Question

Let X₍₁₎ < X₍₂₎ < X₍₃₎ < X₍₄₎ < X₍₅₎ be the order statistics based on a random sample of size 5 from a continuous distribution with the probability density function
\(f(x)=\begin{cases} \frac{1}{x^2}, & 1 < x < \infin\\ 0, & \text{otherwise.} \end{cases}\)
Then the sum of all possible values of r ∈ {1, 2, 3, 4, 5} for which E(X_(r)) is finite equals __________

Answer 1

Correct Answer: 10

1. Determine the Cumulative Distribution Function (CDF)

First, we calculate the CDF, $F(x)$, from the given Probability Density Function (PDF), $f(x)$.

The given PDF is:

$$f(x) = \begin{cases} \frac{1}{x^2}, & 1 < x < \infty \\ 0, & \text{otherwise} \end{cases}$$

The CDF is defined as $F(x) = \int_{-\infty}^{x} f(t) \, dt$.

For $x > 1$:

$$F(x) = \int_{1}^{x} \frac{1}{t^2} \, dt = \left[ -\frac{1}{t} \right]_1^x = \left(-\frac{1}{x}\right) - \left(-\frac{1}{1}\right) = 1 - \frac{1}{x}$$

So,

$$F(x) = 1 - \frac{1}{x}, \quad \text{for } x > 1$$

 

2. Determine the PDF of the $r$-th Order Statistic

For a random sample of size $n$, the PDF of the $r$-th order statistic, denoted as $f_{X_{(r)}}(x)$, is given by the formula:

$$f_{X_{(r)}}(x) = \frac{n!}{(r-1)!(n-r)!} [F(x)]^{r-1} [1 - F(x)]^{n-r} f(x)$$

Here, the sample size is $n=5$. Let's substitute the expressions for $F(x)$ and $f(x)$:

$[F(x)]^{r-1} = \left(1 - \frac{1}{x}\right)^{r-1}$

$[1 - F(x)]^{n-r} = \left[1 - \left(1 - \frac{1}{x}\right)\right]^{5-r} = \left(\frac{1}{x}\right)^{5-r} = x^{-(5-r)}$

$f(x) = \frac{1}{x^2} = x^{-2}$

Substituting these into the formula (ignoring the constant coefficient for now, let's call it $C$):

$$f_{X_{(r)}}(x) = C \cdot \left(1 - \frac{1}{x}\right)^{r-1} \cdot x^{-(5-r)} \cdot x^{-2}$$

$$f_{X_{(r)}}(x) = C \cdot \left(1 - \frac{1}{x}\right)^{r-1} \cdot x^{-(7-r)}$$

3. Analyze the Expectation $E(X_{(r)}))$

The expected value is given by:

$$E(X_{(r)}) = \int_{1}^{\infty} x \cdot f_{X_{(r)}}(x) \, dx$$

Substituting the derived PDF:

$$E(X_{(r)}) = C \int_{1}^{\infty} x \cdot \left(1 - \frac{1}{x}\right)^{r-1} \cdot x^{-(7-r)} \, dx$$

$$E(X_{(r)}) = C \int_{1}^{\infty} \left(1 - \frac{1}{x}\right)^{r-1} \cdot x^{1 - (7-r)} \, dx$$

$$E(X_{(r)}) = C \int_{1}^{\infty} \left(1 - \frac{1}{x}\right)^{r-1} \cdot x^{r-6} \, dx$$

Convergence Analysis:

We need to determine for which values of $r$ this integral converges (is finite).

As $x \to \infty$, the term $\left(1 - \frac{1}{x}\right)^{r-1} \to 1$.

Thus, the convergence of the integral depends entirely on the term $x^{r-6}$.

An integral of the form $\int_{1}^{\infty} x^p \, dx$ converges if and only if $p < -1$.

Here, $p = r - 6$.

So, the condition for finite expectation is:

$$r - 6 < -1$$

$$r < 5$$

4. Identify Valid Values of $r$ and Sum

The problem asks for $r \in \{1, 2, 3, 4, 5\}$.

Based on the condition $r < 5$, the possible values for $r$ are:

$$r \in \{1, 2, 3, 4\}$$

(Note: For $r=5$, the integral behaves like $\int x^{-1} dx$, which results in $\ln(x)$ and diverges to infinity).

Finally, we calculate the sum of these values:

$$\text{Sum} = 1 + 2 + 3 + 4 = 10$$

Answer:

The sum of all possible values of $r$ is 10.