act like a chegg tutor and rewrite the solution Of course! I can walk you through this problem on order statistics. It's a great exercise that combines concepts of probability distributions, calculus,

Question

act like a chegg tutor and rewrite the solution Of course! I can walk you through this problem on order statistics. It's a great exercise that combines concepts of probability distributions, calculus, and expectations.

Let's solve it step-by-step.

### Given Information:
* We have 4 iid (independent and identically distributed) random variables: $X_1, X_2, X_3, X_4$. So, the sample size is $n=4$.
* The probability density function (PDF) is $f_X(x) = 2x$ for $0 < x < 1$.

***

### (a) What is the CDF of $X_1$?

**Concept:** The Cumulative Distribution Function (CDF), denoted as $F_X(x)$, gives the probability that the random variable $X$ is less than or equal to a certain value $x$. We find it by integrating the PDF from the lower bound of its support up to $x$.

**Calculation:**
The CDF is $F_X(x) = P(X \le x) = \int_{-\infty}^{x} f_X(t) \,dt$.
Since our PDF is defined for $0 < x < 1$, we'll integrate from 0 to $x$:

$F_X(x) = \int_{0}^{x} 2t \,dt = [t^2]_0^x = x^2 - 0^2 = x^2$

For a complete definition, the CDF is:
$F_X(x) = \begin{cases} 0 & 	ext{for } x \le 0 \ x^2 & 	ext{for } 0 < x < 1 \ 1 & 	ext{for } x \ge 1 \end{cases}$

**Answer (a):**
The CDF of $X_1$ is $F_X(x) =

VIPSolutions · Accepted Answer

Certainly! Let’s rewrite and explain the solution in a clear, step-by-step Chegg-style manner.

---

## **Problem Restatement**

Let $X_1, X_2, X_3, X_4$ be independent and identically distributed random variables with PDF $f_X(x) = 2x$ for $0 < x < 1$.

Answer the following:

- **(a)** Find the CDF of $X_1$.
- **(b)** Compute the expected value of the second order statistic, $E(X_{(2)})$.
- **(c)** Compute the variance of $X_{(2)}$.
- **(d)** Compute the covariance between $X_{(2)}$ and $X_{(3)}$.

---

## **(a) CDF of $X_1$**

**Definition**: The cumulative distribution function (CDF) is:
$$
F_X(x) = P(X \leq x) = \int_{-\infty}^x f_X(t)\,dt
$$
Since $f_X(x) = 2x$ for $0 < x < 1$, integrate from $0$ to $x$:

$$
F_X(x) = \int_0^x 2t\,dt = [t^2]_0^x = x^2
$$

So, the CDF is fully defined as:
$$
F_X(x) = 
\begin{cases}
0 & x \leq 0 \
x^2 & 0 < x < 1 \
1 & x \geq 1
\end{cases}
$$

**Final answer (a):**
$$
\boxed{F_X(x) = x^2 \quad \text{for } 0 < x < 1}
$$

---

## **(b) $E(X_{(2)})$**

**Step 1: Find the PDF of $X_{(2)}$**

The PDF of the $k$-th order statistic among $n$ i.i.d. samples is:
$$
f_{X_{(k)}}(x) = \frac{n!}{(k-1)!(n-k)!}[F_X(x)]^{k-1}[1-F_X(x)]^{n-k}f_X(x)
$$

Here, $n=4$, $k=2$, $F_X(x) = x^2$, $f_X(x) = 2x$:

$$
f_{X_{(2)}}(x) = \frac{4!}{1!2!}(x^2)^{1}(1-x^2)^{2}(2x)
$$
$$
= \frac{24}{2} x^2 (1-x^2)^2 (2x) = 12 x^2 (1-2x^2 + x^4)(2x)
$$
$$
= 24 x^3 (1-2x^2 + x^4)
$$
$$
= 24x^3 - 48x^5 + 24x^7 \qquad \text{for } 0 < x < 1
$$

---

**Step 2: Compute $E(X_{(2)})$**

$$
E(X_{(2)}) = \int_0^1 x f_{X_{(2)}}(x) dx = \int_0^1 x(24x^3 - 48x^5 + 24x^7) dx
$$
$$
= \int_0^1 (24x^4 - 48x^6 + 24x^8) dx
$$
$$
= 24\int_0^1 x^4 dx - 48\int_0^1 x^6 dx + 24\int_0^1 x^8 dx
$$
$$
= 24 \cdot \frac{1}{5} - 48 \cdot \frac{1}{7} + 24 \cdot \frac{1}{9}
$$
$$
= \frac{24}{5} - \frac{48}{7} + \frac{24}{9}
$$

Compute common denominator ($5 \times 7 \times 9 = 315$):

$$
\frac{24}{5} = \frac{24 \times 63}{315} = \frac{1512}{315}
$$
$$
\frac{48}{7} = \frac{48 \times 45}{315} = \frac{2160}{315}
$$
$$
\frac{24}{9} = \frac{24 \times 35}{315} = \frac{840}{315}
$$
So,
$$
E(X_{(2)}) = \frac{1512 - 2160 + 840}{315} = \frac{192}{315} = \frac{64}{105}
$$

**Final answer (b):**
$$
\boxed{E(X_{(2)}) = \frac{64}{105}}
$$

---

## **(c) $\text{Var}(X_{(2)})$**

**Step 1: Compute $E(X_{(2)}^2)$**

$$
E(X_{(2)}^2) = \int_0^1 x^2 f_{X_{(2)}}(x) dx
= \int_0^1 x^2(24x^3 - 48x^5 + 24x^7) dx
= \int_0^1 (24x^5 - 48x^7 + 24x^9) dx
$$
$$
= 24\int_0^1 x^5 dx - 48\int_0^1 x^7 dx + 24\int_0^1 x^9 dx
$$
$$
= 24 \cdot \frac{1}{6} - 48 \cdot \frac{1}{8} + 24 \cdot \frac{1}{10}
$$
$$
= 4 - 6 + \frac{12}{5}
$$
$$
= -2 + \frac{12}{5} = \frac{-10 + 12}{5} = \frac{2}{5}
$$

---

**Step 2: Find the variance**

$$
\operatorname{Var}(X_{(2)}) = E(X_{(2)}^2) - [E(X_{(2)})]^2 = \frac{2}{5} - \left(\frac{64}{105}\right)^2
$$

Calculate $\left(\frac{64}{105}\right)^2 = \frac{4096}{11025}$.

$$
\operatorname{Var}(X_{(2)}) = \frac{2}{5} - \frac{4096}{11025}
$$
Convert $\frac{2}{5}$ to denominator 11025: $2 \times 2205 = 4410$
$$
= \frac{4410}{11025} - \frac{4096}{11025} = \frac{314}{11025}
$$

**Final answer (c):**
$$
\boxed{\operatorname{Var}(X_{(2)}) = \frac{314}{11025}}
$$

---

## **(d) $\operatorname{Cov}(X_{(2)}, X_{(3)})$**

**Step 1: Compute $E(X_{(3)})$**

PDF of $X_{(3)}$ ($k=3$):
$$
f_{X_{(3)}}(x) = \frac{4!}{2!1!}(x^2)^2(1-x^2)(2x)
= 12 x^4 (1-x^2)(2x)
= 24x^5(1-x^2) = 24x^5 - 24x^7
$$

$$
E(X_{(3)}) = \int_0^1 x (24x^5 - 24x^7) dx = \int_0^1 (24x^6 - 24x^8) dx
= 24 \cdot \frac{1}{7} - 24 \cdot \frac{1}{9}
= \frac{24}{7} - \frac{24}{9}
$$
$$
= \frac{24 \times 9 - 24 \times 7}{63} = \frac{216 - 168}{63} = \frac{48}{63} = \frac{16}{21}
$$

---

**Step 2: Compute $E(X_{(2)}X_{(3)})$**

The joint PDF for $X_{(2)}, X_{(3)}$ ($n=4$, $j=2$, $k=3$):

$$
f_{X_{(2)},X_{(3)}}(x,y) = \frac{4!}{1!0!1!} (x^2)^{1}[y^2 - x^2]^0 [1-y^2]^1 (2x)(2y)
= 24 x^2 (1-y^2) (2x)(2y)
= 96 x^3 y (1-y^2)
$$

Domain: $0 < x < y < 1$.

$$
E(X_{(2)} X_{(3)}) = \iint_{0 < x < y < 1} x y \cdot 96 x^3 y (1-y^2) dy dx
$$
$$
= 96 \int_0^1 x^4 \left[ \int_x^1 y^2 (1 - y^2) dy \right] dx
$$
$$
= 96 \int_0^1 x^4 \left( \left[ \frac{y^3}{3} - \frac{y^5}{5} \right]_x^1 \right) dx
$$
$$
= 96 \int_0^1 x^4 \left( \frac{1}{3} - \frac{1}{5} - \frac{x^3}{3} + \frac{x^5}{5} \right) dx
= 96 \int_0^1 x^4 \left( \frac{2}{15} - \frac{x^3}{3} + \frac{x^5}{5} \right) dx
$$

Expand:
$$
= 96 \left[ \frac{2}{15} \int_0^1 x^4 dx - \frac{1}{3} \int_0^1 x^7 dx + \frac{1}{5} \int_0^1 x^9 dx \right]
$$
$$
= 96 \left[ \frac{2}{15} \cdot \frac{1}{5} - \frac{1}{3} \cdot \frac{1}{8} + \frac{1}{5} \cdot \frac{1}{10} \right]
= 96 \left[ \frac{2}{75} - \frac{1}{24} + \frac{1}{50} \right]
$$

Find common denominator ($600$):
$\frac{2}{75} = \frac{16}{600}$
$\frac{1}{24} = \frac{25}{600}$
$\frac{1}{50} = \frac{12}{600}$

So,
$$
= 96 \left( \frac{16 - 25 + 12}{600} \right)
= 96 \left( \frac{3}{600} \right)
= \frac{288}{600} = \frac{12}{25}
$$

---

**Step 3: Covariance calculation**

\[
\operatorname{Cov}(X_{(2)}, X_{(3)}) =

VIPSolutions · Answer

Certainly! Here's a clear, step-by-step explanation of how to compute the expectation of the second order statistic $E(X_{(2)})$, presented in a style similar to Chegg solutions:

---

### **Step-by-Step Solution for $E(X_{(2)})$**

**Given:**
- Random variables $X_1, X_2, X_3, X_4$ are i.i.d. with probability density function (PDF):
  $$
  f_X(x) = 2x, \quad \text{for } 0 < x < 1
  $$
- The cumulative distribution function (CDF) is:
  $$
  F_X(x) = x^2, \quad \text{for } 0 < x < 1
  $$

---

### **Step 1: Find the PDF of the 2nd order statistic $X_{(2)}$**

The general formula for the PDF of the $k$-th order statistic out of $n$ i.i.d. samples is:

$$
f_{X_{(k)}}(x) = \frac{n!}{(k-1)!(n-k)!} [F_X(x)]^{k-1} [1 - F_X(x)]^{n-k} \cdot f_X(x)
$$

**Plugging in our values:**
- $n = 4$
- $k = 2$
- $F_X(x) = x^2$
- $f_X(x) = 2x$

We get:

$$
f_{X_{(2)}}(x) = \frac{4!}{(2-1)!(4-2)!} \cdot (x^2)^{1} \cdot (1 - x^2)^{2} \cdot 2x
$$

Calculating the combinatorial coefficient:

$$
\frac{4!}{1! \cdot 2!} = \frac{24}{1 \times 2} = 12
$$

So,

$$
f_{X_{(2)}}(x) = 12 \cdot x^2 \cdot (1 - x^2)^2 \cdot 2x
$$

Simplify:

$$
f_{X_{(2)}}(x) = 24 x^3 (1 - x^2)^2
$$

---

### **Step 2: Expand the PDF**

Let's expand $(1 - x^2)^2$:

$$
(1 - x^2)^2 = 1 - 2x^2 + x^4
$$

Therefore,

$$
f_{X_{(2)}}(x) = 24 x^3 (1 - 2x^2 + x^4) = 24 x^3 - 48 x^5 + 24 x^7
$$

valid for $0 < x < 1$.

---

### **Step 3: Compute $E[X_{(2)}]$**

The expectation is:

$$
E[X_{(2)}] = \int_0^1 x \cdot f_{X_{(2)}}(x) \, dx
$$

Substitute the PDF:

$$
E[X_{(2)}] = \int_0^1 x (24 x^3 - 48 x^5 + 24 x^7) dx
$$

Distribute $x$:

$$
E[X_{(2)}] = \int_0^1 (24 x^4 - 48 x^6 + 24 x^8) dx
$$

Now, integrate each term:

$$
\int_0^1 24 x^4 dx = 24 \cdot \frac{1}{5} = \frac{24}{5}
$$
$$
\int_0^1 (-48 x^6) dx = -48 \cdot \frac{1}{7} = -\frac{48}{7}
$$
$$
\int_0^1 24 x^8 dx = 24 \cdot \frac{1}{9} = \frac{24}{9} = \frac{8}{3}
$$

Sum these:

$$
E[X_{(2)}] = \frac{24}{5} - \frac{48}{7} + \frac{8}{3}
$$

Find common denominator: $105$ (since $5 \times 7 = 35$, and $35 \times 3=105$).

Express each term:

$$
\frac{24}{5} = \frac{24 \times 21}{105} = \frac{504}{105}
$$
$$
\frac{48}{7} = \frac{48 \times 15}{105} = \frac{720}{105}
$$
$$
\frac{8}{3} = \frac{8 \times 35}{105} = \frac{280}{105}
$$

Combine:

$$
E[X_{(2)}] = \frac{504 - 720 + 280}{105} = \frac{64}{105}
$$

---

### **### **Final Result:**

$$
\boxed{
\boxed{E[X_{(2)}] = \frac{64}{105}}
}
$$

---

This completes the calculation for the expected value of the second order statistic.

VIPSolutions · Answer

Certainly! Here's a clear, step-by-step explanation of how to compute the expectation of the second order statistic $E(X_{(2)})$, in a style similar to Chegg solutions, with equations formatted in a straightforward manner:

---

### **Step-by-Step Solution for $E(X_{(2)})$**

**Given:**
- Random variables $X_1, X_2, X_3, X_4$ are i.i.d. with PDF:
  
  $f_X(x) = 2x,\quad 0 < x < 1$

- The CDF is:

$F_X(x) = x^2,\quad 0 < x < 1$

---

### **Step 1: Find the PDF of the 2nd order statistic $X_{(2)}$**

The general formula for the PDF of the $k$-th order statistic out of $n$ i.i.d. samples is:

$$
f_{X_{(k)}}(x) = \frac{n!}{(k-1)!(n-k)!} \left[ F_X(x) \right]^{k-1} \left[ 1 - F_X(x) \right]^{n-k} \times f_X(x)
$$

**Plugging in**:

- $n=4$, $k=2$
- $F_X(x) = x^2$
- $f_X(x) = 2x$

we get:

$$
f_{X_{(2)}}(x) = \frac{4!}{1! \times 2!} \times (x^2)^1 \times (1 - x^2)^2 \times 2x
$$

Calculating the coefficient:

$$
\frac{4!}{1! \times 2!} = \frac{24}{2} = 12
$$

Thus,

$$
f_{X_{(2)}}(x) = 12 \times x^2 \times (1 - x^2)^2 \times 2x
$$

which simplifies to:

$$
f_{X_{(2)}}(x) = 24 x^3 (1 - x^2)^2
$$

---

### **Step 2: Expand the expression**

Expand $(1 - x^2)^2$:

$$
(1 - x^2)^2 = 1 - 2x^2 + x^4
$$

Therefore,

$$
f_{X_{(2)}}(x) = 24 x^3 (1 - 2x^2 + x^4) = 24 x^3 - 48 x^5 + 24 x^7
$$

valid for $0 < x < 1$.

---

### **Step 3: Compute $E[X_{(2)}]$**

The expectation is:

$$
E[X_{(2)}] = \int_0^1 x \times f_{X_{(2)}}(x) \, dx
$$

Substituting the PDF:

$$
E[X_{(2)}] = \int_0^1 x (24 x^3 - 48 x^5 + 24 x^7) \, dx
$$

Distribute $x$:

$$
E[X_{(2)}] = \int_0^1 (24 x^4 - 48 x^6 + 24 x^8) \, dx
$$

Integrate term-by-term:

$$
\int_0^1 24 x^4 dx = 24 \times \frac{1}{5} = \frac{24}{5}
$$

$$
\int_0^1 (-48 x^6) dx = -48 \times \frac{1}{7} = -\frac{48}{7}
$$

$$
\int_0^1 24 x^8 dx = 24 \times \frac{1}{9} = \frac{8}{3}
$$

Sum these:

$$
E[X_{(2)}] = \frac{24}{5} - \frac{48}{7} + \frac{8}{3}
$$

Find a common denominator, which is 105:

$$
\frac{24}{5} = \frac{24 \times 21}{105} = \frac{504}{105}
$$
$$
\frac{48}{7} = \frac{48 \times 15}{105} = \frac{720}{105}
$$
$$
\frac{8}{3} = \frac{8 \times 35}{105} = \frac{280}{105}
$$

Combine:

$$
E[X_{(2)}] = \frac{504 - 720 + 280}{105} = \frac{64}{105}
$$

---

### **### **Final Answer:**

$$
\boxed{
E[X_{(2)}] = \frac{64}{105}
}
$$

---

This completes the calculation for the expected value of the second order statistic.

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