Let $X$ and $Y$ be independent random variables with $X \sim \text{Uniform}[0, \theta+3]$, $Y \sim \text{Uniform}[-\theta-5, 0]$, where $\theta \ge -3$. Then the maximum likelihood estimator of $\theta$ based on $(X, Y)$ is
$(-5-Y, X-3)$
Understanding the Distributions
We are given two independent random variables:
- $X \sim \text{Uniform}[0, \theta+3]$. The probability density function (PDF) is $f_X(x) = \frac{1}{\theta+3}$ for $0 \le x \le \theta+3$.
- $Y \sim \text{Uniform}[-\theta-5, 0]$. The PDF is $f_Y(y) = \frac{1}{\theta+5}$ for $-\theta-5 \le y \le 0$.
- We are also given the constraint $\theta \ge -3$.
Deriving the Likelihood Function
Since $X$ and $Y$ are independent, their joint PDF is the product of their individual PDFs:
$f(x, y; \theta) = f_X(x) f_Y(y) = \frac{1}{(\theta+3)} \cdot \frac{1}{(\theta+5)}$
The likelihood function $L(\theta)$ is this joint PDF considered as a function of $\theta$ for the observed values $(x, y)$.
Identifying Constraints on $\theta$
For the likelihood function to be non-zero, the observed values $(x, y)$ must lie within the support of the distributions, which imposes constraints on $\theta$:
- From the support of $X$: $0 \le x \le \theta+3$. This implies $\theta+3 \ge x$, or $\theta \ge x-3$.
- From the support of $Y$: $-\theta-5 \le y \le 0$. This implies $y \ge -\theta-5$, which rearranges to $\theta+5 \ge -y$, or $\theta \ge -y-5$.
- We also have the explicit constraint $\theta \ge -3$.
Combining these, the likelihood is positive only when $\theta \ge \max(x-3, -y-5, -3)$.
Finding the Maximum Likelihood Estimator (MLE)
The likelihood function is $L(\theta) = \frac{1}{(\theta+3)(\theta+5)}$ for $\theta \ge \max(x-3, -y-5, -3)$.
To maximize $L(\theta)$, we need to minimize the denominator $g(\theta) = (\theta+3)(\theta+5)$.
The function $L(\theta)$ is decreasing for $\theta > -3$. Therefore, the maximum likelihood occurs at the smallest possible value of $\theta$ that satisfies all the constraints.
The minimum allowable value for $\theta$ is the maximum of the lower bounds derived:
$\hat{\theta}_{MLE} = \max(x-3, -y-5, -3)$
Matching with Options
The derived constraints provide the lower bounds for $\theta$: $x-3$ (from $X$) and $-y-5$ (from $Y$). The options are presented as pairs, likely highlighting these critical values derived from the data.
Option D, $(-5-Y, X-3)$, corresponds to the terms $-y-5$ and $x-3$. The MLE is the maximum of these terms, considering the constraint $\theta \ge -3$. Thus, this option correctly identifies the key components determining the MLE.
Important Questions from Random Variables
Let $X$, $Y$, and $Z$ be independent Normal random variables with means $-1$, $0$, and $1$, respectively, and variances $1$, $1$, and $3$, respectively. Which of the following random variables has a Cauchy distribution with location parameter $0$ and scale parameter $1$?
- Consider a finite population of size $N = 100$. Let $T_1$ be the sample mean of a study variable based on a sample of size $n$ ($1 < n < N$) under simple random sampling with replacement scheme. Let $T_2$ be the sample mean of the same study variable based on a sample of size $n$ under simple random sampling without replacement scheme. If $Var(T_1) = 9Var(T_2)$, then the sample size $n$ equals
Let $X_1$ and $X_2$ be a random sample from Uniform$[0, \theta]$ distribution, where $\theta > 0$. For testing the hypothesis
$H_0: \theta = 1$ against $H_1: \theta = 2$,
consider a test which rejects $H_0$ if $X_1 + X_2 > \frac{4}{5}$. Then, the probability of type-I error is- Let $\{Y_n: n \ge 1\}$ be a sequence of independent and identically distributed random variables, where $Y_1 \sim \text{Bernoulli}(\frac{1}{2})$. Define $Z = \sum_{n=1}^\infty \frac{4Y_n}{5^n}$. Then, which of the following statements is true?
Let $X$ be a single sample from an absolutely continuous distribution with probability density function
$f(x|\theta) = \begin{cases} \frac{2}{\theta^2}(\theta - x), & \text{if } 0 < x < \theta \\ 0, & \text{otherwise,} \end{cases}$
where $\theta > 0$ is unknown. Which of the following intervals is a $95\%$ confidence interval for $\theta$?
