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Problems on Joint Distribution and Risk Models

Please find my questions from the attachment. I tried to do those questions but failed to get the answers. I really need you guys help! It will be so great to have explanations other than the answers.

Thanks!
Original post by a5a09
Please find my questions from the attachment. I tried to do those questions but failed to get the answers. I really need you guys help! It will be so great to have explanations other than the answers.

Thanks!


What have you done for Q1(b)?

Q2, why are you trying to integrate? What's the question asking, you've only posted part of it.

Q5, no idea.
Reply 2
Original post by ghostwalker
What have you done for Q1(b)?

Q2, why are you trying to integrate? What's the question asking, you've only posted part of it.

Q5, no idea.

Thank you for your reply.

For Q1(b) I tried to integrate x from 0 to 1 and y from 0 to (1-x) but I doubt it is not correct as the result I obtain only depends on 1 variable (in my case it's x).

For Q2, The Fy(y)F_{y}(y) is actually a cdf of the individual claims to the insurer in which the individual loss in excess of 20 is covered by the reinsurer and what is less than or equal to 20 is covered by the insurer. Therefore, as I believe that E(Y)=02(1Fy(y)) dyE(Y) = \int^2_0 (1-F_{y}(y))\ dy, I have to integrate it which gives an error function.......
(edited 9 years ago)
Original post by a5a09
Thank you for your reply.

For Q1(b) I tried to integrate x from 0 to 1 and y from 0 to (1-x) but I doubt it is not correct as the result I obtain only depends on 1 variable (in my case it's x).


There are going to be several parts to this:
For a point inside the region x,y>0, x+y<=1, your integral limits will just be 0,x and 0,y,

Outside the region there will be three areas I think you need to consider.
x+y>1 and x< 1
x+y>1 and y <1 these will overlap, and
x+y>1, x>1, y>1


For Q2, The Fy(y)F_{y}(y) is actually a cdf of the individual claims to the insurer in which the individual loss in excess of 20 is covered by the reinsurer and what is less than or equal to 20 is covered by the insurer. Therefore, as I believe that E(Y)=02(1Fy(y)) dyE(Y) = \int^2_0 (1-F_{y}(y))\ dy, I have to integrate it which gives an error function.......


Can't say I follow that.

I do note that you're integrating a cdf, rather than a pdf. Do you need to differentiate your FY(y)F_Y(y) first to get the pdf, and then just integrate yf(y)?
(edited 9 years ago)
Reply 4
Original post by ghostwalker
There are going to be several parts to this:
For a point inside the region x,y>0, x+y<=1, your integral limits will just be 0,x and 0,y,

Outside the region there will be three areas I think you need to consider.
x+y>1 and x< 1
x+y>1 and y <1 these will overlap, and
x+y>1, x>1, y>1



Can't say I follow that.

I do note that you're integrating a cdf, rather than a pdf. Do you need to differentiate your FY(y)F_Y(y) first to get the pdf, and then just integrate yf(y)?


When it's outside the region h(x,y) is 0. I am wondering if I integrate it with limits [0,x] and [0,y] will be something I am looking for or not....... I am not really sure what I should do for this.......

For Q(2), what I am doing is E(Y)=020xfY(y) dy+20[1Fy(y)]E(Y) = \int^{20}_0 x*f_{Y}(y)\ dy + 20*[1-F_y(y)], which is just equal to the equation given in the previous reply (though I typed a wrong limit).
Original post by a5a09
When it's outside the region h(x,y) is 0. I am wondering if I integrate it with limits [0,x] and [0,y] will be something I am looking for or not....... I am not really sure what I should do for this.......


I don't think I can help you further with this one then.


For Q(2), what I am doing is E(Y)=020xfY(y) dy+20[1Fy(y)]E(Y) = \int^{20}_0 x*f_{Y}(y)\ dy + 20*[1-F_y(y)], which is just equal to the equation given in the previous reply (though I typed a wrong limit).


Well it's not the equation given in the previous reply, as you didn't have that first integral there. I don't follow why you have the second part - what's it meant to represent?

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