The random variable Xn(n=1,2,..) has the density fn(x) given by 
fn(x)=2^n/€ if (i/n)-(€/n2^n)<x<i/n (i=1,2,...n) ,0 otherwise. FIND THE DISTRIBUTION FUNCTION Fn(x)

To find the distribution function \( F_n(x) \) of the random variable \( X_n \) based on the given density function \( f_n(x) \), we proceed with the following steps.

### Given Density Function
The probability density function \( f_n(x) \) is given as:

\[
f_n(x) = 
\begin{cases}
\frac{2^n}{\epsilon} & \text{if } \frac{i}{n} - \frac{\epsilon}{n2^n} < x < \frac{i}{n}, \, i = 1, 2, ..., n, \\
0 & \text{otherwise}.
\end{cases}
\]

This implies that the density function is piecewise, where it takes the value \( \frac{2^n}{\epsilon} \) over certain intervals and is zero otherwise.

### Step 1: Definition of Distribution Function
The distribution function \( F_n(x) \) is defined as the cumulative distribution function (CDF), which is the integral of the probability density function from \( -\infty \) to \( x \):

\[
F_n(x) = \int_{-\infty}^{x} f_n(t) \, dt.
\]

We need to integrate the piecewise function \( f_n(x) \) over the appropriate intervals to find \( F_n(x) \).

### Step 2: Identify Intervals
For each \( i = 1, 2, ..., n \), the probability density function \( f_n(x) \) is non-zero only in the interval \( \left( \frac{i}{n} - \frac{\epsilon}{n2^n}, \frac{i}{n} \right) \).

Thus, the total CDF \( F_n(x) \) will be the sum of contributions from each of these intervals.

### Step 3: Calculate the CDF

1. **For \( x < \frac{1}{n} - \frac{\epsilon}{n2^n} \)**:  
   Since \( f_n(x) = 0 \) in this region, we have:
   \[
   F_n(x) = 0.
   \]

2. **For \( \frac{i}{n} - \frac{\epsilon}{n2^n} < x < \frac{i}{n} \) (for some \( i \))**:  
   In this case, the CDF is obtained by integrating the density function up to \( x \) within the given interval:
   \[
   F_n(x) = \int_{\frac{i}{n} - \frac{\epsilon}{n2^n}}^{x} \frac{2^n}{\epsilon} \, dt.
   \]
   The result of this integral is:
   \[
   F_n(x) = \frac{2^n}{\epsilon} \left( x - \left( \frac{i}{n} - \frac{\epsilon}{n2^n} \right) \right).
   \]

3. **For \( x \geq \frac{n}{n} \)**:  
   Once \( x \) exceeds the upper bound for all intervals, the CDF reaches its maximum value of 1:
   \[
   F_n(x) = 1.
   \]

### Final Answer:
Thus, the cumulative distribution function \( F_n(x) \) is piecewise and can be written as:

\[
F_n(x) =
\begin{cases}
0 & \text{if } x < \frac{1}{n} - \frac{\epsilon}{n2^n}, \\
\frac{2^n}{\epsilon} \left( x - \left( \frac{i}{n} - \frac{\epsilon}{n2^n} \right) \right) & \text{if } \frac{i}{n} - \frac{\epsilon}{n2^n} < x < \frac{i}{n}, \\
1 & \text{if } x \geq \frac{n}{n}.
\end{cases}
\]

This describes the distribution function \( F_n(x) \) based on the given probability density.

Do you have any questions or would you like more details on any of the steps?

### Related Questions:
1. What is the role of the constant \( \epsilon \) in this distribution?
2. How does the density function change as \( n \) increases?
3. What is the interpretation of this distribution in a real-world context?
4. How would you find the mean of \( X_n \) based on this density function?
5. What happens to the distribution as \( n \to \infty \)?

**Tip**: The CDF is always a non-decreasing function that ranges from 0 to 1, with the probability density function being its derivative.

The random variable Xn(n=1,2,..) has the density fn(x) given by fn(x)=2^n/€ if (i/n)-(€/n2^n)<x<i/n (i=1,2,...n), 0 otherwise. FIND THE DISTRIBUTION FUNCTION Fn(x)

Explore how to derive the cumulative distribution function (CDF) F_n(x) from a given piecewise probability density function f_n(x) involving parameters n and ε. This problem covers key concepts in probability theory and is suitable for undergraduate students in statistics or probability theory.

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