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#1




Help plz! Given pdf to estimate prob. of small interval
From an online study material, an example question under AgeAtDeath RV.
An ageatdeath 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




Quote:
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 (ba)*(f(a)+f(b))/2. That's only an estimate of the area, but for small ba 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 (ba)*2f(a)/2=(ba)*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 (ba) and height f(a). Other similar estimates would be (ba)*f(b) or (ba)*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:

#3




Thank you Gandalf!!
I overthink on this, nothing really relates to MLC material lol. But thanks for your explanation, you made me very clear now 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 conceptualbase.. 
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