Maximum Likelihood Estimation for Constant x and Exponential Distribution of z

  • Thread starter Thread starter cutesteph
  • Start date Start date
  • Tags Tags
    Maximum
Click For Summary
SUMMARY

The discussion focuses on finding the Maximum Likelihood Estimation (MLE) for a constant x in the context of an exponential distribution for z, defined by the probability density function (pdf) exp(-v-2) for v ≥ 2. The equation y = x + z is central to the problem, and the condition density f_v(y-x) is derived through integration. The participants clarify the limits of integration, emphasizing the importance of correctly interpreting the integral from -2+x to infinity, which leads to the conclusion that maximizing this function does not yield x = infinity as initially suggested.

PREREQUISITES
  • Understanding of Maximum Likelihood Estimation (MLE)
  • Familiarity with exponential distributions and their probability density functions
  • Knowledge of integration techniques in calculus
  • Experience with conditional densities in statistical analysis
NEXT STEPS
  • Study the properties of exponential distributions and their applications in MLE
  • Learn about integration techniques for evaluating conditional densities
  • Explore advanced topics in statistical inference related to MLE
  • Review examples of MLE in different statistical contexts
USEFUL FOR

Statisticians, data scientists, and students studying statistical inference who are interested in understanding MLE and its application to exponential distributions.

cutesteph
Messages
62
Reaction score
0

Homework Statement


y=x+z
where the pdf of z is exp(-v-2) for v >= 2 and 0 otherwise and x is just an unknown constant

what is the MLE of z which maximizes the condition density f_v(y-x)

Homework Equations


The Attempt at a Solution


f_v(y-x) = integral of -2+x to infinity of exp(-y+x-2) dy = exp(2x) but maximizing this would mean x = infinity?
 
Last edited:
Physics news on Phys.org
cutesteph said:
f_v(y-x) = integral of -2+x to infinity of exp(-y+x-2) dy = exp(2x) but maximizing this would mean x = infinity?
Do you mean "integral from -2+x to infinity of exp(-y+x-2) dy"? I don't get e2x for that.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K