1. Derive the mean, variance, mg and cumulative distribution function for the uniform.

Calculation In proper steps with the final answer.



### 1. **Uniform Distribution Overview**

The uniform distribution is a continuous probability distribution in which all intervals of the same length are equally probable. If \( X \) is a uniformly distributed random variable over the interval \([a, b]\), the probability density function (pdf) is given by:

\[
f_X(x) = \frac{1}{b-a}, \quad \text{for } a \leq x \leq b
\]

### 2. **Mean (Expected Value)**

The mean \( \mu \) of a uniform distribution can be derived as follows:

\[
\mu = \mathbb{E}[X] = \int_{a}^{b} x f_X(x) \, dx
\]

Substituting the pdf \( f_X(x) = \frac{1}{b-a} \):

\[
\mu = \int_{a}^{b} x \cdot \frac{1}{b-a} \, dx
\]

\[
\mu = \frac{1}{b-a} \int_{a}^{b} x \, dx
\]

To integrate \( x \):

\[
\int x \, dx = \frac{x^2}{2}
\]

Evaluating this from \( a \) to \( b \):

\[
\mu = \frac{1}{b-a} \left[ \frac{b^2}{2} - \frac{a^2}{2} \right]
\]

\[
\mu = \frac{1}{b-a} \cdot \frac{b^2 - a^2}{2}
\]

\[
\mu = \frac{b+a}{2}
\]

### 3. **Variance**

The variance \( \sigma^2 \) is given by:

\[
\sigma^2 = \mathbb{E}[X^2] - (\mathbb{E}[X])^2
\]

First, find \( \mathbb{E}[X^2] \):

\[
\mathbb{E}[X^2] = \int_{a}^{b} x^2 f_X(x) \, dx
\]

\[
\mathbb{E}[X^2] = \frac{1}{b-a} \int_{a}^{b} x^2 \, dx
\]

To integrate \( x^2 \):

\[
\int x^2 \, dx = \frac{x^3}{3}
\]

Evaluating from \( a \) to \( b \):

\[
\mathbb{E}[X^2] = \frac{1}{b-a} \left[ \frac{b^3}{3} - \frac{a^3}{3} \right]
\]

\[
\mathbb{E}[X^2] = \frac{1}{b-a} \cdot \frac{b^3 - a^3}{3}
\]

Using the identity \( b^3 - a^3 = (b-a)(b^2 + ab + a^2) \):

\[
\mathbb{E}[X^2] = \frac{b^2 + ab + a^2}{3}
\]

Now, subtract the square of the mean:

\[
\sigma^2 = \mathbb{E}[X^2] - \mu^2
\]

\[
\sigma^2 = \frac{b^2 + ab + a^2}{3} - \left(\frac{b+a}{2}\right)^2
\]

Expanding \( \left(\frac{b+a}{2}\right)^2 \):

\[
\left(\frac{b+a}{2}\right)^2 = \frac{b^2 + 2ab + a^2}{4}
\]

Thus:

\[
\sigma^2 = \frac{b^2 + ab + a^2}{3} - \frac{b^2 + 2ab + a^2}{4}
\]

To simplify, find the common denominator:

\[
\sigma^2 = \frac{4(b^2 + ab + a^2) - 3(b^2 + 2ab + a^2)}{12}
\]

Simplifying further:

\[
\sigma^2 = \frac{b^2 - 2ab + a^2}{12} = \frac{(b-a)^2}{12}
\]

### 4. **Moment-Generating Function (MGF)**

The moment-generating function \( M_X(t) \) is defined as:

\[
M_X(t) = \mathbb{E}[e^{tX}] = \int_{a}^{b} e^{tx} f_X(x) \, dx
\]

Substitute \( f_X(x) = \frac{1}{b-a} \):

\[
M_X(t) = \frac{1}{b-a} \int_{a}^{b} e^{tx} \, dx
\]

The integral of \( e^{tx} \) is:

\[
\int e^{tx} \, dx = \frac{e^{tx}}{t}
\]

Evaluating from \( a \) to \( b \):

\[
M_X(t) = \frac{1}{b-a} \left[ \frac{e^{tb} - e^{ta}}{t} \right]
\]

Thus, the MGF is:

\[
M_X(t) = \frac{e^{tb} - e^{ta}}{t(b-a)}, \quad \text{for } t \neq 0
\]

For \( t = 0 \):

\[
M_X(0) = 1
\]

### 5. **Cumulative Distribution Function (CDF)**

The cumulative distribution function \( F_X(x) \) is:

\[
F_X(x) = \mathbb{P}(X \leq x) = \int_{a}^{x} f_X(t) \, dt
\]

For \( a \leq x \leq b \), substitute \( f_X(t) = \frac{1}{b-a} \):

\[
F_X(x) = \int_{a}^{x} \frac{1}{b-a} \, dt = \frac{x-a}{b-a}
\]

Thus:

\[
F_X(x) =
\begin{cases}
0, & x < a \\
\frac{x-a}{b-a}, & a \leq x \leq b \\
1, & x > b
\end{cases}
\]

### Final Summary:
- **Mean**: \( \mu = \frac{b+a}{2} \)
- **Variance**: \( \sigma^2 = \frac{(b-a)^2}{12} \)
- **MGF**: \( M_X(t) = \frac{e^{tb} - e^{ta}}{t(b-a)} \), for \( t \neq 0 \)
- **CDF**: 
  \[
  F_X(x) =
  \begin{cases}
  0, & x < a \\
  \frac{x-a}{b-a}, & a \leq x \leq b \\
  1, & x > b
  \end{cases}
  \]

---

Do you want further details on any part of the derivation?

### Additional Questions:

1. How does the mean of a uniform distribution relate to the endpoints \( a \) and \( b \)?
2. How would you derive the variance if the distribution was discrete instead of continuous?
3. Can you explain why the moment-generating function (MGF) is useful in probability theory?
4. What happens to the variance if the interval \( [a, b] \) becomes wider?
5. How does the cumulative distribution function (CDF) behave as \( x \) approaches \( b \)?
6. Can the uniform distribution be used to model real-world phenomena? Provide an example.
7. How does the MGF change if \( t = 0 \)?
8. What is the relationship between the MGF and the moments of a distribution?

### Tip:
Remember, the uniform distribution is often used when you have no reason to expect any particular outcome to be more likely than another within a given range.

Learn how to derive the mean, variance, moment-generating function (MGF), and cumulative distribution function (CDF) for the uniform distribution. Understand the formulas and step-by-step calculations involved in solving these key concepts in probability theory.

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