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Conditional probability with marginal and joint density

  • Thread starter Linder88
  • Start date
24
0
1. The problem statement, all variables and given/known data
Determine ##P(X<Y|x>0)##
2. Relevant 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
$$
f_X(x)=2x\\
f_Y(y)=4y^3
$$
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

Science Advisor
Homework Helper
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Gold Member
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1. The problem statement, all variables and given/known data
Determine ##P(X<Y|x>0)##
2. Relevant 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
$$
f_X(x)=2x\\
f_Y(y)=4y^3
$$
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

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1. The problem statement, all variables and given/known data
Determine ##P(X<Y|x>0)##
2. Relevant 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
$$
f_X(x)=2x\\
f_Y(y)=4y^3
$$
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.
 
24
0
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

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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
[tex] f_X(x) = \frac{1}{2} |x| ( 1 - x^2),\; 0 \leq x \leq 1 [/tex]
Unlike your answer, this has ##f_X > 0## for both ##-1 < x < 0## and ##0 < x < 1##.
 

LCKurtz

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Gold Member
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714
For the record: your final answer is correct, but you have computed ##f_X(x)## incorrectly. You should have gotten
[tex] f_X(x) = \frac{1}{2} |x| ( 1 - x^2),\; 0 \leq x \leq 1 [/tex]
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

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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.
 

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