Conditional probability with marginal and joint density

[Linder88]

Homework Statement

Determine $P(X<Y|x>0)$

Homework Equations

X and Y are random variables with the joint density function
$$
f_{XY}(x,y)=
\begin{cases}
4|xy|,-y<x<y,0<y<1\\
0,elsewhere
\end{cases}$$
The marginal densities are given by
$$
f_X(x)=2x\\
f_Y(y)=4y^3
$$

The Attempt at a Solution

The formula for conditional probability is
$$
P(B|A)=\frac{P(B\cap A)}{P(A)}
$$
In this case we have
$$
P(X<Y|x>0)=\frac{P(X<Y\cap x>0)}{P(x>0)}=\frac{F_{XY}(x,y)}{F_X(x)}=\frac{\int_{}4|xy|dxdy}{\int_0^{\infty}2xdx}
$$
This is where I get stuck, I do not know what boundaries I should put for the joint density function, can someone please help me?

 

[LCKurtz]

Homework Statement

Determine $P(X<Y|x>0)$

Homework Equations

X and Y are random variables with the joint density function
$$
f_{XY}(x,y)=
\begin{cases}
4|xy|,-y<x<y,0<y<1\\
0,elsewhere
\end{cases}$$
The marginal densities are given by
$$
f_X(x)=2x\\
f_Y(y)=4y^3
$$

The Attempt at a Solution

The formula for conditional probability is
$$
P(B|A)=\frac{P(B\cap A)}{P(A)}
$$
In this case we have
$$
P(X<Y|x>0)=\frac{P(X<Y\cap x>0)}{P(x>0)}=\frac{F_{XY}(x,y)}{F_X(x)}=\frac{\int_{}4|xy|dxdy}{\int_0^{\infty}2xdx}
$$
This is where I get stuck, I do not know what boundaries I should put for the joint density function, can someone please help me?

Draw the lines $x=y$, $x=-y$, $y=0$, and $y=1$. Shade the part where x and y are between the appropriate lines. The limits are just like in any double integral in calculus. Also, I didn't check your work but surely if the marginal density is $2x$, it wouldn't be valid clear to $\infty$.

 

[Ray Vickson]

Homework Statement

Determine $P(X<Y|x>0)$

Homework Equations

X and Y are random variables with the joint density function
$$
f_{XY}(x,y)=
\begin{cases}
4|xy|,-y<x<y,0<y<1\\
0,elsewhere
\end{cases}$$
The marginal densities are given by
$$
f_X(x)=2x\\
f_Y(y)=4y^3
$$

The Attempt at a Solution

The formula for conditional probability is
$$
P(B|A)=\frac{P(B\cap A)}{P(A)}
$$
In this case we have
$$
P(X<Y|x>0)=\frac{P(X<Y\cap x>0)}{P(x>0)}=\frac{F_{XY}(x,y)}{F_X(x)}=\frac{\int_{}4|xy|dxdy}{\int_0^{\infty}2xdx}
$$
This is where I get stuck, I do not know what boundaries I should put for the joint density function, can someone please help me?

You claimed $f_X(x) = 2x$ cannot possibly be correct. In the sample space, $x$ is allowed to be both $< 0$ and $>0$, and when $x < 0$ your formula delivers a negative probability.

 

[Linder88]

Draw the lines $x=y$, $x=-y$, $y=0$, and $y=1$. Shade the part where x and y are between the appropriate lines. The limits are just like in any double integral in calculus. Also, I didn't check your work but surely if the marginal density is $2x$, it wouldn't be valid clear to $\infty$.

The shaded area is a inverted triangle in the upper half of the plane with y vertical and x horizontal, so we have that
$$
P(X<Y|x>0)=\frac{\int_0^1\int_{-y}^y4|xy|dxdy}{\int_0^12xdx}=\frac{1}{1}=1
$$
I guess this concludes this topic, thanks :)

 

[Ray Vickson]

The shaded area is a inverted triangle in the upper half of the plane with y vertical and x horizontal, so we have that
$$
P(X<Y|x>0)=\frac{\int_0^1\int_{-y}^y4|xy|dxdy}{\int_0^12xdx}=\frac{1}{1}=1
$$
I guess this concludes this topic, thanks :)

For the record: your final answer is correct, but you have computed $f_X(x)$ incorrectly. You should have gotten
$$f_X(x) = \frac{1}{2} |x| ( 1 - x^2),\; 0 \leq x \leq 1$$
Unlike your answer, this has $f_X > 0$ for both $-1 < x < 0$ and $0 < x < 1$.

 

[LCKurtz]

For the record: your final answer is correct, but you have computed $f_X(x)$ incorrectly. You should have gotten
$$f_X(x) = \frac{1}{2} |x| ( 1 - x^2),\; 0 \leq x \leq 1$$
Unlike your answer, this has $f_X > 0$ for both $-1 < x < 0$ and $0 < x < 1$.

Did you mean $$2*|x| ( 1 - x^2),\; 0 \leq x \leq 1\text{ ?}$$

 

[Ray Vickson]

Did you mean $$2*|x| ( 1 - x^2),\; 0 \leq x \leq 1\text{ ?}$$

Yes. I accidentally omitted the factor '4' in front, so got 1/2 instead of 4/2 = 2.