(a) Based on the random sample $Y_1 = 6.3$ , $Y_2 = 1.8$, $Y_3 = 14.2$, and $Y_4 = 7.6$, use the method of maximum likelihood to estimate the parameter $\theta$ in the uniform pdf

$f_Y(y;\theta) = \frac{1}{\theta}$ , $0 \leq y \leq \theta$

(b) Suppose the random sample in part (a) represents the two-parameter uniform pdf

$f_Y(y;\theta_1, \theta_2) = \frac{1}{\theta_2 - \theta_1}$ , $\theta_1 \leq y \leq \theta_2$

Find the maximum likelihood estimates for $\theta_1$ and $\theta_2$.

My attempt

(a) The likelihood function is given by $L(\theta)= \Pi_{i = 0}^{n}\frac{1}{\theta} = \theta^{-n} $

Take natural log both sides, so ln$L(\theta) = - n$ ln($\theta)$ Take derivative and set it equal to zero. Thus, $\frac{d}{d\theta}$ln$L(\theta)$ = -$\frac{n}{\theta}$

if we try to solve for theta equal to zero it wont work. So we have to reduce the the value for $y$ as much as possible. So choose the value of $\theta_e = y_{min} = 1.8$

(b) since the range is dependent on $ \theta$, finding the derivative equal to zero won't work. So the maximun likelihood for $\theta_1e = 14.2$ and the maximun likelihood for $\theta_2e = 1.8$

Can anyone please verify this? If this does not work, can someone please help? Any feedback/hint would help. Thank you in advance.

  • $\begingroup$ I must be doing something wrong... here's my attempt: we have $$L(\theta) = \dfrac{1}{\theta^4} = \theta^{-4}\implies \ell(\theta) = -4\ln(\theta) \text{, }\theta \in [\min_{k}Y_{k}, \infty)\text{.}$$ Notice that there is no maximum value of $\theta$ which maximizes $\ell$... so is there even a MLE for $\theta$? $\endgroup$ – Clarinetist Feb 7 '15 at 2:26
  • $\begingroup$ did you take the derivative and set it equal to zero? and solve for theta? $\endgroup$ – Mahidevran Feb 7 '15 at 2:40
  • $\begingroup$ Yeah, then you get $\ell^{\prime}(\theta) = \dfrac{-4}{\theta}$. If you set this equal to $0$, of course, your equation leads to nonsense. However, notice that you need to maximize $\ell$ but not $\ell^{\prime}$. $\endgroup$ – Clarinetist Feb 7 '15 at 2:45
  • $\begingroup$ MLE of upper limit in case of Uniform Distribution is max(sample values) which is $14.2$ in this case. $\endgroup$ – ethanol Feb 7 '15 at 4:58

Usually to find MLE's for density functions where the domain of the density depends on the parameter is not straightforward in the sense that differentating will not help.

Denote by $\textbf{1}_{[a,b]}(x)$ the indicator function meaning that $$\textbf{1}_{[a,b]}(x)= \begin{cases} 1 \mbox{ if } x\in [a,b] \\ 0 \mbox{ if not.}\end{cases}$$

The density function of a uniform random variable $U(0,\theta)$ is given by $$f_{\theta}(x) = \frac{1}{\theta} \textbf{1}_{[0,1]}(\theta).$$

Given a sample of $n$ random observations: $Y_1,\dots, Y_n$ (it is convenient to write observations with small letters and random variables with big letters, so when I write big $Y$'s means that these are computations before we collect data) the likelihood funtcion is given by: $$L(\theta|Y_1,\dots, Y_n)= \prod_{i=1}^n f_{\theta}(Y_i) = \frac{1}{\theta^n}\prod_{i=1}^n \textbf{1}_{[0,\theta]}(Y_i).$$

Now look at the product of indicator functions. We should try to write this as a function of $\theta$ instead of as a function of all the $Y_i$'s in order to know exactly what is the domain of $L$. This product is not zero only if $0<Y_i<\theta$ for all the $i=1,\dots,n$. In other words, $0<\min_i Y_i < \max_i Y_i < \theta$. So the domain of the likelihood function is only for $\theta>\max_{i=1,\dots,n} Y_i$. That is $$L(\theta|Y_1,\dots, Y_n)= \frac{1}{\theta^n} \textbf{1}_{[\max_{i=1,\dots,n} Y_i, \infty)}(\theta).$$

Now, the function $1/\theta^n$ function is strictly decreasing on $[0,\infty]$. So its maximum must be attained at the left value. In conclusion, the MLE of $\theta$ is given by the maximum values of the sample of $Y_i$'s which is quite intuitive: $$\hat{\theta}_{MLE} = \max_{i=1,\dots,n} Y_i.$$

Now try to reproduce the same computations and ideas when we have a uniform distribution depending on two parameters: $f_{\theta_1, \theta_2} (x) = \frac{1}{\theta_2- \theta_1}$ for $\theta_1 < y < \theta_2$. Intuition tells that if you have a sample of $Y_1,\dots, Y_n$ then the MLE's of $\theta_1$ and $\theta_2$ should be: $$ \hat{\theta_1} = \min_{n=1,\dots,n} Y_i \mbox{ and } \hat{\theta_2} = \max_{n=1,\dots,n} Y_i.$$

Observe that the estimators are expressed with capital letters. So they are random. Each time you collect a new sample you get different estimates. Hence when you collect a sample of observations $y_1,\dots, y_n$ (small letters) the estimates are: $$ \hat{\theta_1} = \min_{n=1,\dots,n} y_i \mbox{ and } \hat{\theta_2} = \max_{n=1,\dots,n} y_i.$$ (Just substitute the values you got in the exercise)

I hope this helped! ;)


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