Question

A random sample of size $5$ is taken from the distribution with density \[ f(x;\theta)= \begin{cases} \dfrac{3x^2}{\theta^3}, & 0[6pt] 0, & \text{elsewhere}, \end{cases} \] where $\theta$ is unknown. If the observations are $3,6,4,7,5$, then the maximum likelihood estimate of the $1/8$ quantile of the distribution (rounded off to one decimal place) is __________.

Hint:

For densities of the form $f(x;\theta)=k x^{k-1}/\theta^k$ on $(0,\theta)$, the MLE is $\widehat{\theta}=\max X_i$, and quantiles scale linearly with $\theta$.

Answer 1

Correct Answer: 3.5

Given

The probability density function is

\[ f(x;\theta)= \begin{cases} \dfrac{3x^2}{\theta^3}, & 0 < x < \theta \\ 0, & \text{elsewhere} \end{cases} \]

Sample size \( n = 5 \) and observations are: \[ 3,\;6,\;4,\;7,\;5 \]

---

Step 1: Find the MLE of \( \theta \)

For this distribution, the likelihood is positive only when \[ \theta \ge \max(X_1, X_2, \dots, X_n) \]

Hence, the maximum likelihood estimator is:

\[ \hat{\theta} = \max\{3,6,4,7,5\} = 7 \]

---

Step 2: Find the distribution function

The cumulative distribution function is:

\[ F(x) = \int_0^x \frac{3t^2}{\theta^3}\,dt = \frac{x^3}{\theta^3}, \quad 0 < x < \theta \]

---

Step 3: Find the \( \tfrac{1}{8} \) quantile

The \( \tfrac{1}{8} \) quantile \( q_{1/8} \) satisfies:

\[ F(q_{1/8}) = \frac{1}{8} \]

\[ \frac{q_{1/8}^3}{\theta^3} = \frac{1}{8} \]

\[ q_{1/8} = \theta \left(\frac{1}{8}\right)^{1/3} = \frac{\theta}{2} \]

---

Step 4: Plug in the MLE of \( \theta \)

\[ \hat{q}_{1/8} = \frac{\hat{\theta}}{2} = \frac{7}{2} = 3.5 \]

---

Final Answer

\[ \boxed{3.5} \]