pawitan2001all
$N$ runners participate in a marathon and they are each assigned a number from 1 to $N$. From one location we record the following participants:
218 88 264 33 368 236 294 116 9
Make a reasonable assumption about the data and draw the likelihood of $N$.
$$ \begin{align} L_{i}(\theta) &= \frac{1}{\theta} \ L(\theta) &= \begin{cases} \prod_{i=1}^{9} L_{i}(\theta) = \left[L_{i}(\theta) \right]^9 \ 0 \quad \text{if } \exists x_{i} > \theta \end{cases} \end{align} $$
The following shows the heart rate (in beats/minute) of a person, measured throughout the day:
73 75 84 76 93 79 85 80 76 78 80
Assume the data are an iid sample from $\mathcal{N}(\theta, \sigma^2)$, where $\sigma$ is known at the observed sample variance. Denote the ordered values by $x_{(1)}, x_{(2)}, ..., x_{(11)}$. Draw and compare the likelihood of $\theta$ if
(a) the whole data $x_{1}, x_{2}, ..., x_{11}$ are reported.
$$ L(\theta) = \prod_{i=1}^{11} \phi\left(\frac{x_{i} - \theta}{\sigma} \right) $$
(b) only the sample mean $\bar{x}$ is reported.
$$ L(\theta) = \phi\left(\frac{\sqrt{n} (\bar{x} - \theta)}{\sigma} \right) $$
(c) only the sample median $x_{(6)}$ is reported.
$$ \begin{align} L(\theta) &= \left[\prod_{i=1}^{5} P_{\theta}(x_{(i)} \leq x_{(6)}) \right] p_{\theta}(x_{(6)}) \left[\prod_{i=7}^{11} P_{\theta}(x_{(i)} \geq x_{(6)}) \right] \ &= \left[\Phi\left(\frac{x_{(6)} - \theta}{\sigma} \right) \right]^5 \phi\left(\frac{x_{(6)} - \theta}{\sigma} \right) \left[1 - \Phi\left(\frac{x_{(6)} - \theta}{\sigma} \right) \right]^5 \end{align} $$
(d) only the minimum $x_{(1)}$ and maximum $x_{(n)}$ are reported.
$$ L(\theta) = \frac{n (n - 1)}{2} \epsilon^2 \phi\left(\frac{x_{(1)} - \theta}{\sigma} \right) \phi\left(\frac{x_{(n)} - \theta}{\sigma} \right) \left[\Phi\left(\frac{x_{(n)} - \theta}{\sigma} \right) - \Phi\left(\frac{x_{(1)} - \theta}{\sigma} \right) \right]^{n-2} $$
(e) only the lowest two values $x_{(1)}$ and $x_{(2)}$ are reported.
$$ L(\theta) = \phi\left(\frac{x_{(1)} - \theta}{\sigma} \right) \phi\left(\frac{x_{(2)} - \theta}{\sigma} \right) \left[1 - \Phi\left(\frac{x_{(2)} - \theta}{\sigma} \right) \right]^9 $$
Given the following data
0.5 -0.32 -0.55 -0.76 -0.07 0.44 -0.48
draw the likelihood of $\theta$ based on each of the following models:
$$ \begin{align} L_{i}(\theta) &= p_{\theta}(x_{i}) \ L(\theta) &= \prod_{i=1}^{n} L_{i}(\theta) \end{align} $$
Ten light bulbs are left on for 30 days. One fails after 6 days, another one after 28 days, but the others are still working after 30 days. Assume the lifetime is exponential with density $p_{\lambda}(x) = \frac{1}{\lambda e^{x / \lambda}}$.
(a) Given $\lambda$, what is the probability of a light bulb working more than 30 days?
$$ \begin{align} P_{\lambda}(x \leq 30) &= 1 - \frac{1}{e^{30 / \lambda}} \ P_{\lambda}(x \geq 30) &= 1 - P_{\lambda}(x \leq 30) \end{align} $$
(b) Derive and draw the likelihood of $\lambda$. (Hint: only the first two order statistics are reported.)
$$ L(\lambda) = p_{\lambda}(6) p_{\lambda}(28) \left[P_{\lambda}(x \geq 30) \right]^8 $$
(c) Derive the MLE of $\lambda$.
$$ \hat{\lambda} = \arg \max_{\lambda} L(\lambda) $$
To get the MLE, the score function and the score equation is used.
$$ \begin{align} \log L(\lambda) &= p_{\lambda}(6) + p_{\lambda}(28) + 8 P_{\lambda}(x \geq 30) \ S(\theta) = \frac{\partial \log L(\lambda)}{\partial \theta} &= 274 - 2 / \lambda \ \hat{\lambda} &= 1 / 137 \end{align} $$
(d) Estimate how long it takes before 90% of the light bulbs will fail.
$$ y = - \hat{\lambda} \ln(1 - 0.9) $$
(e) Suppose the only information available is that two have failed by 30 days, but not their exact failure times. Draw the likelihood of A and compare with (b).
$$ \begin{align} L(\lambda) &= \left[P_{\lambda}(x \leq 30) \right]^2 \left[P_{\lambda}(x \geq 30) \right]^8 \end{align} $$
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