Find Expected Value of X(n) from Uniform Distribution

  • Context: Graduate 
  • Thread starter Thread starter apalmer3
  • Start date Start date
  • Tags Tags
    Maximum
Click For Summary
SUMMARY

The expected value of the largest observation, X(n), from an independent sample of size n drawn from a uniform distribution on the interval [0, θ] can be calculated using its cumulative distribution function (CDF). The CDF is given by F(t) = (t/θ)^n, which leads to the probability density function (PDF) f(t) upon differentiation. The expected value is then computed using the integral ∫₀^θ t f(t) dt, providing a definitive method for determining the expected value of the maximum observation in the sample.

PREREQUISITES
  • Understanding of uniform distribution and its properties
  • Knowledge of cumulative distribution functions (CDF) and probability density functions (PDF)
  • Familiarity with differentiation and integration techniques
  • Basic statistics concepts, particularly related to expected values
NEXT STEPS
  • Study the derivation of cumulative distribution functions for different distributions
  • Learn about the properties of the uniform distribution in depth
  • Explore advanced integration techniques for calculating expected values
  • Investigate the concept of order statistics and their applications in statistics
USEFUL FOR

Statisticians, data analysts, and students studying probability theory who are interested in understanding the behavior of maximum values in random samples from uniform distributions.

apalmer3
Messages
36
Reaction score
0
I swear that I used to know this.

If you have an independent sample of size n, from the uniform distribution (interval [0,[tex]\theta[/tex]]), how do you find the Expected Value of the largest observation(X(n))?
 
Physics news on Phys.org
If [tex]X_{(n)}[/tex] is the maximum in the sample, you first find its distribution. Since you have a random sample of size [tex]n[/tex], you can write

[tex] F(t) = \Pr(X_{(n)} \le t) = \prod_{i=1}^n \Pr(X_i \le t) = \left(\frac{t}{\theta}\right)^n[/tex]

Differentiate this w.r.t. [tex]t[/tex] to find the density [tex]f(t)[/tex], and the expected value is

[tex] \int_0^{\theta} t f(t) \, dt[/tex]
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K