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

• Statement A:
  $X_{(1)} - \theta$ and $\theta - X_{(1)}$ have the same distribution
  This 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.
• Statement B:
  $X_{(1)} - \theta$ and $\theta - X_{(2n-1)}$ have the same distribution
  This 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.
• 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.
• 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.