Maximum Likelihood Estimator for Exponential Density Function

Click For Summary
SUMMARY

The discussion focuses on determining the constant c in the probability density function f(x;λ) = cx²e^{-λx} for x ≥ 0 and finding the maximum likelihood estimator (MLE) for λ. Participants emphasize the need to integrate the function from 0 to infinity and suggest using integration by parts to resolve the integral. The integration process involves recognizing that x e^{-λx} can be expressed as the derivative of e^{-λx} with respect to λ, which simplifies the evaluation of the integral.

PREREQUISITES
  • Understanding of probability density functions and their properties
  • Knowledge of integration techniques, particularly integration by parts
  • Familiarity with the concept of maximum likelihood estimation
  • Basic calculus, including limits and derivatives
NEXT STEPS
  • Study the method of integration by parts in detail
  • Learn about maximum likelihood estimation in the context of exponential distributions
  • Explore the properties of exponential density functions and their applications
  • Review advanced calculus techniques, including differentiation under the integral sign
USEFUL FOR

Students in statistics, mathematicians, and data scientists who are interested in understanding maximum likelihood estimation and its application to exponential distributions.

twoski
Messages
177
Reaction score
2

Homework Statement



Given f(x;λ) = [itex]cx^{2}e^{-λx}[/itex] for x ≥ 0

Determine what c must be (as a function of λ) then determine the maximum likelihood estimator of λ.

The Attempt at a Solution



So I'm supposed to integrate this from 0 to infinity, from what i can gather.

Let u = [itex]x^{2}[/itex], du = 2xdx, dv = [itex]e^{-λx}[/itex] and v = [itex]-e^{-λx} / λ[/itex]

After a bit of work i end up with:

-c/λ [ [itex]x^{2}e^{-λx}|_{0}^{∞} + 2( xe^{-λx}/λ |^{∞}_{0})[/itex] ]

What throws me off is that evaluating this leaves me with -c/λ( 0 ), which has to be wrong...
 
Physics news on Phys.org
twoski said:

Homework Statement



Given f(x;λ) = [itex]cx^{2}e^{-λx}[/itex] for x ≥ 0

Determine what c must be (as a function of λ) then determine the maximum likelihood estimator of λ.

The Attempt at a Solution



So I'm supposed to integrate this from 0 to infinity, from what i can gather.

Let u = [itex]x^{2}[/itex], du = 2xdx, dv = [itex]e^{-λx}[/itex] and v = [itex]-e^{-λx} / λ[/itex]

After a bit of work i end up with:

-c/λ [ [itex]x^{2}e^{-λx}|_{0}^{∞} + 2( xe^{-λx}/λ |^{∞}_{0})[/itex] ]

What throws me off is that evaluating this leaves me with -c/λ( 0 ), which has to be wrong...

You have to integrate the second term from 0 to infinity, not just evaluate it. You'll need to integrate by parts again.
 
twoski said:

Homework Statement



Given f(x;λ) = [itex]cx^{2}e^{-λx}[/itex] for x ≥ 0

Determine what c must be (as a function of λ) then determine the maximum likelihood estimator of λ.

The Attempt at a Solution



So I'm supposed to integrate this from 0 to infinity, from what i can gather.

Let u = [itex]x^{2}[/itex], du = 2xdx, dv = [itex]e^{-λx}[/itex] and v = [itex]-e^{-λx} / λ[/itex]

After a bit of work i end up with:

-c/λ [ [itex]x^{2}e^{-λx}|_{0}^{∞} + 2( xe^{-λx}/λ |^{∞}_{0})[/itex] ]

What throws me off is that evaluating this leaves me with -c/λ( 0 ), which has to be wrong...

It might be easier to recognize that
[tex]x e^{-\lambda x} = - \frac{\partial}{\partial \lambda} e^{- \lambda x},[/tex]
and so forth.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 9 ·
Replies
9
Views
6K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
1
Views
1K