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I've been looking at some statistical distribution work and found a question that I can't solve through standard derivation of an MLE. I've been told to maybe look at an indicator function but I've no idea how to use those for these. The question is to find an MLE for a, when given a random sample $X_1 , X_2 , ... , X_n$ with probability density function $$ f(x) = \frac{53x^{52}}{2a^{53}}$$ for $-a < x < a$.

When I try the standard way of taking the product, log and differential the second differential comes out as a positive, which wouldn't make it a maximum. Any explanation of how to work through this would be much appreciated!

To clarify, the method I was taught was as follows:

$$L(a) = \prod_{i=1}^n\frac{53x^{52}}{2a^{53}} = (\frac{53}{2a^{53}})^n\prod_{i=1}^nX_i^{52}$$ Taking logs

$$Log(L(a)) = nLog(53) -53n(Log(a) + Log(2)) + \sum52Log(X_i)$$

To take the derivative from here wrt to a, set to 0 and then rearrange to find $a$ and call this the mle, but this eliminates $a$ entirely, as I've realised I've done something wrong in my previous work.

kvj121227
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  • what goes wrong when you do that? – user2879934 May 14 '17 at 17:15
  • My initial issue was that the second derivative of the log would be positive, which would make it a local minimum, not the maximum I'm after. I reworked it and found that I'd made a mistake differentiating it though. – kvj121227 May 14 '17 at 20:42

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Remember the likelihood function is a function of $a.$ The likelihood function on data $x_1,\ldots x_n$ is $$L(a) = \left\{\begin{array}{ll}\left(\frac{53}{2a^{53}}\right)^n(x_1x_2\ldots x_n)^{52} & a>\max_i |x_i|\\0&a \le \max_i |x_i|\end{array}\right.$$ since the data is only possible if all of the $x_i$ are in the interval $(-a,a).$ Note that the first piece of this function is always greater than or equal to zero and that it decreases in $a.$ So to maximize, you want to find the smallest value of $a$ such that you are still on the first branch. This is just given by $a=\max_i|x_i|.$

(Technically, the way I wrote the function above (which is from how you wrote the PDF) there is no minimum value of $a$ on the first piece (and $L(\max_i|x_i|)=0$) which is not the maximum value) but in this case its ok to take the limit cause we could have defined the PDF's support to be $[a,a]$ rather than $(a,a).$)

The lesson here is that sometimes we maximize a function by taking derivatives, but not always. But remember when you learned to find global minima, you had to check the endpoints too, as well as the critical points which are only candidate local extrema. That's what's going on here.

spaceisdarkgreen
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  • Sorry, just to clarify, you've stated strictly less than in the likelihood but then taken equal as $a$, how do you get around it not being included in the interval originally? – kvj121227 May 14 '17 at 20:59
  • @kvj121227 If you set the estimator to $\hat a = \max_i|x_i|-\epsilon$ you would see that the likelihood increases as you take $\epsilon\to 0$ (but then drops precipitously to zero when $\epsilon = 0$). So you can think of it like a limit where the maximum likelihood value of $a$ is "just to the left of" $ \max_i|x_i|.$ To see that there's no problem setting $\hat a = \max_i|x_i|$, note that we could have defined the original PDF to have support $[-a,a]$ rather than $(-a,a)$... this is the exact same distribution. In this case $\hat a = \max_i|x_i|$ is literally the maximum likelihood. – spaceisdarkgreen May 15 '17 at 00:54
  • @kvj121227 BTW I don't know why people downvoted your question. Perhaps they didn't understand the problem you were having. It's always good to be more clear about it, but I thought you pointed out where you were having trouble just fine. – spaceisdarkgreen May 15 '17 at 00:59