The Wisdoms Of Examination To Be Actuaries

[actuarymath 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.

Edited September 27, 2008 by actuarymath (see edit history)

[actuarymath 1]

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.PropositionSuppose 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.ProofAs 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 thatf(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 handfor almost any yf(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.

[actuarymath 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.

Let us think the above question in another point of view ("central limit theorem")

(The original Japanese version is

http://d.hatena.ne.jp/actuary_math/20080722" target="_blank"> http://d.hatena.ne.jp/actuary_math/20080722 )

 

The notation of the central limit theorem will be seen in the below URL

https://en.wikipedia.org/wiki/Central_limit_theorem

 

But practically in the actuarial exam

"Central limit theorem"

can be regarded as "We can regard the distribution as a normal distribution."

 

In this question, as 3 distributions are compounded, the distribution can be regarded to be near the normal distribution.

 

The graphs of the normal distributions look like bell form.

(For example see

https://en.wikipedia.org/wiki/Normal_distribution

)

 

 

In this question

(1) The graph of the neighborhood of the mean value is regarded to (upward) convex.

So (G)-(I) are eliminated of the candidates of [2] because their graphs are concave (or downward convex).

(2) The graph is "gathered" to the center. That is the probability P(-1 <= s <=1 ) is larger than P(-3 <= s <=-1 )(=P(1 <= s <=3 ))

So (L) is eliminated from (J)-(L),the rest of (1) because the integral of (L) from -1 to 1 is 1/3.

 

Then (J) and (K) are remained but (J) is out of the question because the integral of (L) from -1 to 1 is 4/3 which is larger than 1.

Then the candidate of [2] is determined to (K).

 

The sides candidates [1] and [3] are set to be the integrals equal to 1/6.So are ( C ) and ( F ).

 

This method can be used when we verify the answer normally calculated.

[actuarymath 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.

 

...

Then 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.