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
- A mobile manufacturing company uses two brands of batteries for its mobiles. The life (in years) of batteries of Brand I follows an exponential distribution with the probability density function
$ f(x) = \begin{cases} e^{-x}, & \text{if } x>0, \\ 0, & \text{otherwise,} \end{cases} $
and that of Brand II follows a gamma distribution with the probability density function
$ g(x) = \begin{cases} \frac{x}{4} e^{-x/2}, & \text{if } x>0, \\ 0, & \text{otherwise.} \end{cases} $
The company uses the batteries of Brands I and II in proportion of $20\%$ and $80\%$ respectively, in its mobiles. The probability that a randomly selected mobile has the battery life more that $2$ years is - Consider a discrete random variable $X$ with the probability mass function
$ P(X = 0) = \frac{\theta}{3}, \ P(X = 1) = 1 - \frac{\theta}{2}, \ P(X = 2) = \frac{\theta}{6}, $
where $\theta \in (0,1)$ is an unknown parameter. In a random sample of size $90$ from this distribution, the observed counts for $X = 0, 1$ and $2$ are $20, 60$ and $10$, respectively. Then, the maximum likelihood estimate of $\theta$ is - Let $X$ be a random sample of size $1$ from the probability density function
$ f(x|\theta) = \begin{cases} \frac{3}{\theta^3} (\theta - x)^2, & \text{if } 0<x<\theta, \\ 0, & \text{otherwise.} \end{cases} $
If $ \left(\frac{X}{1-\lambda_1}, \frac{X}{1-\lambda_2}\right) $ is a confidence interval for $\theta$ with confidence coefficient $1 - \alpha$, where $\lambda_i \in (0,1), \ i = 1,2, \ \lambda_1<\lambda_2$, and $\alpha \in (0,1)$, then which of the following statements is true? - Let $X_1, X_2, . . ., X_n$ be a random sample from a continuous distribution with the common probability density function
$ f(x|\theta) = \begin{cases} \frac{2\theta^2}{x^{\theta+1}}, & \text{if } x>2, \\ 0, & \text{otherwise,} \end{cases} $
where $\theta (> 0)$ is an unknown parameter. Suppose $P(Y>\chi^2_{m,\beta}) = \beta$, where $Y \sim \chi^2_m$. For testing $H_0: \theta = 1$ against $H_1 : \theta>1$, a uniformly most powerful test of size $\alpha, \ 0<\alpha<1$, will reject $H_0$ if - Suppose we want to estimate the population mean $\bar{Y}$ of a variable for a finite population of size $85$, with $34$ Statisticians and $51$ Biologists. We consider the following sampling scheme:
A stratified random sample with $2$ strata of Statisticians (Stratum-1) and Biologists (Stratum-2), where $12$ Statisticians and $15$ Biologists are drawn from Stratum-1 and Stratum-2, respectively, using SRSWOR scheme.
Denote $\bar{y}_S, \bar{y}_B$, and $\bar{y}$ as the mean of the variable among the Statistician sample, Biologist sample, and the combined sample, respectively. Which of the following is an unbiased estimator of $\bar{Y}$?
