Actuarial Outpost p exam problem continuous random variables
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#1
11-15-2017, 08:12 PM
 efast Non-Actuary Join Date: Nov 2017 Posts: 9
p exam problem continuous random variables

Hi guys,

A group insurance policy covers the medical claims of the employees of a small company. The value, V, of the claims made in one year is described by
V = 100000Y
where Y is a random variable with density function f(x) =k(1−y)4 0 < y < 1

0 otherwise
where k is a constant.

What is the conditional probability that V exceeds 40,000, given that V exceeds 10,000?
#2
11-15-2017, 09:01 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,008

Can you calculate the value of k? Can you calculate the probability that V exceeds 40,000? That V exceeds 10,000?
#3
11-16-2017, 11:02 AM
 Kemisola SOA Join Date: May 2015 Location: Lagos, Nigeria Studying for Exam P College: University of Lagos (3rd year) Posts: 8

First, you have to find the value of k and this is done by equating the density function i.e. f(x) to 1. You will get k=5

Then you find P(100,000Y>40,000)|100,000Y>10,000) which is same as P(Y>0.4|Y>0.1)

I hope you can go on from here. Your final answer should be 0.132
#4
11-16-2017, 11:27 PM
 efast Non-Actuary Join Date: Nov 2017 Posts: 9

Quote:
 Originally Posted by Kemisola First, you have to find the value of k and this is done by equating the density function i.e. f(x) to 1. You will get k=5 Then you find P(100,000Y>40,000)|100,000Y>10,000) which is same as P(Y>0.4|Y>0.1) I hope you can go on from here. Your final answer should be 0.132
Thank you, Kemisola. What do you mean by equating the density function to 1?
#5
11-16-2017, 11:29 PM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

Quote:
 Originally Posted by efast Thank you, Kemisola. What do you mean by equating the density function to 1?
The integral of the density function (from 0 to 1 in this case) must be 1.
#6
11-16-2017, 11:52 PM
 efast Non-Actuary Join Date: Nov 2017 Posts: 9

I'm making a mistake but I don't know where.
I find k by saying that the integral of k*(1-y)^4 from 0 to 1 is equal to 1.
That means that the limit of k/5*(1-y)^5 ( y goes towards 1) minus the limit of k/5*(1-y)^5 ( y goes towards 0) is equal to 1.
Since for y goes towards 1 the limit of f(x) is 0, minus the limit of k/5( 1-y)^5 with y goes towards 0 is equal to 1. That gives me k=-5.
Where did I make a mistake?
thank you for the help.

Last edited by efast; 11-17-2017 at 12:03 AM..
#7
11-17-2017, 12:00 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,008

Sign error. You forgot that the (1-y) piece contributes a factor of -1.

I.e., if you differentiated k/5( 1-y)^5, you would get -k(1-y)^4, not k(1-y)^4.
#8
11-17-2017, 02:44 AM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

#9
11-17-2017, 10:23 AM
 efast Non-Actuary Join Date: Nov 2017 Posts: 9

Thank you Gandalf and Z3ta.
what is the rule to integrate k*(1-y)^4?

Last edited by efast; 11-17-2017 at 12:11 PM..
#10
11-17-2017, 12:12 PM
 efast Non-Actuary Join Date: Nov 2017 Posts: 9

Thank God, I got it using the substitution rule.