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I am having trouble understanding how to use the indicator function to help find the likelihood.

Let $Y_1, Y_2, ... , Y_n$ be a random sample from a population with density function $$ f (y | \theta ) = \begin{cases} 2\theta^2 y^{-3}, & \theta < y < \infty \\ 0, & \text{elsewhere} \end{cases} $$ Find the MLE for $\theta$.

I have two questions about this problem:

  1. We are given that Y min is sufficient for $\theta$. I am told that this is enough to state that Y min is then the MLE for $\theta$ but my question is why this is the case. I know that the MLE is always a function of the sufficient statistic, but I am having trouble understanding how we know that the MLE is ismply $\theta$ and not some other function of the sufficient statistic (such as Y min -1) that involves $\theta$. Ah I hope that was worded clearly.

  2. My next question is this. Assuming that we are not given that the Y min is sufficient for $\theta$, I am having some trouble deriving the MLE. Here is what I have so far: $$ L (\theta) = \prod \frac{2\theta^2}{y^3} = \frac{2^n\theta^{2n}}{\prod y^3} $$

Taking the log of this we obtain:

l ($\theta$) = nln(2) + 2nln($\theta$) - 3 ln( $\prod$ y)

Taking the derivative with respect to theta and setting equal to zero results in:

0 = 0 + $\frac{2n}{\theta}$ + 0

This is obviously an issue because the result does not include the data, so clearly I am missing something. In past experience this result means that the MLE will be either a min or max statistic, but I don't know which one or how to obtain it.

Edit:

Is it enough to just take the log:

l ($\theta$) = nln(2) + 2nln($\theta$) - 3 ln( $\prod$ y)

And say that the likelihood is maximized when the value of y is smallest? So, the MLE of $\theta$ is the minimum statistic?

jmoore00
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    Your trouble starts with the PDF, which should instead read $$f(y\mid\theta)=2\theta^2y^{-3}\mathbf 1_{\theta\leqslant y}$$ Thus, your $L(\theta)$ is also incorrect, which should read $$L(\theta)=2^n\theta^{2n}p(y_1,\ldots,y_n)\mathbf 1_{\theta\leqslant m(y_1,\ldots,y_n)}$$ with $$p(y_1,\ldots,y_n)=\prod_{i=1}^ny_i^{-3}\qquad m(y_1,\ldots,y_n)=\min{y_1,\ldots,y_n}$$ Thus, the function $L$ increases from $L(0)=0$ to $$L(\min{y_1,\ldots,y_n})=2^n\min{y_1,\ldots,y_n}^{2n}p(y_1,\ldots,y_n)$$ and $L(\theta)=0$ for every $\theta>\min{y_1,\ldots,y_n}$. Thus, $L(\theta)$ is maximal at... – Did Oct 31 '17 at 20:36
  • Should I always include the indicator function when I write the likelihood? I'm a bit confused because sometimes my professor includes and sometimes she doesn't. Does it matter in cases where you won't need to manipulate it in order to find the MLE? – jmoore00 Nov 01 '17 at 00:20
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    You may omit it if this does not make you lost... All I can tell is that an absurd amount of questions on this site are simply due, one way or another, to its omission. – Did Nov 01 '17 at 07:59
  • Did you find the MLE, based on my comments? – Did Nov 01 '17 at 10:00
  • I did. Thank you! I know that I needed the indicator function for this problem but I was a bit confused on how to use it and it is isn't really introduced in my textbook. I guess my main question now is the first question I asked about using the sufficient statistic to find the MLE. – jmoore00 Nov 01 '17 at 18:26
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    Possible duplicate of Find the MLE of $p$ where $f(y;p)=2p^2y^{-3}$. Also https://math.stackexchange.com/questions/446142/pareto-distribution-mle – StubbornAtom Dec 02 '19 at 12:48

1 Answers1

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Deriving the MLE: You can save yourself some trouble in these problems by simplifying your likelihood function at the outset. Remember that the likelihood function needs only be proportional to the sampling density (in the parameters) and so you can drop any multiplicative term that does not have parameters in it. You should also make sure to keep your range constraint (either as an indicator, or as an explicit constraint on the range) but note that it is now treated as a constraint on the parameter, with the data being treated as fixed.

Hence, you should write your likelihood function in its most simplified form as:

$$L_{\boldsymbol{y}}(\theta) = \theta^{2n} \quad \quad \text{ } \text{ } \text{ } \text{ } \text{for } 0 < \theta \leqslant \min \{ y_1, ..., y_n \}.$$

The corresponding log-likelihood function is:

$$\ell_{\boldsymbol{y}}(\theta) = 2n \ln \theta \quad \quad \text{for } 0 < \theta \leqslant \min \{ y_1, ..., y_n \}.$$

We can immediately see that the log-likelihood is an increasing function of the parameter, which means that the MLE occurs at the upper boundary $\hat{\theta} = \min \{ y_1, ..., y_n \}$.

Adjustment for bias: As with all problems with a monotone log-likelihood function, the MLE is a biased estimator of the true parameter value. With a bit of algebra we can derive the distribution of the minimum observed values, and then show that:

$$\mathbb{E}(\hat{\theta}) = \frac{2n}{2n-1} \cdot \theta.$$

We see from this result that the MLE systematically overestimates the true parameter, which is not surprising, given that $\hat{\theta} \geqslant \theta$ from the boundary condition. We can obtain an unbiased estimator of the parameter by using the adjusted-MLE:

$$\tilde{\theta} = \frac{2n-1}{2n} \cdot \min \{ y_1, ..., y_n \}.$$

This latter estimator is unbiased and is also a consistent estimator of the true parameter value. It is likely that we would use this estimator in this case.

Demonstrating Sufficiency: The simplest way to demonstrate that $T(\boldsymbol{y}) \equiv \min \{ y_1, ..., y_n \}$ is a sufficient statistic is to use the Fisher-Neyman factorisation theorem. In this case we can separate the sampling density into the required multiplicative parts as follows:

$$f_\theta (\boldsymbol{y}) = h(\boldsymbol{y}) g_\theta (T(\boldsymbol{y})) \quad \quad \quad h(\boldsymbol{y}) \equiv \frac{2^n}{\prod y_i^3} \quad \quad g_\theta (t) \equiv \theta^{2n} \mathbb{I}(\theta \leqslant t).$$

Ben
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