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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?

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user136542
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2 Answers2

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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 $\theta_{ML} = \max\{x_1,\ldots,x_n\}$.

Marc
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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.

Reverend_Dude
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