Compute the $Cov(X,Y)$ of the two random variables
$begingroup$
Let $X,Y$ be absolutely continuous random variables. $X$ is $Uniform[0,12]$ and $f_{Y|X}(y|x)=frac{1}{x}$, for $yin [0,x]$ and $0$ otherwise.
We are asked to find $Cov(X,Y)$.
$Cov(X,Y)=E(XY)-E(X)E(Y)$.
Usually in these types of problems, I'm given $Y=X^2, X+3$, or a function of $X$. Here I am given the conditional function.
So, $f_{Y|X}=dfrac{f_{XY}(x,y)}{f_{X}(x)}$, by definition.
We know the $f_X(x)=dfrac{1}{12}; 0leq x leq 12$, as it is uniform.
Then, we get that $dfrac{1}{x}=dfrac{f_{XY}(x,y)}{(1/12)}to f_{XY}(x,y)=dfrac{1}{12x}$
This is as far as I got.
To find $f_Y$, I need to integrate $f_{XY}$ over a range ($x$ values). What is this range?
We have firstly that $0 leq y leq x$, and then we have $0 leq x leq 12$. How can I combine these two?
I don't know what to integrate $f_Y$ over.
probability probability-theory statistics random-variables covariance
asked Dec 6 '18 at 0:37
K Split XK Split X
4,20611132
$endgroup$
add a comment |
$begingroup$
Let $X,Y$ be absolutely continuous random variables. $X$ is $Uniform[0,12]$ and $f_{Y|X}(y|x)=frac{1}{x}$, for $yin [0,x]$ and $0$ otherwise.
We are asked to find $Cov(X,Y)$.
$Cov(X,Y)=E(XY)-E(X)E(Y)$.
Usually in these types of problems, I'm given $Y=X^2, X+3$, or a function of $X$. Here I am given the conditional function.
So, $f_{Y|X}=dfrac{f_{XY}(x,y)}{f_{X}(x)}$, by definition.
We know the $f_X(x)=dfrac{1}{12}; 0leq x leq 12$, as it is uniform.
Then, we get that $dfrac{1}{x}=dfrac{f_{XY}(x,y)}{(1/12)}to f_{XY}(x,y)=dfrac{1}{12x}$
This is as far as I got.
To find $f_Y$, I need to integrate $f_{XY}$ over a range ($x$ values). What is this range?
We have firstly that $0 leq y leq x$, and then we have $0 leq x leq 12$. How can I combine these two?
I don't know what to integrate $f_Y$ over.
probability probability-theory statistics random-variables covariance
asked Dec 6 '18 at 0:37
K Split XK Split X
4,20611132
$endgroup$
$begingroup$
Not a duplicate, mse accidently posted it twice, I don't even know how thats possible
$endgroup$
– K Split X
Dec 6 '18 at 0:59
$begingroup$
Don't forget the condition that $0 le y le x$ in your formula for $f_{XY}(x,y)$.
$endgroup$
– zoidberg
Dec 6 '18 at 1:48
add a comment |
$begingroup$
Let $X,Y$ be absolutely continuous random variables. $X$ is $Uniform[0,12]$ and $f_{Y|X}(y|x)=frac{1}{x}$, for $yin [0,x]$ and $0$ otherwise.
We are asked to find $Cov(X,Y)$.
$Cov(X,Y)=E(XY)-E(X)E(Y)$.
Usually in these types of problems, I'm given $Y=X^2, X+3$, or a function of $X$. Here I am given the conditional function.
So, $f_{Y|X}=dfrac{f_{XY}(x,y)}{f_{X}(x)}$, by definition.
We know the $f_X(x)=dfrac{1}{12}; 0leq x leq 12$, as it is uniform.
Then, we get that $dfrac{1}{x}=dfrac{f_{XY}(x,y)}{(1/12)}to f_{XY}(x,y)=dfrac{1}{12x}$
This is as far as I got.
To find $f_Y$, I need to integrate $f_{XY}$ over a range ($x$ values). What is this range?
We have firstly that $0 leq y leq x$, and then we have $0 leq x leq 12$. How can I combine these two?
I don't know what to integrate $f_Y$ over.
probability probability-theory statistics random-variables covariance
asked Dec 6 '18 at 0:37
K Split XK Split X
4,20611132
$endgroup$
Let $X,Y$ be absolutely continuous random variables. $X$ is $Uniform[0,12]$ and $f_{Y|X}(y|x)=frac{1}{x}$, for $yin [0,x]$ and $0$ otherwise.
We are asked to find $Cov(X,Y)$.
$Cov(X,Y)=E(XY)-E(X)E(Y)$.
Usually in these types of problems, I'm given $Y=X^2, X+3$, or a function of $X$. Here I am given the conditional function.
So, $f_{Y|X}=dfrac{f_{XY}(x,y)}{f_{X}(x)}$, by definition.
We know the $f_X(x)=dfrac{1}{12}; 0leq x leq 12$, as it is uniform.
Then, we get that $dfrac{1}{x}=dfrac{f_{XY}(x,y)}{(1/12)}to f_{XY}(x,y)=dfrac{1}{12x}$
This is as far as I got.
To find $f_Y$, I need to integrate $f_{XY}$ over a range ($x$ values). What is this range?
We have firstly that $0 leq y leq x$, and then we have $0 leq x leq 12$. How can I combine these two?
I don't know what to integrate $f_Y$ over.
probability probability-theory statistics random-variables covariance
probability probability-theory statistics random-variables covariance
asked Dec 6 '18 at 0:37
K Split XK Split X
4,20611132
asked Dec 6 '18 at 0:37
K Split XK Split X
4,20611132
edited Dec 6 '18 at 1:53
K Split X
asked Dec 6 '18 at 0:37
K Split XK Split X
4,20611132
asked Dec 6 '18 at 0:37
K Split XK Split X
4,20611132
asked Dec 6 '18 at 0:37
K Split XK Split X
4,20611132
4,20611132
$begingroup$
Not a duplicate, mse accidently posted it twice, I don't even know how thats possible
$endgroup$
– K Split X
Dec 6 '18 at 0:59
$begingroup$
Don't forget the condition that $0 le y le x$ in your formula for $f_{XY}(x,y)$.
$endgroup$
– zoidberg
Dec 6 '18 at 1:48
add a comment |
$begingroup$
Not a duplicate, mse accidently posted it twice, I don't even know how thats possible
$endgroup$
– K Split X
Dec 6 '18 at 0:59
$begingroup$
Don't forget the condition that $0 le y le x$ in your formula for $f_{XY}(x,y)$.
$endgroup$
– zoidberg
Dec 6 '18 at 1:48
$begingroup$
Not a duplicate, mse accidently posted it twice, I don't even know how thats possible
$endgroup$
– K Split X
Dec 6 '18 at 0:59
$begingroup$
Not a duplicate, mse accidently posted it twice, I don't even know how thats possible
$endgroup$
– K Split X
Dec 6 '18 at 0:59
$begingroup$
Don't forget the condition that $0 le y le x$ in your formula for $f_{XY}(x,y)$.
$endgroup$
– zoidberg
Dec 6 '18 at 1:48
$begingroup$
Don't forget the condition that $0 le y le x$ in your formula for $f_{XY}(x,y)$.
$endgroup$
– zoidberg
Dec 6 '18 at 1:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
You know $0 le x le 12$ and $0 le y le x$, which you can combine into $0 le y le x le 12$
Write the joint density with an indicator function taking the value $1$ when the condition is met and $0$ otherwise
$$f_{XY}(x,y)=dfrac{1}{12x} I{[0 le y le x le 12]}$$
Then you want to integrate this over $x$ to get the marginal density for $Y$ and then adjust the limits of integration to respect the indicator function
$$f_{Y}(y)=int_{x=-infty}^{infty} dfrac{1}{12x} I{[0 le y le x le 12]} , dx= int_{x=y}^{12} dfrac{1}{12x} I{[0 le y le 12]} , dx$$
and that integral is not particularly difficult: you will be left with a function of $y$ multiplied by an indicator function depending on $y$ showing where the marginal density is non-zero
Alternative hint:
You know that $E[Y mid X=x]=frac{x}{2}$ so $E[Y mid X]=frac{X}{2}$
This will give you $E[Y]=frac12E[X]$ and $E[XY]=frac12E[X^2]$
So $text{Cov}(X,Y)=frac12E[X^2]-frac12(E[X])^2 = frac12text{Var}(X)$ which you can calculate
answered Dec 7 '18 at 20:07
HenryHenry
100k480165
$endgroup$
$begingroup$
I really liked Henry's Alternative hint. Similarly, one could define $Z= Y-X/2$. It's not too hard to see that Cov(Z,X)=0, so Cov(X,Y) = Cov(X,X/2+Z) = Cov(X,X/2) = $sigma_X^2/2$.
$endgroup$
– irchans
Dec 7 '18 at 21:15
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint:
You know $0 le x le 12$ and $0 le y le x$, which you can combine into $0 le y le x le 12$
Write the joint density with an indicator function taking the value $1$ when the condition is met and $0$ otherwise
$$f_{XY}(x,y)=dfrac{1}{12x} I{[0 le y le x le 12]}$$
Then you want to integrate this over $x$ to get the marginal density for $Y$ and then adjust the limits of integration to respect the indicator function
$$f_{Y}(y)=int_{x=-infty}^{infty} dfrac{1}{12x} I{[0 le y le x le 12]} , dx= int_{x=y}^{12} dfrac{1}{12x} I{[0 le y le 12]} , dx$$
and that integral is not particularly difficult: you will be left with a function of $y$ multiplied by an indicator function depending on $y$ showing where the marginal density is non-zero
Alternative hint:
You know that $E[Y mid X=x]=frac{x}{2}$ so $E[Y mid X]=frac{X}{2}$
This will give you $E[Y]=frac12E[X]$ and $E[XY]=frac12E[X^2]$
So $text{Cov}(X,Y)=frac12E[X^2]-frac12(E[X])^2 = frac12text{Var}(X)$ which you can calculate
answered Dec 7 '18 at 20:07
HenryHenry
100k480165
$endgroup$
$begingroup$
I really liked Henry's Alternative hint. Similarly, one could define $Z= Y-X/2$. It's not too hard to see that Cov(Z,X)=0, so Cov(X,Y) = Cov(X,X/2+Z) = Cov(X,X/2) = $sigma_X^2/2$.
$endgroup$
– irchans
Dec 7 '18 at 21:15
add a comment |
$begingroup$
Hint:
You know $0 le x le 12$ and $0 le y le x$, which you can combine into $0 le y le x le 12$
Write the joint density with an indicator function taking the value $1$ when the condition is met and $0$ otherwise
$$f_{XY}(x,y)=dfrac{1}{12x} I{[0 le y le x le 12]}$$
Then you want to integrate this over $x$ to get the marginal density for $Y$ and then adjust the limits of integration to respect the indicator function
$$f_{Y}(y)=int_{x=-infty}^{infty} dfrac{1}{12x} I{[0 le y le x le 12]} , dx= int_{x=y}^{12} dfrac{1}{12x} I{[0 le y le 12]} , dx$$
and that integral is not particularly difficult: you will be left with a function of $y$ multiplied by an indicator function depending on $y$ showing where the marginal density is non-zero
Alternative hint:
You know that $E[Y mid X=x]=frac{x}{2}$ so $E[Y mid X]=frac{X}{2}$
This will give you $E[Y]=frac12E[X]$ and $E[XY]=frac12E[X^2]$
So $text{Cov}(X,Y)=frac12E[X^2]-frac12(E[X])^2 = frac12text{Var}(X)$ which you can calculate
answered Dec 7 '18 at 20:07
HenryHenry
100k480165
$endgroup$
$begingroup$
I really liked Henry's Alternative hint. Similarly, one could define $Z= Y-X/2$. It's not too hard to see that Cov(Z,X)=0, so Cov(X,Y) = Cov(X,X/2+Z) = Cov(X,X/2) = $sigma_X^2/2$.
$endgroup$
– irchans
Dec 7 '18 at 21:15
add a comment |
$begingroup$
Hint:
You know $0 le x le 12$ and $0 le y le x$, which you can combine into $0 le y le x le 12$
Write the joint density with an indicator function taking the value $1$ when the condition is met and $0$ otherwise
$$f_{XY}(x,y)=dfrac{1}{12x} I{[0 le y le x le 12]}$$
Then you want to integrate this over $x$ to get the marginal density for $Y$ and then adjust the limits of integration to respect the indicator function
$$f_{Y}(y)=int_{x=-infty}^{infty} dfrac{1}{12x} I{[0 le y le x le 12]} , dx= int_{x=y}^{12} dfrac{1}{12x} I{[0 le y le 12]} , dx$$
and that integral is not particularly difficult: you will be left with a function of $y$ multiplied by an indicator function depending on $y$ showing where the marginal density is non-zero
Alternative hint:
You know that $E[Y mid X=x]=frac{x}{2}$ so $E[Y mid X]=frac{X}{2}$
This will give you $E[Y]=frac12E[X]$ and $E[XY]=frac12E[X^2]$
So $text{Cov}(X,Y)=frac12E[X^2]-frac12(E[X])^2 = frac12text{Var}(X)$ which you can calculate
answered Dec 7 '18 at 20:07
HenryHenry
100k480165
$endgroup$
Hint:
You know $0 le x le 12$ and $0 le y le x$, which you can combine into $0 le y le x le 12$
Write the joint density with an indicator function taking the value $1$ when the condition is met and $0$ otherwise
$$f_{XY}(x,y)=dfrac{1}{12x} I{[0 le y le x le 12]}$$
Then you want to integrate this over $x$ to get the marginal density for $Y$ and then adjust the limits of integration to respect the indicator function
$$f_{Y}(y)=int_{x=-infty}^{infty} dfrac{1}{12x} I{[0 le y le x le 12]} , dx= int_{x=y}^{12} dfrac{1}{12x} I{[0 le y le 12]} , dx$$
and that integral is not particularly difficult: you will be left with a function of $y$ multiplied by an indicator function depending on $y$ showing where the marginal density is non-zero
Alternative hint:
You know that $E[Y mid X=x]=frac{x}{2}$ so $E[Y mid X]=frac{X}{2}$
This will give you $E[Y]=frac12E[X]$ and $E[XY]=frac12E[X^2]$
So $text{Cov}(X,Y)=frac12E[X^2]-frac12(E[X])^2 = frac12text{Var}(X)$ which you can calculate
answered Dec 7 '18 at 20:07
HenryHenry
100k480165
edited Dec 7 '18 at 20:54
answered Dec 7 '18 at 20:07
HenryHenry
100k480165
answered Dec 7 '18 at 20:07
HenryHenry
100k480165
answered Dec 7 '18 at 20:07
HenryHenry
100k480165
100k480165
$begingroup$
I really liked Henry's Alternative hint. Similarly, one could define $Z= Y-X/2$. It's not too hard to see that Cov(Z,X)=0, so Cov(X,Y) = Cov(X,X/2+Z) = Cov(X,X/2) = $sigma_X^2/2$.
$endgroup$
– irchans
Dec 7 '18 at 21:15
add a comment |
$begingroup$
I really liked Henry's Alternative hint. Similarly, one could define $Z= Y-X/2$. It's not too hard to see that Cov(Z,X)=0, so Cov(X,Y) = Cov(X,X/2+Z) = Cov(X,X/2) = $sigma_X^2/2$.
$endgroup$
– irchans
Dec 7 '18 at 21:15
$begingroup$
I really liked Henry's Alternative hint. Similarly, one could define $Z= Y-X/2$. It's not too hard to see that Cov(Z,X)=0, so Cov(X,Y) = Cov(X,X/2+Z) = Cov(X,X/2) = $sigma_X^2/2$.
$endgroup$
– irchans
Dec 7 '18 at 21:15
$begingroup$
I really liked Henry's Alternative hint. Similarly, one could define $Z= Y-X/2$. It's not too hard to see that Cov(Z,X)=0, so Cov(X,Y) = Cov(X,X/2+Z) = Cov(X,X/2) = $sigma_X^2/2$.
$endgroup$
– irchans
Dec 7 '18 at 21:15
add a comment |
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$begingroup$
Not a duplicate, mse accidently posted it twice, I don't even know how thats possible
$endgroup$
– K Split X
Dec 6 '18 at 0:59
$begingroup$
Don't forget the condition that $0 le y le x$ in your formula for $f_{XY}(x,y)$.
$endgroup$
– zoidberg
Dec 6 '18 at 1:48