Independent continuous random variables problem
$begingroup$
I have some problem at how to determine if two random variables are independent or not. If X and Y are two continuous random variables and their join probability density function is listed below
$$
f(x,y) = left{begin{matrix}
frac{1}{2}x^{3}e^{-xy-x}, x > 0, y > 0\
0, otherwise
end{matrix}right.
$$
My Question is that intuitively that random variable X is independent of Y since X won't affect the value of Y, but I calculated $f_{X}(x)$ and $f_{Y}(y)$ and found that $f(x,y) neq f_{X}(x)f_{Y}(y)$
I want to why random variables X and Y are look independent and either value won't affect the value of the other random variable, but they are dependent actually?
Thank you!!!!!
probability random-variables independence
asked Jan 7 at 9:30
Haohao ChangHaohao Chang
81
$endgroup$
add a comment |
$begingroup$
I have some problem at how to determine if two random variables are independent or not. If X and Y are two continuous random variables and their join probability density function is listed below
$$
f(x,y) = left{begin{matrix}
frac{1}{2}x^{3}e^{-xy-x}, x > 0, y > 0\
0, otherwise
end{matrix}right.
$$
My Question is that intuitively that random variable X is independent of Y since X won't affect the value of Y, but I calculated $f_{X}(x)$ and $f_{Y}(y)$ and found that $f(x,y) neq f_{X}(x)f_{Y}(y)$
I want to why random variables X and Y are look independent and either value won't affect the value of the other random variable, but they are dependent actually?
Thank you!!!!!
probability random-variables independence
asked Jan 7 at 9:30
Haohao ChangHaohao Chang
81
$endgroup$
2
$begingroup$
Why do you think that the value of $X$ will not affect the value of $Y$?
$endgroup$
– drhab
Jan 7 at 9:35
$begingroup$
@drhab Since the the value of X doesn't bother which value of Y, Y can be any value.
$endgroup$
– Haohao Chang
Jan 7 at 10:08
$begingroup$
@HaohaoChang $$f(x,y)=underbrace{xe^{-xy}mathbf1_{y>0}}_{f_{Ymid X}(y)},underbrace{frac{1}{2}x^2e^{-x}mathbf1_{x>0}}_{f_X(x)}$$ Clearly distribution of $Ymid X$ depends on $X$, so no question of independence.
$endgroup$
– StubbornAtom
Jan 7 at 10:58
$begingroup$
Actually you are not answering my question, but just repeat what you think. Anyway, if $X=Y$ (ultimate dependence) and $X$ can take any value then also $Y$ can take any value. So the ability of $Y$ to take any value does not exclude the possibility of dependence. If there is no PDF $f(x,y)$ that can be written as a product $g(x)h(y)$ then there is no independence.
$endgroup$
– drhab
Jan 7 at 11:18
$begingroup$
@drhab Thank you for your answer. But I still get confused why X and Y are dependent in this case, can you point it out more generally, thank you for your help. I agree with you about X and Y is not independent if the PDF $f(x,y)$ can't be written as a product $g(x)h(y)$.
$endgroup$
– Haohao Chang
Jan 7 at 11:48
add a comment |
$begingroup$
I have some problem at how to determine if two random variables are independent or not. If X and Y are two continuous random variables and their join probability density function is listed below
$$
f(x,y) = left{begin{matrix}
frac{1}{2}x^{3}e^{-xy-x}, x > 0, y > 0\
0, otherwise
end{matrix}right.
$$
My Question is that intuitively that random variable X is independent of Y since X won't affect the value of Y, but I calculated $f_{X}(x)$ and $f_{Y}(y)$ and found that $f(x,y) neq f_{X}(x)f_{Y}(y)$
I want to why random variables X and Y are look independent and either value won't affect the value of the other random variable, but they are dependent actually?
Thank you!!!!!
probability random-variables independence
asked Jan 7 at 9:30
Haohao ChangHaohao Chang
81
$endgroup$
I have some problem at how to determine if two random variables are independent or not. If X and Y are two continuous random variables and their join probability density function is listed below
$$
f(x,y) = left{begin{matrix}
frac{1}{2}x^{3}e^{-xy-x}, x > 0, y > 0\
0, otherwise
end{matrix}right.
$$
My Question is that intuitively that random variable X is independent of Y since X won't affect the value of Y, but I calculated $f_{X}(x)$ and $f_{Y}(y)$ and found that $f(x,y) neq f_{X}(x)f_{Y}(y)$
I want to why random variables X and Y are look independent and either value won't affect the value of the other random variable, but they are dependent actually?
Thank you!!!!!
probability random-variables independence
probability random-variables independence
asked Jan 7 at 9:30
Haohao ChangHaohao Chang
81
asked Jan 7 at 9:30
Haohao ChangHaohao Chang
81
asked Jan 7 at 9:30
Haohao ChangHaohao Chang
81
asked Jan 7 at 9:30
Haohao ChangHaohao Chang
81
asked Jan 7 at 9:30
Haohao ChangHaohao Chang
81
81
2
$begingroup$
Why do you think that the value of $X$ will not affect the value of $Y$?
$endgroup$
– drhab
Jan 7 at 9:35
$begingroup$
@drhab Since the the value of X doesn't bother which value of Y, Y can be any value.
$endgroup$
– Haohao Chang
Jan 7 at 10:08
$begingroup$
@HaohaoChang $$f(x,y)=underbrace{xe^{-xy}mathbf1_{y>0}}_{f_{Ymid X}(y)},underbrace{frac{1}{2}x^2e^{-x}mathbf1_{x>0}}_{f_X(x)}$$ Clearly distribution of $Ymid X$ depends on $X$, so no question of independence.
$endgroup$
– StubbornAtom
Jan 7 at 10:58
$begingroup$
Actually you are not answering my question, but just repeat what you think. Anyway, if $X=Y$ (ultimate dependence) and $X$ can take any value then also $Y$ can take any value. So the ability of $Y$ to take any value does not exclude the possibility of dependence. If there is no PDF $f(x,y)$ that can be written as a product $g(x)h(y)$ then there is no independence.
$endgroup$
– drhab
Jan 7 at 11:18
$begingroup$
@drhab Thank you for your answer. But I still get confused why X and Y are dependent in this case, can you point it out more generally, thank you for your help. I agree with you about X and Y is not independent if the PDF $f(x,y)$ can't be written as a product $g(x)h(y)$.
$endgroup$
– Haohao Chang
Jan 7 at 11:48
add a comment |
2
$begingroup$
Why do you think that the value of $X$ will not affect the value of $Y$?
$endgroup$
– drhab
Jan 7 at 9:35
$begingroup$
@drhab Since the the value of X doesn't bother which value of Y, Y can be any value.
$endgroup$
– Haohao Chang
Jan 7 at 10:08
$begingroup$
@HaohaoChang $$f(x,y)=underbrace{xe^{-xy}mathbf1_{y>0}}_{f_{Ymid X}(y)},underbrace{frac{1}{2}x^2e^{-x}mathbf1_{x>0}}_{f_X(x)}$$ Clearly distribution of $Ymid X$ depends on $X$, so no question of independence.
$endgroup$
– StubbornAtom
Jan 7 at 10:58
$begingroup$
Actually you are not answering my question, but just repeat what you think. Anyway, if $X=Y$ (ultimate dependence) and $X$ can take any value then also $Y$ can take any value. So the ability of $Y$ to take any value does not exclude the possibility of dependence. If there is no PDF $f(x,y)$ that can be written as a product $g(x)h(y)$ then there is no independence.
$endgroup$
– drhab
Jan 7 at 11:18
$begingroup$
@drhab Thank you for your answer. But I still get confused why X and Y are dependent in this case, can you point it out more generally, thank you for your help. I agree with you about X and Y is not independent if the PDF $f(x,y)$ can't be written as a product $g(x)h(y)$.
$endgroup$
– Haohao Chang
Jan 7 at 11:48
2
2
$begingroup$
Why do you think that the value of $X$ will not affect the value of $Y$?
$endgroup$
– drhab
Jan 7 at 9:35
$begingroup$
Why do you think that the value of $X$ will not affect the value of $Y$?
$endgroup$
– drhab
Jan 7 at 9:35
$begingroup$
@drhab Since the the value of X doesn't bother which value of Y, Y can be any value.
$endgroup$
– Haohao Chang
Jan 7 at 10:08
$begingroup$
@drhab Since the the value of X doesn't bother which value of Y, Y can be any value.
$endgroup$
– Haohao Chang
Jan 7 at 10:08
$begingroup$
@HaohaoChang $$f(x,y)=underbrace{xe^{-xy}mathbf1_{y>0}}_{f_{Ymid X}(y)},underbrace{frac{1}{2}x^2e^{-x}mathbf1_{x>0}}_{f_X(x)}$$ Clearly distribution of $Ymid X$ depends on $X$, so no question of independence.
$endgroup$
– StubbornAtom
Jan 7 at 10:58
$begingroup$
@HaohaoChang $$f(x,y)=underbrace{xe^{-xy}mathbf1_{y>0}}_{f_{Ymid X}(y)},underbrace{frac{1}{2}x^2e^{-x}mathbf1_{x>0}}_{f_X(x)}$$ Clearly distribution of $Ymid X$ depends on $X$, so no question of independence.
$endgroup$
– StubbornAtom
Jan 7 at 10:58
$begingroup$
Actually you are not answering my question, but just repeat what you think. Anyway, if $X=Y$ (ultimate dependence) and $X$ can take any value then also $Y$ can take any value. So the ability of $Y$ to take any value does not exclude the possibility of dependence. If there is no PDF $f(x,y)$ that can be written as a product $g(x)h(y)$ then there is no independence.
$endgroup$
– drhab
Jan 7 at 11:18
$begingroup$
Actually you are not answering my question, but just repeat what you think. Anyway, if $X=Y$ (ultimate dependence) and $X$ can take any value then also $Y$ can take any value. So the ability of $Y$ to take any value does not exclude the possibility of dependence. If there is no PDF $f(x,y)$ that can be written as a product $g(x)h(y)$ then there is no independence.
$endgroup$
– drhab
Jan 7 at 11:18
$begingroup$
@drhab Thank you for your answer. But I still get confused why X and Y are dependent in this case, can you point it out more generally, thank you for your help. I agree with you about X and Y is not independent if the PDF $f(x,y)$ can't be written as a product $g(x)h(y)$.
$endgroup$
– Haohao Chang
Jan 7 at 11:48
$begingroup$
@drhab Thank you for your answer. But I still get confused why X and Y are dependent in this case, can you point it out more generally, thank you for your help. I agree with you about X and Y is not independent if the PDF $f(x,y)$ can't be written as a product $g(x)h(y)$.
$endgroup$
– Haohao Chang
Jan 7 at 11:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Fixing e.g. $x=1$ we get $f(1,y)=frac12e^{-y-1}$ for $y>0$ and $f(1,y)=0$ otherwise.
For some constant $c$ this function equalizes $ccdot f_Y(ymid X=1)$ where $f_Y(ymid X=1)$ denotes the PDF of $Y$ under condition $X=1$.
If we do the same for $x=2$ we get $f(2,y)=8e^{-2y-2}$ for $y>0$ and $f(1,y)=0$ otherwise.
For some constant $d$ this function equalizes $dcdot f_Y(ymid X=2)$ where $f_Y(ymid X=2)$ denotes the PDF of $Y$ under condition $X=2$.
Evidently the functions are of different shape which makes clear that $f_Y(ymid X=1)$ and $f_Y(ymid X=2)$ are definitely not the same.
Apparantly under condition $X=1$ the distribution of $Y$ differs from the distribution of $Y$ under condition $X=2$.
So the value that is taken by $X$ affects the distribution of $Y$.
This observation is enough to conclude that $Y$ and $X$ are not independent.
answered Jan 7 at 12:14
drhabdrhab
103k545136
$endgroup$
add a comment |
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$begingroup$
Fixing e.g. $x=1$ we get $f(1,y)=frac12e^{-y-1}$ for $y>0$ and $f(1,y)=0$ otherwise.
For some constant $c$ this function equalizes $ccdot f_Y(ymid X=1)$ where $f_Y(ymid X=1)$ denotes the PDF of $Y$ under condition $X=1$.
If we do the same for $x=2$ we get $f(2,y)=8e^{-2y-2}$ for $y>0$ and $f(1,y)=0$ otherwise.
For some constant $d$ this function equalizes $dcdot f_Y(ymid X=2)$ where $f_Y(ymid X=2)$ denotes the PDF of $Y$ under condition $X=2$.
Evidently the functions are of different shape which makes clear that $f_Y(ymid X=1)$ and $f_Y(ymid X=2)$ are definitely not the same.
Apparantly under condition $X=1$ the distribution of $Y$ differs from the distribution of $Y$ under condition $X=2$.
So the value that is taken by $X$ affects the distribution of $Y$.
This observation is enough to conclude that $Y$ and $X$ are not independent.
answered Jan 7 at 12:14
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Fixing e.g. $x=1$ we get $f(1,y)=frac12e^{-y-1}$ for $y>0$ and $f(1,y)=0$ otherwise.
For some constant $c$ this function equalizes $ccdot f_Y(ymid X=1)$ where $f_Y(ymid X=1)$ denotes the PDF of $Y$ under condition $X=1$.
If we do the same for $x=2$ we get $f(2,y)=8e^{-2y-2}$ for $y>0$ and $f(1,y)=0$ otherwise.
For some constant $d$ this function equalizes $dcdot f_Y(ymid X=2)$ where $f_Y(ymid X=2)$ denotes the PDF of $Y$ under condition $X=2$.
Evidently the functions are of different shape which makes clear that $f_Y(ymid X=1)$ and $f_Y(ymid X=2)$ are definitely not the same.
Apparantly under condition $X=1$ the distribution of $Y$ differs from the distribution of $Y$ under condition $X=2$.
So the value that is taken by $X$ affects the distribution of $Y$.
This observation is enough to conclude that $Y$ and $X$ are not independent.
answered Jan 7 at 12:14
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Fixing e.g. $x=1$ we get $f(1,y)=frac12e^{-y-1}$ for $y>0$ and $f(1,y)=0$ otherwise.
For some constant $c$ this function equalizes $ccdot f_Y(ymid X=1)$ where $f_Y(ymid X=1)$ denotes the PDF of $Y$ under condition $X=1$.
If we do the same for $x=2$ we get $f(2,y)=8e^{-2y-2}$ for $y>0$ and $f(1,y)=0$ otherwise.
For some constant $d$ this function equalizes $dcdot f_Y(ymid X=2)$ where $f_Y(ymid X=2)$ denotes the PDF of $Y$ under condition $X=2$.
Evidently the functions are of different shape which makes clear that $f_Y(ymid X=1)$ and $f_Y(ymid X=2)$ are definitely not the same.
Apparantly under condition $X=1$ the distribution of $Y$ differs from the distribution of $Y$ under condition $X=2$.
So the value that is taken by $X$ affects the distribution of $Y$.
This observation is enough to conclude that $Y$ and $X$ are not independent.
answered Jan 7 at 12:14
drhabdrhab
103k545136
$endgroup$
Fixing e.g. $x=1$ we get $f(1,y)=frac12e^{-y-1}$ for $y>0$ and $f(1,y)=0$ otherwise.
For some constant $c$ this function equalizes $ccdot f_Y(ymid X=1)$ where $f_Y(ymid X=1)$ denotes the PDF of $Y$ under condition $X=1$.
If we do the same for $x=2$ we get $f(2,y)=8e^{-2y-2}$ for $y>0$ and $f(1,y)=0$ otherwise.
For some constant $d$ this function equalizes $dcdot f_Y(ymid X=2)$ where $f_Y(ymid X=2)$ denotes the PDF of $Y$ under condition $X=2$.
Evidently the functions are of different shape which makes clear that $f_Y(ymid X=1)$ and $f_Y(ymid X=2)$ are definitely not the same.
Apparantly under condition $X=1$ the distribution of $Y$ differs from the distribution of $Y$ under condition $X=2$.
So the value that is taken by $X$ affects the distribution of $Y$.
This observation is enough to conclude that $Y$ and $X$ are not independent.
answered Jan 7 at 12:14
drhabdrhab
103k545136
answered Jan 7 at 12:14
drhabdrhab
103k545136
answered Jan 7 at 12:14
drhabdrhab
103k545136
answered Jan 7 at 12:14
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
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2
$begingroup$
Why do you think that the value of $X$ will not affect the value of $Y$?
$endgroup$
– drhab
Jan 7 at 9:35
$begingroup$
@drhab Since the the value of X doesn't bother which value of Y, Y can be any value.
$endgroup$
– Haohao Chang
Jan 7 at 10:08
$begingroup$
@HaohaoChang $$f(x,y)=underbrace{xe^{-xy}mathbf1_{y>0}}_{f_{Ymid X}(y)},underbrace{frac{1}{2}x^2e^{-x}mathbf1_{x>0}}_{f_X(x)}$$ Clearly distribution of $Ymid X$ depends on $X$, so no question of independence.
$endgroup$
– StubbornAtom
Jan 7 at 10:58
$begingroup$
Actually you are not answering my question, but just repeat what you think. Anyway, if $X=Y$ (ultimate dependence) and $X$ can take any value then also $Y$ can take any value. So the ability of $Y$ to take any value does not exclude the possibility of dependence. If there is no PDF $f(x,y)$ that can be written as a product $g(x)h(y)$ then there is no independence.
$endgroup$
– drhab
Jan 7 at 11:18
$begingroup$
@drhab Thank you for your answer. But I still get confused why X and Y are dependent in this case, can you point it out more generally, thank you for your help. I agree with you about X and Y is not independent if the PDF $f(x,y)$ can't be written as a product $g(x)h(y)$.
$endgroup$
– Haohao Chang
Jan 7 at 11:48