actuarymath

I understand 000webhost storage is 250MB.

I am in insurance industry and I am interested in pet insurance.In Japan some insurers launched on pet insurance but some of them are said to use the foreign (outside Japan) statistics.If some of you know the statistics of pet insurance or those of pet sickness and injuries, would you teach me where the statistics are?

Would someone teach me the lauguage usage in Xisto free web hosting? In the TOS(terms of services) of Xisto free web hosting, http://forums.xisto.com/topic/5769-terms-of-service-and-acceptable-use-policy/ "All websites must be in English to qualify for free hosting. We do this to ensure that Accounts are legal and have permissable contents." But I found at a review site http://forums.xisto.com/no_longer_exists/ that "They are amazing. Non-English sites ARE allowed, except you have to have an English overlook page detailing your site so they know what it is about." Which would be right?

Forbez Thank you for your reply. As you will find when you visit the site http://www.business-opportunities.biz/2005/11/16/how-much-is-my-blog-worth-now/ It seems to depend on http://technorati.com/ I think the points would depend on the linkage from technorati. I tried on some (Japanese) blogs I know but many of them which had more posts than mine gained NO points. This may be because the blogs have no linkage from technorati.

I've found a site as below that shows you the "value" of your blog. http://www.business-opportunities.biz/2005/11/16/how-much-is-my-blog-worth-now/ My blog http://d.hatena.ne.jp/actuary_math/ is worth $1,129.08.

wwilliams Thank you very much for your comment. I also think FAQs would contain such information about SSH. By the way will you tell me the below thing if you know? I understand once we are hosted the points correspond to the survival days (1 point=1 day survival). But I am afraid that my survival days seem to decrease more faster than 1point/1day. Do you know the rule of "decreasing survival days"?

Thank you for your reply. I have thought that Qupis allows ssh because it has SSH menu. I would have not been confusued if your Qupis cpanel menu have not shown the SSH menu!

Thank you very much for your reply.By the way is this same as Qupis? That is, do I have to send an application email to qupis support like the above format to use SSH on Qupis? (If I have to post the matter at the Qupis corner, please indicate me so)

I was allowed to have an account http://forums.xisto.com/no_longer_exists/ and I would like to use SSH access that I believe is a salespoint of Xisto. At the video tutorial, it seems that we can use Java SSH but I could not find the icon about it. And I used the PuTTy and tried to connect but I was refused. Of course I have authorized the public key at my cpanel. My setting was as follows. host name:actuary.trap17.com account name:actuary password:(password I use at cpanel) Port No 22 If I made some mistakes, will some of you please let me know?

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 isN_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' theoremwill 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)soP follows to B(n+1,N+1)(beta distribution).

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.

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.

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.

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.

I owned a Japanese blog at http://d.hatena.ne.jp/actuary_math/ Once I posted taking about pension mathematics at http://d.hatena.ne.jp/actuary_math/20080819 Now I translate it into English as follows because it will somehow server those who studies pension mathematics. Because I think that I become wrong perhaps if it digs more than this down about corporate pension accounting/retirement benefit accounting because it is unprofessional, could you pardon the respect? ----

How about internet video converter? http://internet-video-converter.en.softonic.com/ It seems that this tool can convert .swf or .flv to .mpg which I think can be read by QT.