Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Two random variables X and Y have a constant joint density function $f(x,y)=c$ in the domain ${x<4,y>0,x>y}$ (and zero elsewhere).

a)Find c.

b) Find $P(X-Y<2)$

c)Find $P(X^2+Y^2<1)$

This is probably a stupid question, but given the domain does it make a triangle? If not, what shape does it form?

For (a) isnt it just 1/area of the domain? I am unsure of what region (b) makes in the domain and (c) is the unit circle, but what is holding me back on this one is not knowing the overall shape of the domain. Studying for a test so any help would be great.

share|improve this question
2  
The easiest way of getting at the answers you need, and also the one that is almost universally resisted by students - they refuse to follow it - is to draw a sketch of the plane with coordinate axes $x$ and $y$ and figure out where the density is nonzero. In this case, you are told that $x<4$, so draw a vertical line at $x=4$ and note that $(X,Y)$ must be to the left of the line, and also above the $x$ axis since you are also told that $y > 0$. Now figure out which points correspond to $x > y$. For (a) you are correct. For (b), sketch the line $y = 2+x$, for (c), the circle $x^2+y^2=1$. –  Dilip Sarwate Nov 2 '12 at 2:33
add comment

2 Answers

up vote 2 down vote accepted

You're right, the domain is a triangle. You can make a diagram of the domain by plotting the lines that bound it, whose equations you get by replacing the comparison operators by equalities. For instance, the boundary of the half-plane specified by $x\lt4$ is the line with equation $x=4$.

Your answer to (a) is right. For (b), apply the same approach as above for the boundaries of the domain. For (c), you're right, or more precisely, the boundary is the unit circle and the region it bounds is the unit disk.

share|improve this answer
add comment

I hope the following visualization is useful:

enter image description here

Probabilities can be computed geometrically as ratios of areas with the answers: $$ \mathbb{P}(X-Y<2) = 1-\frac{1}{4} = \frac{3}{4} $$ $$ \mathbb{P}(X^2+Y^2<1) = \frac{\frac{\pi}{8}}{ \frac{4 \cdot 4}{2} } = \frac{\pi}{64} $$

share|improve this answer
    
@Sasha-For the first one why is it 3/4? I see how you did it, but why is that so? Cant you subtract out the smaller triangle from the larger and get 4-2=2? –  Sprock Nov 2 '12 at 2:49
1  
The area of the purple trapezoid is the difference in areas of large triangle and the small blue triangle. The blue triangle has twice smaller side-length, thus its area is 1/4 of the area of the whole region. Thus the probability is $\frac{A-A/4}{A} = \frac{3}{4}$. –  Sasha Nov 2 '12 at 2:51
    
Now I see. Thanks –  Sprock Nov 2 '12 at 2:54
2  
@Sprock I think an even easier geometric argument is that the big triangle is divided into $4$ triangles congruent to the small blue triangle and so the desired probability (that the random point lies in the small blue triangle) is $\frac{1}{4}$. Don't even need to know the formula for the area of a triangle to get the answer! –  Dilip Sarwate Nov 2 '12 at 3:05
1  
Yes, the intersection of region $X^2+Y^2<1$ and the support of your random vector is $1/8$ of a circle, hence the area is $\pi/8$. Then divide over the area of the triangle. –  Sasha Nov 2 '12 at 3:05
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.