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Find the MLE of $p$ where $f(y;p)=2p^2y^{-3}$.

Attempt:

Method: find the likelihood function, differentiate with respect to $p$ then set to zero and solve for $p$.

$L(p;y)=\prod\limits_{i=1}^n [2p^2y^{-3}_i] =\prod\limits_{i=1}^n[2] \prod\limits_{i=1}^n[p^2] \prod\limits_{i=1}^n [y^{-3}_i] = 2^n p^{2n} \prod\limits_{i=1}^n [y^{-3}_i]$

Differentiating the log likelihood gives $2n/p$ (i think) and the $y_i$ all disappear. So what happens when you set this to zero? Because if $0=2n/p$ then surely $0=2n$ and $n=0$.

Have I misstated $L(p;y)$? Or is something going wrong with my differentiation? (Or both?)

Did
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Maths Barry
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    Previously asked: https://math.stackexchange.com/questions/446142/pareto-distribution-mle. Without mentioning support of the distribution, your $f(y;p)$ is incomplete. And when you mention the support, say by using indicator functions, you will see that differentiation is not valid to derive the MLE because the likelihood is not differentiable everywhere in the first place. See the linked threads here. – StubbornAtom Jan 06 '19 at 14:16
  • THANK YOU! I didn't realise this was a Pareto distribution so I am extremely grateful for this. – Maths Barry Jan 06 '19 at 14:36

1 Answers1

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Both the PDF and the likelihood in your post are incorrect, due to the fact that you forget to include indicator functions.

In fact, the PDF is $$f(y;p)=2p^2y^{-3}\mathbf 1_{y\geqslant p}$$ hence, the likelihood of some i.i.d. sample $\mathbf y=(y_i)_{1\leqslant i\leqslant n}$ for the PDF $f(\ ;p)$ is $$L(p;\mathbf y)=2^np^{2n}\left(\prod_iy_i^{-3}\right)\mathbf 1_{m(\mathbf y)\geqslant p}$$ where $$m(\mathbf y)=\min_{1\leqslant i\leqslant n} y_i$$ One sees readily that $$L(p;\mathbf y)=c(\mathbf y)p^{2n}\mathbf 1_{m(\mathbf y)\geqslant p}$$ for some positive constant $c(\mathbf y)$ independent of $p$, hence $L(\ ;\mathbf y)$ is maximum at $$\hat p=m(\mathbf y)$$ No differentiation is involved, rather a precise understanding of the situation.

Did
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  • What if we assume $Y \geq p$? – Maths Barry Jan 06 '19 at 13:14
  • Sorry, but what if what? – Did Jan 06 '19 at 13:22
  • What if the range is $0<p\leq y<\infty$. The indicator function is not still needed then is it? And differentation would be the only way to get the MLE? – Maths Barry Jan 06 '19 at 13:33
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    Yes the range of each $f(\ ;p)$ is $[p,\infty)$. This is exactly why the indicator function is (much) needed. "And different[i]ation would be the only way to get the MLE?" Sorry but what are you talking about? If you can present a proof based on solving $$\frac{\partial L(p;\mathbf y)}{\partial p}=0$$ please do so (there are none...). – Did Jan 06 '19 at 14:13