Order Statistics Symmetry MCQs Analysis
This question concerns the properties of order statistics derived from independent and identically distributed (i.i.d.) random variables ($X_1, \dots, X_{2n-1}$) whose probability density function (p.d.f.), $f_{\theta}$, is symmetric about a parameter $\theta$ and has bounded support.
Let $X_{(1)} < X_{(2)} < \dots < X_{(2n-1)}$ denote the order statistics. We analyze the given statements based on the symmetry property.
Define a transformation $Y_i = X_i - \theta$. Since $f_{\theta}(x)$ is symmetric about $\theta$, the p.d.f. of $Y_i$, denoted $g(y) = f_{\theta}(y+\theta)$, is symmetric about 0. Let the bounded support of $X_i$ be $[L, U]$. Then the support of $Y_i$ is $[L-\theta, U-\theta]$, which is symmetric about 0 (i.e., $[-c, c]$ for some $c > 0$).
Let $Y_{(1)}, \dots, Y_{(m)}$ be the order statistics of $Y_1, \dots, Y_m$, where $m = 2n-1$ (an odd number). A key property derived from the symmetry of $g(y)$ is that the vector $(Y_{(1)}, \dots, Y_{(m)})$ has the same distribution as $(-Y_{(m)}, \dots, -Y_{(1)})$. This implies $Y_{(k)}$ has the same distribution as $-Y_{(m+1-k)}$ for $k=1, \dots, m$.
Analyzing the Statements
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Statement A: $X_{(1)} - \theta$ and $\theta - X_{(1)}$ have the same distributionThis statement is equivalent to checking if $Y_{(1)}$ and $-Y_{(1)}$ have the same distribution. While $Y_{(1)}$ has the same distribution as $-Y_{(2n-1)}$ (from the property above), its own distribution is not necessarily symmetric about 0. For example, the minimum of standard normal variables does not have a distribution symmetric about 0. Thus, Statement A is incorrect.
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Statement B: $X_{(1)} - \theta$ and $\theta - X_{(2n-1)}$ have the same distributionThis is equivalent to checking if $Y_{(1)}$ and $-(X_{(2n-1)} - \theta) = -Y_{(2n-1)}$ have the same distribution. Here $m = 2n-1$. The property $Y_{(k)} \sim \mathcal{D} -Y_{(m+1-k)}$ holds. For $k=1$, we get $Y_{(1)} \sim \mathcal{D} -Y_{(m+1-1)} = -Y_{(m)} = -Y_{(2n-1)}$. Thus, Statement B is correct.
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Statement C: The distribution of $X_{(n)}$ is symmetric about $\theta$This statement is equivalent to checking if $Y_{(n)}$ has a distribution symmetric about 0. Since $m = 2n-1$ is odd, the median order statistic is $Y_{(n)}$ where $n = (m+1)/2$. Using the property $Y_{(k)} \sim \mathcal{D} -Y_{(m+1-k)}$ with $k=n$, we get $Y_{(n)} \sim \mathcal{D} -Y_{(2n-1+1-n)} = -Y_{(n)}$. This means the distribution of $Y_{(n)}$ is symmetric about 0. Thus, the distribution of $X_{(n)}$ is symmetric about $\theta$. Statement C is correct.
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Statement D: $E[X_{(k)} + X_{(2n-k)}]$ is same for all $k = 1, 2, \dots, n$We have $X_{(k)} = Y_{(k)} + \theta$ and $X_{(2n-k)} = Y_{(2n-k)} + \theta$. So, $X_{(k)} + X_{(2n-k)} = Y_{(k)} + \theta + Y_{(2n-k)} + \theta = Y_{(k)} + Y_{(2n-k)} + 2\theta$. Taking the expectation: $E[X_{(k)} + X_{(2n-k)}] = E[Y_{(k)}] + E[Y_{(2n-k)}] + 2\theta$. From Statement B, we know $Y_{(k)} \sim \mathcal{D} -Y_{(2n-k)}$. Therefore, their expectations are related: $E[Y_{(k)}] = E[-Y_{(2n-k)}] = -E[Y_{(2n-k)}]$. This implies $E[Y_{(k)}] + E[Y_{(2n-k)}] = 0$. Substituting this back, we get $E[X_{(k)} + X_{(2n-k)}] = 0 + 2\theta = 2\theta$. Since the expectation is $2\theta$ for all values of $k$ from 1 to $n$, it is constant. Thus, Statement D is correct.
Based on the analysis, statements B, C, and D are correct.
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 $\bar{Y}$?
