Mixing
Given a pretty complex joint PDF of X,Y, what can I derive about them?
$begingroup$
this is the kind of questions where from just looking at you realize that the answer might not be straightforward:
$$f_{X,Y}(x,y)begin{cases}
e^{-y}e^{-xy}y^2 & y>0 ,xgeq0\
0 & else
end{cases}
$$
And the questions is: What can you conclude about X,Y? (Are they independent? $Cov(X,Y)>0$ ? $COV(X,Y)<0$? are they coordinated? etc.
Well, as a newbie, I started by calculating $ f_X(x)$. It didn't work out. It was too complicated and I got stuck. I then turned to $f_Y(y)$ which gave me $f_Y(y) = ye^{-y}$, but I couldn't figure out what to do with that data.
So I then though that this is probably deeper than just a technically solvable problem, there's probably a concept behind it.
Could you please enlighten me?
Thanks!
probability random-variables
asked Jan 23 at 15:33
superuser123superuser123
48628
$endgroup$
add a comment |
$begingroup$
this is the kind of questions where from just looking at you realize that the answer might not be straightforward:
$$f_{X,Y}(x,y)begin{cases}
e^{-y}e^{-xy}y^2 & y>0 ,xgeq0\
0 & else
end{cases}
$$
And the questions is: What can you conclude about X,Y? (Are they independent? $Cov(X,Y)>0$ ? $COV(X,Y)<0$? are they coordinated? etc.
Well, as a newbie, I started by calculating $ f_X(x)$. It didn't work out. It was too complicated and I got stuck. I then turned to $f_Y(y)$ which gave me $f_Y(y) = ye^{-y}$, but I couldn't figure out what to do with that data.
So I then though that this is probably deeper than just a technically solvable problem, there's probably a concept behind it.
Could you please enlighten me?
Thanks!
probability random-variables
asked Jan 23 at 15:33
superuser123superuser123
48628
$endgroup$
$begingroup$
begin{align} f_{X,Y}(x,y)&=y^2e^{-(1+x)y}mathbf1_{y>0,x>0} \&=underbrace{frac{(1+x)^3}{2}y^2e^{-(1+x)y}mathbf1_{y>0}}_{f_{Ymid X=x}(y)}frac{2}{(1+x)^3}mathbf1_{x>0} end{align} It is clear that $Ymid X$ has a Gamma density which depends on $X$, hence $X,Y$ are not independent.
$endgroup$
– StubbornAtom
Jan 24 at 7:17
add a comment |
$begingroup$
this is the kind of questions where from just looking at you realize that the answer might not be straightforward:
$$f_{X,Y}(x,y)begin{cases}
e^{-y}e^{-xy}y^2 & y>0 ,xgeq0\
0 & else
end{cases}
$$
And the questions is: What can you conclude about X,Y? (Are they independent? $Cov(X,Y)>0$ ? $COV(X,Y)<0$? are they coordinated? etc.
Well, as a newbie, I started by calculating $ f_X(x)$. It didn't work out. It was too complicated and I got stuck. I then turned to $f_Y(y)$ which gave me $f_Y(y) = ye^{-y}$, but I couldn't figure out what to do with that data.
So I then though that this is probably deeper than just a technically solvable problem, there's probably a concept behind it.
Could you please enlighten me?
Thanks!
probability random-variables
asked Jan 23 at 15:33
superuser123superuser123
48628
$endgroup$
this is the kind of questions where from just looking at you realize that the answer might not be straightforward:
$$f_{X,Y}(x,y)begin{cases}
e^{-y}e^{-xy}y^2 & y>0 ,xgeq0\
0 & else
end{cases}
$$
And the questions is: What can you conclude about X,Y? (Are they independent? $Cov(X,Y)>0$ ? $COV(X,Y)<0$? are they coordinated? etc.
Well, as a newbie, I started by calculating $ f_X(x)$. It didn't work out. It was too complicated and I got stuck. I then turned to $f_Y(y)$ which gave me $f_Y(y) = ye^{-y}$, but I couldn't figure out what to do with that data.
So I then though that this is probably deeper than just a technically solvable problem, there's probably a concept behind it.
Could you please enlighten me?
Thanks!
probability random-variables
probability random-variables
asked Jan 23 at 15:33
superuser123superuser123
48628
asked Jan 23 at 15:33
superuser123superuser123
48628
asked Jan 23 at 15:33
superuser123superuser123
48628
asked Jan 23 at 15:33
superuser123superuser123
48628
asked Jan 23 at 15:33
superuser123superuser123
48628
48628
$begingroup$
begin{align} f_{X,Y}(x,y)&=y^2e^{-(1+x)y}mathbf1_{y>0,x>0} \&=underbrace{frac{(1+x)^3}{2}y^2e^{-(1+x)y}mathbf1_{y>0}}_{f_{Ymid X=x}(y)}frac{2}{(1+x)^3}mathbf1_{x>0} end{align} It is clear that $Ymid X$ has a Gamma density which depends on $X$, hence $X,Y$ are not independent.
$endgroup$
– StubbornAtom
Jan 24 at 7:17
add a comment |
$begingroup$
begin{align} f_{X,Y}(x,y)&=y^2e^{-(1+x)y}mathbf1_{y>0,x>0} \&=underbrace{frac{(1+x)^3}{2}y^2e^{-(1+x)y}mathbf1_{y>0}}_{f_{Ymid X=x}(y)}frac{2}{(1+x)^3}mathbf1_{x>0} end{align} It is clear that $Ymid X$ has a Gamma density which depends on $X$, hence $X,Y$ are not independent.
$endgroup$
– StubbornAtom
Jan 24 at 7:17
$begingroup$
begin{align} f_{X,Y}(x,y)&=y^2e^{-(1+x)y}mathbf1_{y>0,x>0} \&=underbrace{frac{(1+x)^3}{2}y^2e^{-(1+x)y}mathbf1_{y>0}}_{f_{Ymid X=x}(y)}frac{2}{(1+x)^3}mathbf1_{x>0} end{align} It is clear that $Ymid X$ has a Gamma density which depends on $X$, hence $X,Y$ are not independent.
$endgroup$
– StubbornAtom
Jan 24 at 7:17
$begingroup$
begin{align} f_{X,Y}(x,y)&=y^2e^{-(1+x)y}mathbf1_{y>0,x>0} \&=underbrace{frac{(1+x)^3}{2}y^2e^{-(1+x)y}mathbf1_{y>0}}_{f_{Ymid X=x}(y)}frac{2}{(1+x)^3}mathbf1_{x>0} end{align} It is clear that $Ymid X$ has a Gamma density which depends on $X$, hence $X,Y$ are not independent.
$endgroup$
– StubbornAtom
Jan 24 at 7:17
add a comment |
1 Answer
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$begingroup$
The random variables $X$ and $Y$ are independent iff the joint density is the product of the marginal densities. You have found the density of $Y$. For the density of $X$, note that
$$
f_{X}(x)=int_0^infty f(x,y), dy=frac{2}{(x+1)^3}int_{0}^infty frac{(x+1)^3}{2}y^{2}e^{-y(x+1)}, dy=frac{2}{(x+1)^3}quad (xgeq0)
$$
where the integral equals one since it is the integral of a gamma density with shape parameter $3$ and rate parameter $x+1$.
It follows that $X$ and $Y$ are not independent.
answered Jan 23 at 15:45
Foobaz JohnFoobaz John
22.6k41452
$endgroup$
add a comment |
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1 Answer
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$begingroup$
The random variables $X$ and $Y$ are independent iff the joint density is the product of the marginal densities. You have found the density of $Y$. For the density of $X$, note that
$$
f_{X}(x)=int_0^infty f(x,y), dy=frac{2}{(x+1)^3}int_{0}^infty frac{(x+1)^3}{2}y^{2}e^{-y(x+1)}, dy=frac{2}{(x+1)^3}quad (xgeq0)
$$
where the integral equals one since it is the integral of a gamma density with shape parameter $3$ and rate parameter $x+1$.
It follows that $X$ and $Y$ are not independent.
answered Jan 23 at 15:45
Foobaz JohnFoobaz John
22.6k41452
$endgroup$
add a comment |
$begingroup$
The random variables $X$ and $Y$ are independent iff the joint density is the product of the marginal densities. You have found the density of $Y$. For the density of $X$, note that
$$
f_{X}(x)=int_0^infty f(x,y), dy=frac{2}{(x+1)^3}int_{0}^infty frac{(x+1)^3}{2}y^{2}e^{-y(x+1)}, dy=frac{2}{(x+1)^3}quad (xgeq0)
$$
where the integral equals one since it is the integral of a gamma density with shape parameter $3$ and rate parameter $x+1$.
It follows that $X$ and $Y$ are not independent.
answered Jan 23 at 15:45
Foobaz JohnFoobaz John
22.6k41452
$endgroup$
add a comment |
$begingroup$
The random variables $X$ and $Y$ are independent iff the joint density is the product of the marginal densities. You have found the density of $Y$. For the density of $X$, note that
$$
f_{X}(x)=int_0^infty f(x,y), dy=frac{2}{(x+1)^3}int_{0}^infty frac{(x+1)^3}{2}y^{2}e^{-y(x+1)}, dy=frac{2}{(x+1)^3}quad (xgeq0)
$$
where the integral equals one since it is the integral of a gamma density with shape parameter $3$ and rate parameter $x+1$.
It follows that $X$ and $Y$ are not independent.
answered Jan 23 at 15:45
Foobaz JohnFoobaz John
22.6k41452
$endgroup$
The random variables $X$ and $Y$ are independent iff the joint density is the product of the marginal densities. You have found the density of $Y$. For the density of $X$, note that
$$
f_{X}(x)=int_0^infty f(x,y), dy=frac{2}{(x+1)^3}int_{0}^infty frac{(x+1)^3}{2}y^{2}e^{-y(x+1)}, dy=frac{2}{(x+1)^3}quad (xgeq0)
$$
where the integral equals one since it is the integral of a gamma density with shape parameter $3$ and rate parameter $x+1$.
It follows that $X$ and $Y$ are not independent.
answered Jan 23 at 15:45
Foobaz JohnFoobaz John
22.6k41452
answered Jan 23 at 15:45
Foobaz JohnFoobaz John
22.6k41452
answered Jan 23 at 15:45
Foobaz JohnFoobaz John
22.6k41452
answered Jan 23 at 15:45
Foobaz JohnFoobaz John
22.6k41452
22.6k41452
add a comment |
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$begingroup$
begin{align} f_{X,Y}(x,y)&=y^2e^{-(1+x)y}mathbf1_{y>0,x>0} \&=underbrace{frac{(1+x)^3}{2}y^2e^{-(1+x)y}mathbf1_{y>0}}_{f_{Ymid X=x}(y)}frac{2}{(1+x)^3}mathbf1_{x>0} end{align} It is clear that $Ymid X$ has a Gamma density which depends on $X$, hence $X,Y$ are not independent.
$endgroup$
– StubbornAtom
Jan 24 at 7:17