The joint density of $(X,Y)$ is $f_{XY}(x,y)=k(1-\sqrt{\frac{y}{x}}),0<x<1,0<y<x$.
- Find the value of $k$ and say if $X$ and $Y$ are independent or not.
$\rightarrow k=6 \Rightarrow f_{XY}(x,y)=6(1-\sqrt{\frac{y}{x}})$; $f_X(x)=2x$ and $f_Y(y)=6(1+y-2^{\frac{1}{2}})$, so $f_X(x)f_Y(y)\neq f_{XY}(x,y)\Rightarrow X,Y$ are not independent
- Let $V=X$ and $Z=\sqrt{\frac{Y}{X}}$. Say if $V$ and $Z$ are independent or not.
$\rightarrow cov(V,Z)=\mathbb{E}(VZ)-\mathbb{E}(V)\mathbb{E}(Z)=\mathbb{E}(X^{\frac{1}{2}}Y^{\frac{1}{2}})-\mathbb{E}(X)\mathbb{E}(X^{-\frac{1}{2}}Y^{\frac{1}{2}})=\frac{1}{3}-\frac{2}{3}\cdot \frac{1}{3}=0 \Rightarrow V \perp Z$
- Find the joint density of $(V,Z)$.
Well, I'm having a bit of a problem with this: indeed I know that if $X \perp Y \Rightarrow f_{VZ}(v,z)=f_X(x)f_Y(y)|\det(J)|$, but in this case $X$ and $Y$ are not independent. So I thought I'd do in this way:
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$F_{V,Z}(v,z)=\mathbb{P}(V\geq v,Z\geq z)=\mathbb{P}(X\geq x, \sqrt{\frac{Y}{X}\geq z})$ $=\mathbb{P}(X\geq x,X>\frac{y}{z^2})=\mathbb{P}(\frac{y}{z^2}<X<x)$;
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$f_{X|Y}(x|y):=\frac{f_{XY}(x,y)}{f_Y(y)}=\frac{x^{\frac{1}{2}}-y^{\frac{1}{2}}}{x^{\frac{1}{2}}(1+y-2y^{\frac{1}{2}})}$;
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$\mathbb{P}(\frac{y}{z^2}<X<x|Y=y)=\int_{\frac{y}{z^2}}^{x}f_{X|Y}(x|y)dx$ and then $\mathbb{P}(\frac{y}{z^2}<X<x)=\int_{\frac{y}{z^2}}^{x}\mathbb{P}(\frac{y}{z^2}<X<x|Y=y)f_X(x)dx$
Is it correct? Thanks in advance.
Best Answer
To say if $X,Y$ are independent or not no calculation is needed. In fact, Necessary Condition for independence is that the joint domain is rectangular....you have a triangle so they are NOT independent
To calculate the joint density of $(Z,V)$ let's use the Fundamental Transformation Theorem, setting
$$ \begin{cases} v=x \\ z=\sqrt{\frac{y}{x}} \end{cases}$$
$$ \begin{cases} x=v \\ y=z^2v \end{cases}$$
The jacobian is $|J|=2zv$
So the joint density is
$$f_{ZV}(z,v)=12(1-z)zv\cdot\mathbb{1}_{(0;1)}(z)\cdot\mathbb{1}_{(0;1)}(v)$$
It is very simple to see that
$$f_{ZV}(z,v)=6z(1-z)\mathbb{1}_{(0;1)}(z)\times 2v\mathbb{1}_{(0;1)}(v)=f_V(v)f_Z(z)$$
Thus the rv's V and Z are independent.
IMPORTANT OBSERVATION
The calculation of the covariance $Cov(Z,V)$ is not requested and finding $Cov(Z,V)=0$ says nothing about independence.
Further explanation
Fact: $X$ and $Y$ are not independent as their join domain is a triangle:
$$\mathbb{1}_{(0;1)}(x)\mathbb{1}_{(0;x)}(y)$$
using the above system
$$ \begin{cases} x=v \\ y=z^2v \end{cases}$$
we calculate the jacobian
$$ |J|= \begin{bmatrix} \frac{\partial x}{\partial v} & \frac{\partial x}{\partial z} \\ \frac{\partial y}{\partial v}&\frac{\partial x}{\partial z} \\ \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ z^2&2zv \\ \end{bmatrix}=2zv $$
Now simply changing the variables (this is called Fundamental Transformation theorem but it is just a variables' change) you get
$$f_{ZV}(z,v)=f_{XY}(z,v)\times |J|=6(1-z)\times 2zv=12zv(1-z)$$
When I look at this density, defined on the unit square $[0;1]\times[0;1]$ it is immediately to identify that
$f_{ZV}=f_Z\times f_V=6z(1-z)\times 2v$
If this is an issue for you, no problem at all! integrate with respect to z to derive $f_V$ and viceversa:
$f_V=\int_0^1 f_{ZV}(z,v)dz=2v$
$f_Z=\int_0^1 f_{ZV}(z,v)dv=6z(1-z)$
Further question in same exercise
It is also requested to calculate $\mathbb{P}[V^2+Z^2>\frac{1}{2}]$
Rmemebering that $z^2+v^2=\frac{1}{2}$ is a circumference with center in the origin and radius $r=\frac{\sqrt{2}}{2}$ we get
$$\mathbb{P}[V^2+Z^2<\frac{1}{2}]=\int_0^{\frac{\sqrt{2}}{2}}6z(1-z)dz\int_0^{\sqrt{\frac{1}{2}-z^2}}2v dv\approx 0.234$$
Thus
$$\mathbb{P}[V^2+Z^2>\frac{1}{2}]\approx 0.760$$