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Please let me know whether I am wrong or right regarding the calculation of mle for uniform distribution

[ in both the cases below $\theta > 0$ ]

  1. let $X_1,X_2, . . .X_n $ be a random sample from uniform (-$\theta , 3\theta$)

then the joint density $\frac{1}{(4\theta)^n}$ must be max at $\frac{max|x_i|}{3}$ ( and this should be the m.l.e)

  1. let $X_1,X_2, . . .X_n $ be a random sample from uniform ($\theta , 3\theta +7$)

then the joint density $\frac{1}{(2\theta + 7)^n}$ must be max at $\frac{X_{(n)}-7}{3}$ ( and this should be the m.l.e)

Please let me know whether i am wrong or right ?

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    $\begingroup$ No they are not correct. These solutions would be valid if the boundary of the support from below was $0$. Now that it is not, the situation is more complicated. Do you know the step-by-step logic based on which your solutions would be correct in that case? Can you extend this logic for the case you are examining? $\endgroup$ – Alecos Papadopoulos Mar 30 '17 at 17:13
  • $\begingroup$ Study the link provided in this answer math.stackexchange.com/a/49546/87400 for the usual case, if you have doubts. $\endgroup$ – Alecos Papadopoulos Mar 30 '17 at 17:21
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I will solve the first case, and leave the second for the OP's self-study.

It is evident that $\theta >0$, and the problem is not trivial if $n>1$.

For the first case, the likelihood function is

$$L=\frac {1}{(4\theta)^n}\cdot I\{-\theta \leq x_{(1)} \leq x_{(n)} \leq 3\theta\}$$

We have two conditions here that must hold at the same time, for the likelihood to be non-zero. We can write

$$I\{-\theta \leq x_{(1)} \leq x_{(n)} \leq 3\theta\} = I\{-\theta \leq x_{(1)} \}\cdot I\{ x_{(n)} \leq 3\theta\}$$

Note that in principle the fact that the support extends to negative values does not guarantee that the available sample at hand will contain negative values also, or even that some of the values will be positive. So we are not guaranteed that the minimum order statistic will be negative, nor that the maximum order statistic will be positive.

So to cover all possible samples, the likelihood will be non-zero if $\theta \geq -x_{(1)}$ and $\theta \geq x_{(n)}/3 \implies \theta \geq \max \{-x_{(1)},x_{(n)}/3\}$.

At the same time, the likelihood under this condition is a decreasing function of $\theta$. So the best we can do is to select the minimum positive value for $\theta$ that at the same times ensures that the likelihood will be non-zero. So we obtain

$$\hat \theta_{MLE} = \max \{-x_{(1)},x_{(n)}/3\}$$

Then we have cases (I ignore for clarity the special cases of equality between the minimum and the maximum order statistic, which also includes the pathological case where both are equal to zero).

A) $x_{(1)} , x_{(n)} < 0$.
Here the only positive value and so the MLE is $\hat \theta = |x_{(1)}|$. The intuition is that if the maximum order statistic is negative, then the minimum order statistic will be closer to the unknown parameter.

B) $0< x_{(1)} , x_{(n)}$.
Here too we have only one positive candidate in the $\max$ condition, so $\hat \theta = x_{(n)}/3$

C) $ x_{(1)} < 0 < x_{(n)}$. Here both candidate values in the $\max$ condition will be positive so $\hat \theta = \max \{|x_{(1)}|,x_{(n)}/3\}$.

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  • $\begingroup$ (oops! yes , i forgot to mention that $\theta$ > 0 ) ,well thanks 4 answering part 1 , but in the second part i am still getting the same answer , $\endgroup$ – ANUJ NAIN Mar 31 '17 at 3:39
  • $\begingroup$ @ANUJNAIN If the result for the second case satisfies the general conditions laid out in my answer, then it is correct. $\endgroup$ – Alecos Papadopoulos Mar 31 '17 at 9:24

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