))=frac14(g(-2)+g(-1)+g(1)+g(2))$$
$endgroup$
– Did
Dec 2 '18 at 23:00
$begingroup$
thanks a lot, that was super simple.
$endgroup$
– pin_r
Dec 2 '18 at 23:33
$begingroup$
thanks a lot, that was super simple.
$endgroup$
– pin_r
Dec 2 '18 at 23:33
$begingroup$
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 $(X,Y)$.
$endgroup$
– Song
Dec 2 '18 at 23:45
$begingroup$
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 $(X,Y)$.
$endgroup$
– Song
Dec 2 '18 at 23:45
add a comment |
1 Answer
1
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votes
$begingroup$
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.
answered Dec 2 '18 at 23:21
Kavi Rama MurthyKavi Rama Murthy
56k42158
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add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
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oldest
votes
active
oldest
votes
$begingroup$
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.
answered Dec 2 '18 at 23:21
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 2 '18 at 23:21
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 2 '18 at 23:21
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
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.
answered Dec 2 '18 at 23:21
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Dec 2 '18 at 23:21
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Dec 2 '18 at 23:21
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Dec 2 '18 at 23:21
Kavi Rama MurthyKavi Rama Murthy
56k42158
56k42158
add a comment |
add a comment |
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$begingroup$
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 $g$, $$E(g(X))=frac14(g(-2)+g(-1)+g(1)+g(2))$$
$endgroup$
– Did
Dec 2 '18 at 23:00
$begingroup$
thanks a lot, that was super simple.
$endgroup$
– pin_r
Dec 2 '18 at 23:33
$begingroup$
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 $(X,Y)$.
$endgroup$
– Song
Dec 2 '18 at 23:45