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#1




SOA sample #204
For , why is the integral from 0 to , instead of from 0 to ?

#3




I'm really bad at calculus and math and sometimes struggle with the posterior math we encounter on Bayesian credibility questions.
So can someone verify, is this all true? Are there other situations beyond these that anyone has seen in their studies? Let X be the counting variable, and let X's distribution have a parameter Y which has its own continuous probability distribution where Y > 0. Then these are the 4 types of situations that are likely to show up on exam since they were in the sample questions: Situation 1: "solve the probability X is less than some number," but no information given on what values X has taken previously: in which F(XY) is the conditional CDF of X given Y, and f(y) is the prior distribution of y Situation 2: "solve for the probability Y is between a and b," no information given on what value X has been previously: In which Fy is the CDF of Y Situation 3: From sample #43 and #157, given X was some number z previously, what's the probability parameter Y is between a and b is: in which f(yx=z) is the posterior distribution Situation 4: From sample #76, given X was some number z, whats the probability X is between a and b next period is: and we would replace the probability that X is between a and b given y, with the probability of the event of whatever information we were asked to solve for for X possibly being next period. 
#4




I inverted the gamma's theta to get an exp/inverse gamma conjugate prior and used the predictive pareto to find the correct answer choice.
Is this a correct approach or just a coincidence? Thanks 
#5




Quote:
in this case there is no "n" to add to alpha or "x" to add to theta, right? so then just use CDF of pareto with alpha = 2 and theta = 0.5. at this point, I think I would use first principles for a problem like this on the exam. that might change over the next few weeks. I'm not great at calculus, but I am becoming comfortable with the integral of 0 to infinity of (y^1)*(e^(0.5y)) dy = (1/0.5^2) * (21)! 
#6




It could certainly change with exam time constraints. The algebra on some of these questions really chews up my time.
While we're on the subject of gammas here's another one probably worth knowing: poisson expansion of Gamma(a;y) = 1sum[i=0 to a1] (y^i)(e^y)/i! where y=x/theta 
#7




Quote:

#8




I know this is an old thread, but I thought I'd ask this.
If the length of time X is modeled by an Exponential with parameter theta = 1/Y, can we say that length of time X is modeled by an Inverse Exponential with parameter Theta = Y? I'm trying to do the problem this way, with no luck. Edit: Nope, that's not right. If we invert, then the variable inverts. 1/X would be Inverse Exponential with 1/Y. Last edited by Demo99; 10072016 at 01:10 PM.. 
#9




I found the easiest way for me to solve it was replacing x=1/2 right away, then the integral for all values of y is just the difference of 2 easy ones. I used the basic result of an Erlang that gives
(integral from 0 to infinite) (x^a * e^(x/b) dx) = a! * b^(a+1)
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