Steven G
Elite Member
- Joined
- Dec 30, 2014
- Messages
- 14,379
Let X be a loss random variable with cdf
\(\displaystyle F(x) = 1 - (\dfrac {u}{u+x})^\alpha\ \ , \ \ x\geq 0 \ \ and \ \ 0 \ \ otherwise\)
The 10th percentile is u-k and the 90th percentile is 5u-3k.
Determine the value of \(\displaystyle \alpha\)
Here is what I did. I found the pdf function f(x) by computing F'(x).
I got f(x) = \(\displaystyle \dfrac{\alpha u^\alpha}{(u + x)^{\alpha + 1}}\)
Since \(\displaystyle f(x) = 0 \ for \ x \ < 0\), I get that \(\displaystyle F(0) = f(0) = \dfrac {\alpha}{u} = 0. \ \ So \ \ \alpha = 0\)
But apparently this is wrong and clearly did not take into account that the 10th percentile is u-k and the 90th percentile is 5u-3k.
What is wrong in my logic? I think this mistake is a big one and I need to learn where I'm going wrong.
Also a hint on how to do this will be helpful. For now I only see two equations in three unknowns (F(u-k)=.1 and F(5u-3k) = .9)
\(\displaystyle F(x) = 1 - (\dfrac {u}{u+x})^\alpha\ \ , \ \ x\geq 0 \ \ and \ \ 0 \ \ otherwise\)
The 10th percentile is u-k and the 90th percentile is 5u-3k.
Determine the value of \(\displaystyle \alpha\)
Here is what I did. I found the pdf function f(x) by computing F'(x).
I got f(x) = \(\displaystyle \dfrac{\alpha u^\alpha}{(u + x)^{\alpha + 1}}\)
Since \(\displaystyle f(x) = 0 \ for \ x \ < 0\), I get that \(\displaystyle F(0) = f(0) = \dfrac {\alpha}{u} = 0. \ \ So \ \ \alpha = 0\)
But apparently this is wrong and clearly did not take into account that the 10th percentile is u-k and the 90th percentile is 5u-3k.
What is wrong in my logic? I think this mistake is a big one and I need to learn where I'm going wrong.
Also a hint on how to do this will be helpful. For now I only see two equations in three unknowns (F(u-k)=.1 and F(5u-3k) = .9)
Last edited: