Maximum Likelihood Estimator

[bitty]

Homework Statement

An observation X has density function: f(x,/theta)=6x/(t^3)*(t-x) where t is a parameter: 0<x<t.

Given the single observation X, determine the maximum likelihood estimator for t.

Homework Equations

Included below

The Attempt at a Solution

For a sample size of n, the likelihood function is
L(t)=product[6x_i/(t^3)*(t-x_i)] from i=1 to n.
To maximize a product, we take the log L[t] and differentiate with respect to t and equate to 0.

d/dt [Log[t]=3/t+Sum[1/(t-x_i)] for i=1 to n.
However, I don't know how to solve 3/t+Sum[1/(t-x_i)]=0 for t since there is a t on the denominator in the sum. Is there a way to do this?

Or have I misinterpreted the question? Since the question says, "Given a single observation X" am I not supposed to take the product over n samples but rather set n=1 when finding my likelihood function L(t)=6x_i/(t^3)*(t-x_i?

 

[nate808]

You are correct in your interpretation of the question. Since the question specifies a single observation X, you do not need to consider a sample size of n. Your likelihood function should be L(t)=6x/(t^3)*(t-x).

To find the maximum likelihood estimator for t, we can take the log of the likelihood function and differentiate with respect to t:

Log[L(t)]=Log[6x/(t^3)*(t-x)]
=Log[6x]-3Log[t]+Log[t-x]

Taking the derivative with respect to t and setting it equal to 0, we get:

d/dt [Log[L(t)]] = -3/t + 1/(t-x) = 0

Solving for t, we get:

3/t = 1/(t-x)
3(t-x) = t
3t - 3x = t
2t = 3x
t = 3x/2

Therefore, the maximum likelihood estimator for t is t=3x/2.

I hope this helps clarify your understanding. Keep up the good work in your studies!