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So I worked out how to find the maximum likelihood estimator of

$f(x, \alpha) = \alpha x^{-\alpha-1}$ for $x > 0,$ where $\theta > 0$ is a parameter is:

$$L(\alpha) = \prod_{i=1}^n \alpha x_i^{-\alpha-1} = \alpha^n \prod_{i=1}^n x_i^{-\alpha-1}$$

Then:

$$\log L(\alpha) = n\log(\alpha) + \sum_{i=1}^n (-\alpha-1) \log(x_i)$$

So:

$$n\log(\alpha) + (-\alpha-1) \sum_{i=1}^n \log(x_i)$$

However I'm not sure how to find the likelihood estimator for

$$f(x;\theta) = \frac{1}{2(\theta x)^{1/2}} \text{ for } 0<x\le\theta$$

Any Advice would be very appreciated!

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  • $\begingroup$ Welcome to Math.SE! I have tried to improve the readability of your question by improving $\rm \LaTeX$. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 23 '18 at 20:39
  • $\begingroup$ BEGIN QUOTE $f(x, \alpha) = \alpha x^{-\alpha-1}$ for $x > 0,$ where $\theta > 0$ is a parameter is: END QUOTE Did you mean "where $\alpha>0$ is a paramter"? $\qquad$ $\endgroup$ – Michael Hardy Mar 24 '18 at 17:58
  • $\begingroup$ Please read the answer to this. $\endgroup$ – Did Mar 26 '18 at 16:04
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Note that for the second problem if $\theta < x$, then $f(x,\theta)=0$. In particular $$ \sup_{\theta>0}f(x;\theta)=\sup_{\theta\geq x} f(x,\theta)=\frac{1}{2\sqrt{x}}\sup_{\theta\geq x} \frac{1}{\sqrt{\theta}}=\frac{1}{2\sqrt{x}}\frac{1}{\sqrt{x}}=f(x,x) $$ Thus $\hat{\theta}(x)= x$ for a sample of size $1$. For a sample of size $n$, a similar approach can be taken. In fact in that case $\hat{\theta}(\mathbf{x})=\max(x_1,\dotsc,x_n)$.

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  • $\begingroup$ thank you, but $\theta$ is not smaller that 0.I just need a bit of help in the first step of writing $\f(x, \theta)$ in the new form. I know that you need to differentiate that form. I'm just not confident in how to write it. @FoobazJohn $\endgroup$ – abigalpeters Mar 24 '18 at 9:00
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\begin{align} L(\theta) & = \left.\begin{cases} \displaystyle \prod_{i=1}^n \frac 1 {2(\theta x_i)^{1/2}} & \text{if } \theta \ge \text{all of }x_1,\ldots,x_n \\[8pt] 0 & \text{otherwise} \end{cases}\right\} \\[15pt] & = \left. \begin{cases} \displaystyle \frac 1 {2^n\theta^{n/2}} \cdot \frac 1 {\prod_{i=1}^n x_i^{1/2}} & \text{if } \theta\ge\max\{x_1,\ldots,x_n\} \\[8pt] 0 & \text{otherwise} \end{cases} \right\}. \end{align}

As $\theta$ decreases, $L(\theta)$ increases until $\theta$ gets down to $\max\{x_1,\ldots,x_n\}.$ So the MLE for $\theta$ is $\max\{x_1,\ldots,x_n\}.$

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  • $\begingroup$ in the second bracketed equation, where the xi is. why is it not 1/(big pie)* xi^(1/2)? $\endgroup$ – abigalpeters Mar 26 '18 at 11:29
  • $\begingroup$ @abigalpeters : Typo. I've added that. $\endgroup$ – Michael Hardy Mar 26 '18 at 15:41
  • $\begingroup$ thank you so much! I understand now. There is a follow on question that asks to find the her 90% confidence interval for θ, i there a specific technique i should use for this question? $\endgroup$ – abigalpeters Mar 26 '18 at 17:38
  • $\begingroup$ @abigalpeters : Note that you can "accept" this answer. To find a confidence interval, you probably want to use the probability distribution of $\max\{X_1,\ldots,X_n\}.$ Note that $\Pr(X_1 \le x) = \sqrt{x/\theta\,\,}$ for $0\le x\le\theta.$ This implies that $\Pr(\max \le x) = \sqrt{ x/\theta \,\,}^{\,n}$ for $0\le x\le\theta. \qquad$ $\endgroup$ – Michael Hardy Mar 26 '18 at 21:40
  • $\begingroup$ Notice that $$ f(x_1,\ldots,x_n) = \frac 1 {2^n \theta^{n/2}} \left( \prod_{i=1}^n x_i^{-1/2} \right) \cdot 1_{[0,\,\theta]} (\max\{ x_1,\ldots, x_n\}), $$ and here you have $(1)$ a factor that does not depend on $\theta,$ namely the product from $i=1$ to $n$, and $(2)$ a factor that depends on $(x_1,\ldots,x_n)$ only through $\max,$ namely everything else. By Fisher's factorization theorem, that means the maximum is sufficient for this family of distributions, i.e. the conditional distribution of $X_1,\ldots,X_n$ given their maximum does not depend on $\theta.$ That being the case$\,\ldots$ $\endgroup$ – Michael Hardy Mar 26 '18 at 21:47

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