actuary_math's blog

parameter risk

I believe that how few we consider the "parameter risk".

Suppose that the probability of occurring a accident is P (0<P<1).
For N polices, the probability of occurring n accidents is
N_C_n * P^n * (1-P)^(N-n)
(where N_C_n means the combinatorial number which is chosen n things form N things)

Suppose that the p.d.f.(probability density function) of the prior distribution is g(p),
f(p),the p.d.f. of the posterior distribution by Bayes' theorem
will be proportional to
P^n * (1-P)^(N-n) *g(p)
(because N_C_n is a constant which has no relations with p)

Now if the prior distribution is a uniform distribution, that is
g(p)=1(0<p<1)
f(p) is proportional to P^n*(1-P)^(N-n)
so
P follows to B(n+1,N+1)(beta distribution).

Translation from my Japanese blog (2)

In the last post I left proof of a proposition.
So I will prove it as below.

We must use the Lebesgue's integral theory to prove the proposition.

Proposition
Suppose that X and Y are independent random variables with probable density functions f(x) and g(y) respectively.
And suppose that f(x) and g(y) are continuous almost everywhere, and that a least one of f(x) and g(y) is bounded.
Then h(s),the p.d.f. of S=X+Y is continuous at ANY real number s.

Proof
As
h(s)=\int f(s-y)g(y)dy=\int f(x)g(s-x)dx
(In the above formula "\int" means integral from -infinity to +infinity.)
it is sufficient that we suppose that f(x) is bounded.
That is there is a positive real number K that f(x)<K for all x.

And it is sufficient for us to prove that for any sequence {s_n} that converges to s
\int f(s_n-y)g(y)dy =h(s)

Note that
f(s_n-y)g(y) <= K*g(y)
and that
\int K*g(y)=K*1=K<+infinity
(because h(s)=\int g(y)dy=1)

On the other hand
for almost any y
f(s_n-y)g(y) converge to f(s-y)g(y).
(because f(x) and g(y) are continuous almost everywhere)

So we get
\int f(s_n-y)g(y)dy =h(s)
by dominated convergence theorem(or Lebesgue's convergence theorem)

q.e.d.

Translation from my Japanese blog (1)

I am running the blog about "Examination to be Actuaries"
at
http://d.hatena.ne.jp/actuary_math/

The blog is written in Japanese, but I will introduce you some articles by translation in English.

Let's brgin with the use of the "elimination" this time.
(The oringinal url is
http://d.hatena.ne.jp/actuary_math/20080718
)

I think you often use naturally the method of "elimination" in other selection type questions and quizzes etc.

The problem below is taken up in the past exams.

(Question)
Let X,Y, and Z are independent random variables with uniform distributions of U(0,2), U(-1,1), and U(-2,0) respectively.
The probable density function(p.d.f.) of S=X+Y+Z is
[1] (-3 <= s <= -1)
[2] (-1 <= s <= 1)
[3] (1 <= s <= 3)
0 (otherwise)

(select the blanks [1] to [3] by the following choices (A) to (L) )

(A){(s+3)^2}/4
( B ){(s+3)^2}/8
( C ){(s+3)^2}/16
(D){(s-3)^2}/4
(E){(s-3)^2}/8
(F){(s-3)^2}/16
(G)(3+s^2)/4
(H)(3+s^2)/8
(I)(3+s^2)/16
(J)(3-s^2)/4
(K)(3-s^2)/8
(L)(3-s^2)/16
-----------------

This question can be solved by the compound of 2 or 3 random variables.
But the situation division is unexpectedly troublesome.
Here, we will squeeze the candidate of the answer is squeezed by the elimination without calculating like this.

If we choose without no thought than not choosing the same one twice
the number of the candidates are 1,320( permutations 12P3 ,that is selecting 3 in a row from 12).
However it is possible to narrow up to two candidates by the following discussion and only one remains with high possibility of the two. (It is actually the correct answer.)

1.
First of all, let us clear the first hurdle installed in the problem sentence.
We put X'=X-1 and Z'=Z+1 respectively.
The variables X',Y and Z' are the same and independent variables which follow uniform distributions U(-1,1).
Moreover, it is understood that the answer becomes symmetric because S=X'+Y+Z' and because U(-1,1) is symmetric distribution, too.

So, the combination of candidates of [1]-[3] are the following 3 pairs
(A)-(D),
( B )-(E)
and
( C )-(F)

And the candidates of [2] are the 6 choices from (G) to (L).

At this point, the number of the candidates becomes 3*6=18.

2.
Next,
We will use the fact that
"We get 1 when we integrate a probability density function (p.d.f.) where the value of the p.d.f is positive (in this case only with -3 <= s <=3)".

(i)
The integral from -3 to -1 of (A) is
\int_{-3}^{-1} \frac{(s+3)^2}{4}ds=[{(s+3)^3}/{12}]_{-3}^{-1}=2/3
(I mean \int_a^b f(x) dx and [F(x)]_a^b by the integral f(x) from a to b and F( b )-F(a) respectively)

So is the integral of (D) from 1 to 3 (because (A) and (D) are symmetric)

We get the integral from -3 to -1 of ( B ) (= the integral from 1 to 3 of (E) ) and the integral from -3 to -1 of ( C ) (= the integral from 1 to 3 of (F) ) are respectively 1/3 and 1/6

(ii)
We get the integrals of (G) - (H) from -1 to 1 are respectively
(G) 5/3,
(H) 5/6,
(I) 5/12,
(J) 4/3,
(K) 2/3
and
(L) 1/3

So
The candidates of three combinations of [1]-[2]-[3] are
(1)( B )-(L)-(E)@(1/3 + 1/3 + 1/3 =1)
(2)( C )-(K)-(F)@(1/6 + 2/3 + 1/6 =1)

Well, which of the 2 candidates is the correct answer?
When two graphs are drawn, they are
(1)
http://f.hatena.ne.jp/actuary_math/20080718194236
And,
(2)
http://f.hatena.ne.jp/actuary_math/20080718194235
respectively.

Here, the candidate (2) of which the graph is "Continuous" seems to be correct.

Because of the following proposition, it is actually correct.

(Proposition)
"Let X and Y are independent probability density functions f(x) and g(y) respectively which are almost everywhere continuous(*). At least one of the functions (f(x) and/or g(y)) is(are) bounded. So h(s),the probability density function of S=X+Y is continuous at all s"
(This proposition will proved at another moment because it is not easy. )
(*) I will not define "almost everywhere continuous." But even if it is discontinuous at limited points, the definition of "It is almost continuous everywhere" holds.

The point is that even if there are limited "discontinuous" points of f(x) and g(y) the h(s) is continuous for ALL points.

Therefore, by the "elimination" we can not set the combination of [1]-[2]-[3]
other than
( C )-(K)-(F).

Of course things do not go well like every time
But we can often narrow by the elimination.

Japanese post · Japanese post in Actuary_math's blog with 2 comments on Mon Sep 01, 2008 1:15 am

I have found that Japanese post are NOT allowed at the blog

About my blog · About my blog in Actuary_math's blog with 0 comments on Sat Aug 30, 2008 7:24 am

The blog
http://d.hatena.ne.jp/actuary_math
is intended to the examinee of IAJ(institute of Actuaries of Japan) exam