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Can you show me what I'm doing wrong here? This is the homework problem:

Consider a random sample $Y_1, \ldots , Y_n$ from the pdf $f_Y(y;\theta) = 2y\theta^2$ where $0\le y \le \frac{1}{\theta}$. Find the maximum likelihood estimator for $\theta$.

The likelihood function for this pdf is $\prod_{i=1}^{n} f_Y(y;\theta) = \prod_{i=1}^{n} 2y_{i}\theta^2 = 2^n \theta^{2n} \prod_{i=1}^{n} y_i$.

So to me, it looks like I can maximize the likelihood function by choosing $y_{max}$ for the MLE. But then what do I do about the condition $0\le y \le \frac{1}{\theta}$?

I tried looking up how to do this but the only example I can find is the one about the uniform distribution, but that one is much more straightforward because the condition on y is just $0 \le y \le \theta$ and not $\frac{1}{\theta}$. To maximize $\frac{1}{\theta}$ I would choose my MLE to be $y_{min}$ but that wouldn't maximize the likelihood function.

Any help would be GREATLY appreciated.

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I suggest you draw the likelihood as a function of $\theta$, without forgetting that $1/\theta$ must be greater than any observation (i.e. what are the bounds on $\theta$?).

Keep in mind that everything but $\theta^{2n}$ in the likelihood is a constant, and so you can write it as $c.\theta^{2n}$; so just draw $\cal{L}/c$ over the domain of $\theta$. You may find it more convenient to deal with the log-likelihood, or you may not.

It should help you clarify what you're doing.

If you're still stuck, consider thinking in terms of $\psi = 1/\theta$ and then go back to doing it in terms of $\theta$.

(Alternatively - look back at that uniform example. What would the MLE of $\theta$ be if the data were uniform on $[0,1/\theta]$? Can you see how to do the original problem now?)

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  • $\begingroup$ When I think of the alternate uniform example and of your idea of $\psi = 1/\theta$, I come up with the MLE being 1/$y_{max}$. Am I on the right track? $\endgroup$
    – smootie
    Commented Mar 18, 2013 at 0:58
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    $\begingroup$ It's right in that case, so perhaps you're on the right track; let's make sure you're vizualizing it correctly. What does the likelihood curve look like for that example (in general, descriptive terms)? $\endgroup$
    – Glen_b
    Commented Mar 18, 2013 at 1:00
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In this kind of problem, it helps a lot to write explicitly the indicators in the definitions of the densities. You have $Y_1,\dots,Y_n$, conditionally IID given $\Theta=\theta$, such that $$ f_{Y_i\mid\Theta}(y_i\mid\theta) = 2y_i\theta^2\, I_{[0,\,1/\theta]}(y_i) \, . $$ Since $0\leq y_i\leq 1/\theta$ if and only if $0\leq \theta\leq 1/y_i$, the indicator can be rewritten as $$ I_{[0,\,1/\theta]}(y_i) = I_{[0,\,1/y_i]}(\theta) \, . $$ Introducing the notations $Y=(Y_1,\dots,Y_n)$ and $y=(y_1,\dots,y_n)$, the likelihood is $$ L_y(\theta) = f_{Y\mid\Theta}(y\mid\theta) = \prod_{i=1}^n f_{Y_i\mid\Theta}(y_i\mid\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} \left( \prod_{i=1}^n I_{[0,\,1/y_i]}(\theta) \right) \, . $$ But a product of indicators is the indicator of the intersection (why?), so $$ L_y(\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} I_{\cap_{i=1}^n [0,\,1/y_i]}(\theta) \, , $$ and (why?) $$ I_{\cap_{i=1}^n [0,\,1/y_i]}(\theta) = I_{[0,\,1/y_{(n)}]}(\theta) \, , $$ in which $y_{(n)}=\max \{y_1,\dots,y_n\}$. Hence, $$ L_y(\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} I_{[0,\,1/y_{(n)}]}(\theta) \, , $$ and this is maximized (why? draw the graph for some fixed sample $y$) when $\theta=1/y_{(n)}$, yielding the maximum likelihood estimator $$ \hat{\Theta}_{ML}=\frac{1}{Y_{(n)}} \, . $$

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    $\begingroup$ Zen, please read the discussion of the self-study tag here. I feel your answer goes some what beyond giving 'helpful hints', and jumps right into 'doing someone's homework for them'. You do make an error partway through, so it may not matter so much for the OP's learning this time. $\endgroup$
    – Glen_b
    Commented Mar 18, 2013 at 1:35
  • $\begingroup$ Hi, Glen_b. Normally I would give a good beginning and add details later, but I'm in a hurry here and won't have the opportunity to check the development of this question. Although I've given a lot of details, the OP will still have to think carefully to understand exactly how it is done. Anyway, I'll flag the mods, and check if this answer should be deleted. $\endgroup$
    – Zen
    Commented Mar 18, 2013 at 1:48
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    $\begingroup$ I see you corrected the error. I'd really rather you didn't delete the whole thing - especially not the recommendation about using indicators, which is an important tool for keeping bounds straight, and given that, some of your algebra is essential for showing how to use them. At most I'd ask that you maybe didn't carry it the whole way through (though the OP has probably already read it by now, so it's likely moot). On the whole I think it's an excellent answer and it would be a shame to lose it. If it didn't polish the thing off, I'd have upvoted it already. $\endgroup$
    – Glen_b
    Commented Mar 18, 2013 at 2:20
  • $\begingroup$ To clarify, if the options are deletion or leaving it as is, I'd say leave it stand. $\endgroup$
    – Glen_b
    Commented Mar 18, 2013 at 2:26
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    $\begingroup$ "because even though Y(n) would maximize the original pdf, there could be y's that fall outside of the range [0, 1/Y(n)], thus making the likelihood function equal zero. Instead, we choose an MLE that will allow us to "accept" the most y's" -- Smootie, you don't seem to correctly characterize how it works there. You simply maximize the likelihood function. There's nothing fancy going on. You are leading yourself into misunderstanding. I suggest that you go back to absolute basics - I'd start by attempting to respond to the question I asked you under my answer. $\endgroup$
    – Glen_b
    Commented Mar 18, 2013 at 5:25

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