Actuarial Outpost a single point in a continuous distribution?
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#1
10-23-2009, 05:08 PM
 uabutterfly Member Join Date: Oct 2007 Posts: 106
a single point in a continuous distribution?

I am suddenly very confused about a single point in a continuous distribution.

I know P(X=x)=0 for any continuous dist. based on the definition.

But in a lot of questions or reality, let's model loss size using 2-pareto dist. altha=2, theta=1,000.

What if I ask 'what is the probabilyt of the loss size = 500?

May I use f(x)=*** to get the probablity? Thanks!
#2
10-23-2009, 05:28 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,458

If the loss distribution is continuous (including, for example, 2 parameter theta), then the probability (Loss=500) = 0.

If there are coverage modifications, a continuous loss distribution could create some point masses of payments. In particular, if there is an ordinary deductible, then P(Payment=0) can be positive. If there is a maximum payment amount, then P(Payment=maximum) can be positive.
#3
10-23-2009, 05:36 PM
 badmaj5 Member CAS Join Date: Sep 2005 Location: Chicago, IL Posts: 454

Quote:
 Originally Posted by uabutterfly I am suddenly very confused about a single point in a continuous distribution. I know P(X=x)=0 for any continuous dist. based on the definition. But in a lot of questions or reality, let's model loss size using 2-pareto dist. altha=2, theta=1,000. What if I ask 'what is the probabilyt of the loss size = 500? May I use f(x)=*** to get the probablity? Thanks!
Or for example when you do a discrete Bayesian problem where losses follow a Pareto. You observe 2 losses of 100 and 200, you would take f(100)f(200) to get the probability of observing that outcome for the given individual.
#4
10-23-2009, 05:38 PM
 eagles418 Member Join Date: Aug 2009 Studying for waiting on FA results! College: Grad of a CAE Posts: 1,792

I think the OP is thinking about Bayesian credibility where the model distribution can easily be pareto and they give you 1 or 2 losses and you need a f(x|theta)>0 somewhere in that mess..
#5
10-23-2009, 05:43 PM
 uabutterfly Member Join Date: Oct 2007 Posts: 106

Thsi is my concern actually. Why can you use f(100) and f(200) here?

Quote:
 Originally Posted by badmaj5 Or for example when you do a discrete Bayesian problem where losses follow a Pareto. You observe 2 losses of 100 and 200, you would take f(100)f(200) to get the probability of observing that outcome for the given individual.
#6
10-23-2009, 06:31 PM
 hoosier91 Member Join Date: Aug 2007 Posts: 38

My undertanding of this is that when you plug a number into f(x) say 100 or 200 and get .2932 or .3431 or whatever; you are not truly getting the probability of exactly 100 (in other words not 100.01 or 100.001 or 100.00001 or etc or 99.9 or 99.99 or 99.9999 but exactly 100.0000000000000000....).

But instead what you are getting the probability density at point 100 and if it's .2932 then the density at 99.9 and 100.01 are probably pretty close to .2932 (maybe a little more, maybe a little less).

But no when you plug 100 or 200 into f(x) it is not the same thing as rolling a dice and having a 1/6 probability of getting exactly a 3.
#7
10-23-2009, 07:56 PM
 JavaGeek Member Join Date: Feb 2007 Posts: 336

$P(X=c)=\lim_{t\rightarrow 0} \int_{c-t}^{c+t} f(x)dx$, which is 0 [unless there's a point mass]. (This is called a probability function)

$f(x)$ is defined as $f(x)=\frac{dF(x)}{dx}$ (probability density function).
#8
10-24-2009, 02:55 PM
 uabutterfly Member Join Date: Oct 2007 Posts: 106

Quote:
 Originally Posted by JavaGeek $P(X=c)=\lim_{t\rightarrow 0} \int_{c-t}^{c+t} f(x)dx$, which is 0 [unless there's a point mass]. (This is called a probability function) $f(x)$ is defined as $f(x)=\frac{dF(x)}{dx}$ (probability density function).

In other words, f(x) ^=P(X=x), right?
#9
10-24-2009, 03:04 PM
 JavaGeek Member Join Date: Feb 2007 Posts: 336

Quote:
 Originally Posted by uabutterfly In other words, f(x) ^=P(X=x), right?
Let's try x=1 for [0,1] (uniform on [0,1]).
f(x)=1;
$P(X=0.52) = \int_{0.52}^{0.52} 1 dx = x|_{0.52}^{0.52} = 0.$
$f(0.52)=1$
#10
10-24-2009, 03:13 PM
 JavaGeek Member Join Date: Feb 2007 Posts: 336

If you really want a relationship. I believe it is.
$f(c)=\lim_{t\rightarrow 0}\frac{ \int_{c-t}^{c+t} f(x)dx}{2t}=\lim_{t\rightarrow 0}\frac{ P(X=c)}{2t}=\frac{F(c+t)-F(c-t)}{2t} = F'(c)$.

Last edited by JavaGeek; 10-25-2009 at 11:42 AM..

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