Let me introduce some wisdom of actuary exam(s) by several post after this.

I have maintained a Japanese blog "The research of the actuary exam mathematics."
(in Japanese "Actuary shiken suugaku no kenkyu")
http://d.hatena.ne.jp/actuary_math

Of course the blog is intended for the examinee of IAJ(the Institute of Actuaries of Japan) but the techniques written there will be served to other examinees for example SOA(Society of Actuaries),CAS(Casualty Actuary Society), etc.

First, I will introduce
http://d.hatena.ne.jp/actuary_math/20080705
by translating into English.

The following question is a little alternation from one seen in the past exam.

Question

In the above picture, AB is parallel to x-axis and AC is parallel to y-axis.
(X,Y) are 2 dimensional random variables which follow uniform distributions on triangle ABC.
The correlation coefficient X,Y rho(X,Y) is [ ]
(fill in the blank)
----------

In this question neither the position (x,y) of A the nor the length of AB or AC are not given.
So some of you may begin to present the position and the length as some variances.

However, we will make the perceptional change.
That is
"The answer is the same even if the position (x,y) of A the or the length of AB or AC.
Therefore, we can let them whatever we like.｣(*)

So we will let A(0,0),B(1,0),C(0,1).

E(X)=E(Y)
=2{\int_0^{1} xdx \int_0^{1-x}dy}
(I will mean "\int_a^b" by integral from a to b as TEX usage)
=2 \int_0^1 (x-x^2)dx
=2(1/2 - 1/3)=1/3

E(X^2)
=2 \int_0^1 x^2dx \int_0^{1-x}dy
=2 \int_0^1 (x^2-x^3)dx
=2(1/3 - 1/4)=1/6

V(X)=V(Y)
=E(X^2)-{E(X)} ^2
=1/6-(1/3)^2=1/18

E(XY)
=2 \int_0^1 xdx \int_0^{1-x} ydy
= \int_0^1 x(1-x)^2 dx
=\int_0^1 (x-2x^2+x^3) dx
=1/2 - 2/3+ 1/4=1/12

Cov(X,Y)
=E(XY)-E(X)*E(Y)
=1/12 - 1/3*(1/3)=- 1/36

rho (X,Y)
=Cov(X,Y)/sqrt{V(X)*V(Y)}
=(-1/36)/(1/18)=-1/2

(*)The ideas can be authorized as follows.
First we put the position of A is (a,b ) and the length of AB and AC are c and d respectively.
Then we put X'=(X-a)/c, Y'=(Y-b )/d
as rho (X,Y)=rho (X',Y')
(Because the correlation coefficient does not vary by addition, subtraction, multiplication and division of constants.)

In the original question A is the Original point(0,0) and the length of AB=a and AC=b were given.
But according to the above thought, we can easily find that the answers with a,b are all WRONG.

The idea can not be limited to the actuary examination, and to be applicable to all objective type examinations.