0
$\begingroup$

Give IID Data Samples $X_n = $ {$x_1, x_2, ..., x_n$} generated from a uniform distribution $U(x|0,\theta)$.

$p(x|\theta) = U(x|0,θ) = ${$ \frac{1}{\theta}$ for $0 \leq x \leq θ$ and $0$ otherwise}.

Now assuming $X_2 = [1, 3, 2, 4]$ have been observed.

What is the maximum likelihood arg $max \theta, p(X_2 | \theta)$?

The $p(X|\theta) = p(x_1, ..., x_n | \theta) = \theta^{-n}$

Taking log both sides we get $\log p(X|θ) = n\cdot \log(\frac{1}{θ})$ Taking derivative we get $\frac{-n}{\theta} $which is less than $0$.

Could you tell me how to proceed for the next?

$\endgroup$
  • $\begingroup$ If the (log of the) likelihood is decreasing, at which end is the maximum? $\endgroup$ – Henry Apr 9 '14 at 0:14
2
$\begingroup$

This is a problem which is difficult to solve with the general technique. Note that the likelihood $$ \mathcal{L}(\theta|x_1,\ldots,x_n) = p(x_1,\ldots,x_n|\theta) = \theta^{-n}, $$ however you need to be careful here. The likelihood is the probability distribution of $\theta$ given the sample $\{x_1,\ldots,x_n\}$, so note that the above equality is valid only for $\theta>\max\{x_1,\ldots,x_n\}$, since otherwise the likelihood is zero. Now we need to find $\theta$ that maximizes the likelihood. From its easy form (decreasing from $\max\{x_1,\ldots,x_n\}$ onwards), you immedeatly see that the maximizer is $\theta_{ML} = \max\{x_1,\ldots,x_n\}$.

$\endgroup$
0
$\begingroup$

When you deal with the uniform distribution you are best off just thinking in general about what the problem is asking.

The answer in this case is 4. That is because it is the largest number in the X2 set. however this is often an underestimate since theta could be much larger, but the most likly is just that largest of the group.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.