Little help needed

[Shaun2357]

An insurance company insures a large number of drivers. Let X be the
random variable representing the company's losses under collision insurance,
and let Y represent the company's losses under liability insurance. X and Y
have joint density function
fXY (x; y) =
 (2x +2 - y)/4 0 < x < 1; 0 < y < 2
0 otherwise
What is the probability that the total loss is at least 1 ?

They want P(X + Y >= 1)?

[Sinkingfun]

draw a graph. X + Y > 1 is the same as y > 1-x. Then you just do calculus for the shaded area above the triangle in the graph. When I did it, I got 17/24.

Bounds can be y going from 1-x to 2 and x going from 0 to 1.

Or you could use the bounds x going from 0 to 1 and y going from 0 to 1-x. Once you arrive at the answer, do 1 - the answer.

[Shaun2357]

Originally Posted by Sinkingfun

draw a graph. X + Y > 1 is the same as y > 1-x. Then you just do calculus for the shaded area above the triangle in the graph. When I did it, I got 17/24.

Bounds can be y going from 1-x to 2 and x going from 0 to 1.

Or you could use the bounds x going from 0 to 1 and y going from 0 to 1-x. Once you arrive at the answer, do 1 - the answer.

That's exactly what I did. Then I guess it's an adding mistake. Thanks though

[Sinkingfun]

it's best to not rush the calculus unless you are a speed demon.

[Shaun2357]

True. Found out it was an adding mistake.

[Shaun2357]

Problem 30.3
Let X and Y be random variables with joint pdf given by
fxy(x,y) =

6(1−y) 0≤x≤y≤1
0 otherwise
(a) Find P(X ≤ 3/4,Y ≥ 1/2)

I am getting a little confused here with respect to finding out a. A little help is all I need. I don't expect someone to work out the integral, that I can do myself. But what should I integrate from?

[nonlnear]

If the answer isn't jumping off the page at you, it's usually best to draw a diagram of the domain of integration. When doing this forget about the calculus and just focus on getting the domain right. Once you have the diagram complete, then evaluate the integral.

Could you clarify what the lower bound on x is? Is it 0, or 6(1-y), the product (which is strictly speaking what you wrote, but likely not what you meant to write), or both?

[Gandalf]

Originally Posted by nonlnear

If the answer isn't jumping off the page at you, it's usually best to draw a diagram of the domain of integration. When doing this forget about the calculus and just focus on getting the domain right. Once you have the diagram complete, then evaluate the integral.

In particular, here you should consider whether these points should be included in your integration. If you classify all these correctly, you probably have the correct region: (.4,.7), (.7,.4), (.6,.7), (.7,.6), (.4,.8),(.8,.4) I have 3 of those inside the region.

Could you clarify what the lower bound on x is? Is it 0, or 6(1-y), the product (which is strictly speaking what you wrote, but likely not what you meant to write), or both?

I interpret what he wrote as 0 is the lower limit. 6(1-y) is the density, for combinations of x and y where there is density.

[nonlnear]

Originally Posted by Gandalf

I interpret what he wrote as 0 is the lower limit. 6(1-y) is the density, for combinations of x and y where there is density.

Now that you mention it, that seems the best interpretation. I got a funny character after f(x,y) which made a line break, making everything look like the bounds expression. If 6(1-y) is the density function that simplifies things a bit. If the bounds are 0 < x < y < 1 then you can construct a diagram as follow. For each inequality (look at the list below and make sure you understand why each of them is contained in that one bounds statement) set up a separate inequality:
0 ≤ x
x ≤ y
0 ≤ y
x ≤ 1
y ≤ 1
All of these must be satisfied simultaneously, so after plotting all of them there is only one region in that mess of lines that represents your domain.
Plot each of these lines, and as you plot each one shade the part that is not part of your domain. Whatever is left unshaded when you're done is your domain. You're probably used to shading the region of integration during setup, but if you shade the desired part of each inequality then it's not as simple as just picking out the shaded region; you'll have to pick out the most darkly shaded region, which could be problematic. It's safer to shade the region that you aren't integrating over. After you're done identifying the region of integration you can redraw your diagram and shade the domain if you so prefer.

[Shaun2357]

Originally Posted by nonlnear

Now that you mention it, that seems the best interpretation. I got a funny character after f(x,y) which made a line break, making everything look like the bounds expression. If 6(1-y) is the density function that simplifies things a bit. If the bounds are 0 < x < y < 1 then you can construct a diagram as follow. For each inequality (look at the list below and make sure you understand why each of them is contained in that one bounds statement) set up a separate inequality:
0 ≤ x
x ≤ y
0 ≤ y
x ≤ 1
y ≤ 1
All of these must be satisfied simultaneously, so after plotting all of them there is only one region in that mess of lines that represents your domain.
Plot each of these lines, and as you plot each one shade the part that is not part of your domain. Whatever is left unshaded when you're done is your domain. You're probably used to shading the region of integration during setup, but if you shade the desired part of each inequality then it's not as simple as just picking out the shaded region; you'll have to pick out the most darkly shaded region, which could be problematic. It's safer to shade the region that you aren't integrating over. After you're done identifying the region of integration you can redraw your diagram and shade the domain if you so prefer.

Sorry about that. I copied and pasted from a pdf file and didn't realize that it doesn't copy correctly.