NoMoreExams - the only reason I am questioning this problem is that I cannot see why the interval of integration for P (premiums) is infinity. In comparing this problem to #108 - I see that you have to solve in terms of one of the variables to see the limit, and when going to solve in terms of the 2nd variable set that first one equal to 0, so that's why t1 (in #108) goes up to X.
However when I tried to use this theory of mine lol in 109 I could see 0 < C < xp but then in setting c=0 and solving for the limit for p you have 0/p < x. This means 0 < px which does not hold because we just said c < px. So is that why it is infinity?
If this is confusing can you explain how to get the limits for 108 and 109, and maybe the difference between the two?
Thanks a million.....
-
Ok so you are asked for the pdf of X, i.e. f(x), the easiest way to get is to first find F(x) (the cdf) and then differentiate it to get to f(x). You are also told that X = C/P (where C and P are claims and premiums respectively), on top of that you are told the distribution C and P follow from which you can figure out their respective pdfs and cdfs. Want me to go on?Leave a comment:
-
WOOPS, SORRY; I did leave out part of the problem 109, the last line. That's what you are asking for, duh!
A company offers earthquake insurance. Annual premiums are modeled by an exponential random variable and mean 2. Annual claims are modeled by an exponential random variable with mean 1. Premiums and claims are independent. Let X denote the ratio of claims to premiums.
What is the density function of X?
When it says what is the density function of X... I have a hard time visualizing what is really being done or asked.
ok, let me try thinking about it backwards. Lets say f(x) is the density function of X.
Then F(x) would be the distribution function {F'(x)=f(x)} ?? Honestly I don't know what distribution function means. I assume it is the integral of the density function because that's how they are related in this problem.Leave a comment:
-
I'm too lazy to go look for sample SOA problems for 1/P but you haven't even asked the question that was asked in the problem (unless I missed it).
As far as your other question, come on, you should know what Pr(X = x) for cont. dist. is...Leave a comment:
-
I don't understand what is going on in the problem to be able to even ask a specific question.
What I see is: If A happens, then do B.
I have no idea why we are doing B, I just know that we must do it.
The idea of making a joint probability function doesn't make sense to me. I imagine it has something to do with calculating the infinite probabilities that can come up with the infinite combinations of claims and premiums. Then use P[X<=x] = P[u/v<=x]=P[u<=vx] to find the range of the inner part of the integral.
Why use P[X<=x] and not P[X=x] or P[X>x]? All I know is that every type of problem like this uses P[X<=x] and nothing else.
I think in order to understand this problem. I need to see an easier problem of the same type.Leave a comment:
-
X is the ratio of claims to premiums, not the ratio of the respective pdfs...
I'm not sure what your actual question is. Can you maybe summarize it.Leave a comment:
-
SOA released problem #109
A company offers earthquake insurance. Annual premiums are modeled by an exponential random variable and mean 2. Annual claims are modeled by an exponential random variable with mean 1. Premiums and claims are independent. Let X denote the ratio of claims to premiums.
Premium f(t) = (1/2)e^(-t/2)
Claim g(t) = e^(-t)
X = g(t)/f(t) 'the ratio?'
^anyways, that's what I was thinking... then I saw the solutions...
u = annual claims
v = annual premiums
"next you need to make a joint density function of U and V... "
g(u,v) = joint density function of U and V = [e^(-u)][(1/2)e^(-v/2)] with ranges 0<u<inf and 0<v<inf
"next you must create the distribution function of X; in order to make the pdf of X"
F(x) = Pr[X<=x] = P[u/v <= x] "ok, there's the ratio"
^this I've seen a lot, and I keep forgetting about it.
= P[U<=Vx] "I don't get this part"
What does this mean? The probability that U is less than or equal to V times X? and that equals the integral of g(u,v)dudv...... I'm stumped where this comes from.... Is there a similar problem to this that is easier to understand?? Oh, the range of U is from 0 < u < vx; that's smurfy because it's the probability of U<=Vx and also u>0.
Ok, some recap.. when solving for the distribution function and you have some function f(a,b). You may start with something like P[X<=x] or P[Y<=y]. Then, you want to solve the Pr[??<=??] to be either Pr[a<=??] -OR- Pr[b<=??] and respectively the range of a would be 0<a<?? and the range of b would be 0<b<??.
finding f(x) is somewhat easy after getting F(x)
F'(x) = f(x) = [(-1)(2x+1)^(-1) + 1]' = 2(-1)(-1)(2x+1)^(-2) = 2*(2x+1)^(-2)
1) What would you call this type of problem? Or what key words would I use to research info on this type of problem.
2) Is there an easier way to explain the steps that I'm confused on? All my explanations were derived using the solution and some logic.
Leave a comment: