Have a question about a particular formula or about the actuarial exam itself? Email us your question! Here are some of our previous responses:

Question 1 : I was curious if you could show how to do the Example discussed on page 54 (of the manual). It seems to me that you would need to sum up the 30 q¹s since UDD only applies from 30 to 31, from 31 to 32, etc. Or do you treat it as UDD all the way from 30 to 65? (Submitted by Scott Chupack)

Answer 1 : Some explanation for this one is contained in this Adobe pdf file: Answer to example on page 54



Question 2 : What are the cdf and pdf? What is the difference between them? I lack a statistics background, so I have a hard time distinguishing between these two. (Submitted by Mark)

Answer 2 : Here is a partial answer to your question: The ``cdf '' of a random variable X is the cumulative distribution function F(x). For any real number x, the cdf F(x) gives the probability that (X <= x). Note that X here is a random variable, while x is just some real number.
For example, if X is the random variable representing my future lifetime, then F(20.5) is the probability that my future lifetime will be less than or equal to 20.5 years. In other words, it is the probability that I will die in the next 20.5 years.

The ``pdf'' of a random variable is the probability distribution function, f(x). When we are thinking of continouous random variables, I think of it as being the derivative of the cumulative distribution function ----> f(x) = F'(x). Then for any real numbers x and y with x less than y, the probability that X is greater than x but less than y is the integral from x to y of f(x). So if X is the continuous random variable for my future lifetime, the probability that I will die between 20 and 21 years from now is the integral from 20 to 21 of f(x). One other important point: If you have a curve representing the p.d.f. of a random variable X, then F(x) is just the area under f that is to the left of x (see page 2 of the manual).

When X is a discrete random variable, f(x) is just the probability that X = x. The definition of F(x) is the same as in the continuous case.





Question 3 : Probability Models, number 56(c) (your solutions, page 11), I did the problem a different way and came up with a different solution. Could you tell me where I went wrong? This is what I did: 50*.05 = 2.5 = expected number of vans, so lambda = 2.5. Then plug into the Poisson formula the probability of 5 vans = (e^-2.5)*(2.5^5)/5! = .0668. The answer is very close to what you have, but I don't think rounding is the difference, because I plugged both ways of doing the problem into Excel and the answers still come out as .0658 and .0668.

Answer 3 : The answer is a little bit subtle but this is a great question! It all depends on whether you know how many cars actually passed. The question your formula answers is the following: ``If the number of cars passing in the next hour is a Poisson process with lambda = 50.0 and 5% of cars are vans, what is the probability exactly 5 vans pass in the next hour." Note that we don't know how many cars total will pass (it might be zero), all we know is that it is a poisson process - and so we know which formula to apply for the vans.
Once we do *know* that exactly 50 cars actually pass in an hour, it is a very different question (in fact it really doesn't matter then that it is a Poisson Process). Now we know there were 50 `events' and each of them has probability 0.05 of being a van. So it's really just a binomial probability problem, like flipping an unfair coin. Which gives rise to the solution on the website.

Here is another example that is a lot more extreme but makes the same point:
Suppose the same Poisson Process applies but half of the cars are vans and we know that exactly 1 car passed in the last hour. What is the probability that it was a van. The intuitive answer is 1/2 (and that is what the binomial formula gives), but if you try to phrase it as a Poisson process where we don't know how many will pass -- using your method, we get that the van process is Poisson with lambda = 0.5 so Pr[1 van] = (e^-0.5)(0.5)/1! = 0.476 and you would get the same answer for the probability that that the car was not a van! (Since that is also a Poisson process with the same lambda).