Solving a Probability Question with Poisson Distribution

Click For Summary
SUMMARY

The discussion focuses on solving a probability question using the Poisson distribution, specifically the equation 1.67 = e^-a (1/1-a). The user struggles to isolate the variable 'a' and attempts to expand e^-a using series without success. The solution involves the product-log function, expressed as 1 - a = W_n(-a/e), indicating that there is no straightforward analytical solution to the equation. The conversation highlights the complexity of solving such equations analytically.

PREREQUISITES
  • Understanding of Poisson distribution and its applications
  • Familiarity with exponential functions and their properties
  • Knowledge of the product-log function (Lambert W function)
  • Basic skills in solving equations involving transcendental functions
NEXT STEPS
  • Research the properties and applications of the product-log function (Lambert W function)
  • Learn about numerical methods for solving transcendental equations
  • Explore series expansions of exponential functions in greater detail
  • Study the Poisson distribution and its relationship with other probability distributions
USEFUL FOR

Students studying probability theory, mathematicians dealing with complex equations, and anyone interested in the applications of the Poisson distribution in statistical analysis.

persist911
Messages
7
Reaction score
0

Homework Statement


I am solving a particular probability question using Poission distributin after the Solving I get an equation

Homework Equations




Code:
 1.67 =  e^-a (1/1-a)
I ought to get the value of a from the equation but I was Unable to go further from here .

The Attempt at a Solution



I have tried expaning e^-a using series but no luck . I am totally stuck.
 
Physics news on Phys.org
Assuming you meant
1.67 = \frac{e^{-a}}{1 - a}
there is no "nice" solution.
It can be written as
1 - a = W_n(- a / e) \qquad (n \in \mathbb{Z})
where Wn the so-called product-log function, i.e. the solution for w of z = w ew. So that doesn't bring you much further, I'm afraid... are you sure you need to solve this analytically and - if so - the equation is correct?
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K