Parameter Estimation for N(theta, theta)
This problem requires finding the Maximum Likelihood Estimate (MLE) of the parameter $\theta$ for a random sample $X_1, \ldots, X_n$ from a normal distribution $N(\theta, \theta)$. The parameter space is constrained to $\theta > 0$.
Deriving the Log-Likelihood Function
The probability density function (PDF) for a single observation from $N(\theta, \theta)$ is: $f(x; \theta) = \frac{1}{\sqrt{2\pi\theta}} \exp\left(-\frac{(x-\theta)^2}{2\theta}\right)$
The likelihood function $L(\theta)$ for the sample is the product of the PDFs: $L(\theta) = \prod_{i=1}^n f(X_i; \theta) = \left(\frac{1}{\sqrt{2\pi\theta}}\right)^n \exp\left(-\frac{1}{2\theta} \sum_{i=1}^n (X_i-\theta)^2\right)$
Working with the log-likelihood function $\ell(\theta) = \log L(\theta)$ simplifies differentiation: $\ell(\theta) = -\frac{n}{2} \log(2\pi) - \frac{n}{2} \log(\theta) - \frac{1}{2\theta} \sum_{i=1}^n (X_i-\theta)^2$
Maximizing the Log-Likelihood
To find the MLE, we compute the derivative of $\ell(\theta)$ with respect to $\theta$ and set it to zero. Expanding the sum term $\sum_{i=1}^n (X_i-\theta)^2 = \sum X_i^2 - 2\theta \sum X_i + n\theta^2$. The derivative $\frac{d\ell}{d\theta}$ is calculated as: $\frac{d\ell}{d\theta} = -\frac{n}{2\theta} - \frac{1}{2} \frac{d}{d\theta}\left(\frac{\sum X_i^2}{\theta} - 2\sum X_i + n\theta\right)$ $\frac{d\ell}{d\theta} = -\frac{n}{2\theta} - \frac{1}{2} \left(-\frac{\sum X_i^2}{\theta^2} + n\right)$ $\frac{d\ell}{d\theta} = -\frac{n}{2\theta} + \frac{\sum X_i^2}{2\theta^2} - \frac{n}{2}$
Setting the derivative to zero gives the equation for critical points: $-\frac{n}{2\theta} + \frac{\sum X_i^2}{2\theta^2} - \frac{n}{2} = 0$
Solving the Quadratic Equation for $\theta$
Multiply the equation by $2\theta^2$ to eliminate denominators: $-n\theta + \sum X_i^2 - n\theta^2 = 0$ Rearrange into the standard quadratic form $n\theta^2 + n\theta - \sum X_i^2 = 0$. Divide by $n$: $\theta^2 + \theta - \frac{1}{n} \sum X_i^2 = 0$
Using the quadratic formula $\theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1, b=1, c = -\frac{1}{n} \sum X_i^2$: $\theta = \frac{-1 \pm \sqrt{1^2 - 4(1)(-\frac{1}{n} \sum X_i^2)}}{2(1)}$ $\theta = \frac{-1 \pm \sqrt{1 + \frac{4}{n} \sum X_i^2}}{2}$
Selecting the Valid MLE Solution
This yields two potential solutions for $\theta$: $\theta_1 = \frac{-1 - \sqrt{1 + \frac{4}{n} \sum X_i^2}}{2}$ $\theta_2 = \frac{-1 + \sqrt{1 + \frac{4}{n} \sum X_i^2}}{2}$
The problem specifies that $\theta > 0$. Since $\sqrt{1 + \frac{4}{n} \sum X_i^2}$ is always greater than 1 (because $n>0$ and $\sum X_i^2 \ge 0$), the first solution $\theta_1$ is always negative. The second solution $\theta_2$ is always positive.
Therefore, the Maximum Likelihood Estimate (MLE) of $\theta$ is the positive root: $\hat{\theta} = \frac{-1 + \sqrt{1 + \frac{4}{n} \sum X_i^2}}{2}$
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 $\bar{Y}$?
