CDF question: F(x) = 1 + (u/(u+x))^a, x>=0; 0 otherwise

Steven G

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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)
 
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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 \(\displaystyle F(0) = f(0) = \dfrac {\alpha}{u^\alpha} = 0. \ \ So \ \ \alpha = 0\)

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)

Right now all I see is the two equations in three unknowns which one can write by rearranging F(u-k)=.1 and F(5u-3k) = .9
\(\displaystyle \alpha\) = \(\displaystyle \dfrac {ln(0.9)}{ln(u)\, -\, ln(2u-k)}\)
\(\displaystyle \alpha\) = \(\displaystyle \dfrac {ln(0.1)}{ln(u)\, -\, ln(6u-3k)}\)
That should give you a simple linear solution between u and k and thus maybe a variety of \(\displaystyle \alpha\)'s should satisfy the conditions.
 
Right now all I see is the two equations in three unknowns which one can write by rearranging F(u-k)=.1 and F(5u-3k) = .9
\(\displaystyle \alpha\) = \(\displaystyle \dfrac {ln(0.9)}{ln(u)\, -\, ln(2u-k)}\)
\(\displaystyle \alpha\) = \(\displaystyle \dfrac {ln(0.1)}{ln(u)\, -\, ln(6u-3k)}\)
That should give you a simple linear solution between u and k and thus maybe a variety of \(\displaystyle \alpha\)'s should satisfy the conditions.


Since \(\displaystyle f(u-k)=1- \dfrac{\alpha u^\alpha}{(2u-k)^{\alpha}}= .1\)
and \(\displaystyle f(5u-3k)=1- \dfrac{\alpha u^\alpha}{(6u-3k)^{\alpha}}=.9\)

Then \(\displaystyle \dfrac{\alpha u^\alpha}{(2u-k)^{\alpha}}= .9\)
and \(\displaystyle \dfrac{\alpha u^\alpha}{(6u-3k)^{\alpha}}=.1\)

You can factor out \(\displaystyle \dfrac{1}{3^{\alpha}}\) from that last equation. Then dividing the two expressions will remove u and k. At that point it is easy to find \(\displaystyle \alpha\)

I still have my initial question--where was my initial mistake (see beginning of thread) in getting \(\displaystyle \alpha = 0?\)
 
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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)

Of course I got the wrong answer as you can't find probability of a continuous pdf at a single point!!!
 
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