Actuarial Outpost Help plz! Given pdf to estimate prob. of small interval
 Register Blogs Wiki FAQ Calendar Search Today's Posts Mark Forums Read
 FlashChat Actuarial Discussion Preliminary Exams CAS/SOA Exams Cyberchat Around the World Suggestions

Search Actuarial Jobs by State @ DWSimpson.com:
AL AK AR AZ CA CO CT DE FL GA HI ID IL IN IA KS KY LA
ME MD MA MI MN MS MO MT NE NH NJ NM NY NV NC ND
OH OK OR PA RI SC SD TN TX UT VT VA WA WV WI WY

 Long-Term Actuarial Math Old Exam MLC Forum

#1
03-08-2014, 05:50 AM
 ScorpioAB CAS SOA Join Date: Jan 2013 College: University of Washington Posts: 16
Help plz! Given pdf to estimate prob. of small interval

From an online study material, an example question under Age-At-Death RV.

An age-at-death random variable is modeled by an exponential random variable
with PDF f(x) = 0.34e^0.34x; x >= 0. Use the given PDF to estimate
Pr(1 < X < 1:02).

The answer is just one line:
Pr(1 < X < 1.02)  0.02*f(1) = 0.005

Could anyone tell me how this is derived???

Thanks!

BTW it's interesting that exam MLC has been revised for several times, from pencil & paper MC to almost CBT, and now it contains both MC and WA.
#2
03-08-2014, 06:09 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,366

Quote:
 Originally Posted by ScorpioAB From an online study material, an example question under Age-At-Death RV. An age-at-death random variable is modeled by an exponential random variable with PDF f(x) = 0.34e^0.34x; x >= 0. Use the given PDF to estimate Pr(1 < X < 1:02). The answer is just one line: Pr(1 < X < 1.02) 0.02*f(1) = 0.005 Could anyone tell me how this is derived??? Thanks!
This is just probability and calculus, nothing unique to MLC. For any continuous random variable X, the probability that X is between a and b is $\int_a^b f(x)dx$

From calculus, an integral is an area under the graph. As long as a and b are close together and the value of f(x) does not vary much over the integral, that area is pretty close to the area of the trapezoid with corners (a,0), (b,0),(a,f(a)), (b,f(b)) whose area is (b-a)*(f(a)+f(b))/2. That's only an estimate of the area, but for small b-a it's usually a pretty good estimate.

It's not the estimate used here, though. Here's it's even more of an estimate.

Since a and b are close, f(a) and f(b) are close, so f(b) is pretty close to f(a). Therefore (f(a)+f(b)) is approximately equal to f(a)+f(a)=2f(a), and our first estimate is pretty close to (b-a)*2f(a)/2=(b-a)*f(a). That's the estimate the author uses. Thinking geometrically, that's saying the integral is approximately the area of a rectangle with width (b-a) and height f(a).

Other similar estimates would be (b-a)*f(b) or (b-a)*f((a+b)/2). Those are still estimating the integral as the area of a rectangle, just different estimates of the height of the rectangle. If you tried them with your pdf (which is probably missing a - sign), all four estimates would be pretty close.

Quote:
 BTW it's interesting that exam MLC has been revised for several times, from pencil & paper MC to almost CBT, and now it contains both MC and WA.
What is "almost CBT" and when was it that?
#3
03-08-2014, 06:30 AM
 ScorpioAB CAS SOA Join Date: Jan 2013 College: University of Washington Posts: 16

Thank you Gandalf!!

I overthink on this, nothing really relates to MLC material lol.
and sorry for the missing sign there :P

almost CBT was just I think they shouldve made MC to CBT :P
but now they added WA portion I think it will be very conceptual-base..