Apologies if the title is not well formatted. First time posting a formula in a title.
My issue comes from the following situation. I want to define a probability function by regions so that:
$p(x) = \left\{ \begin{array}{ll} \frac{0.4}{10-2} & \mbox{if } 2 \leq x \leq 10\\ \lambda e^{-\lambda x} & \mbox{if } x \geq 11 \end{array} \right.$
$0$ otherwise.
The total $p$ should sum to $1$. So if i set the uniform distribution region to sum $0.4$, the exponential part should sum $0.6$
So I do the following maths: $0.6 = \sum_{11}^{\infty}\lambda e^{-\lambda x} = \lambda e^{-11\lambda}\sum_0^{\infty}e^{-\lambda x}$
Knowing that $\sum_{n=0}^{\infty}x^n = \frac{1}{1-x}$ then if $e^{-\lambda x} = x^n, \quad x = e^{-\lambda} \quad and \quad n = x$
I get: $0.6 = \frac{\lambda e^{-11\lambda}}{1-e^{-\lambda}}$.
After that I am not sure how to continue since I am incapable of solving after taking logarithms: $log\left(\frac{\lambda e^{-11\lambda}}{1-e^{-\lambda}}\right) = log(0.6)$
So esentially I have 3 questions.
- Is the approach correct?
- If it is indeed correct. How would I go for solving the final equation?
- The interpretation of the solution means that in order for the probability density function to be correct. The exponential expression should have a fixed $\lambda$ which is found through the logarithmic equation. Is that correct?
Thanks!
