Obtain the m.g.f. of a random variable X having p.d.f f(x) = {

x, 0 ≤ x ≤ 1
2 − x, 1 ≤ x ≤ 2
0, otherwise

To obtain the **moment generating function (m.g.f.)** of a random variable \( X \) with probability density function (p.d.f.) \( f(x) \), we proceed as follows:

The moment generating function \( M_X(t) \) of a random variable \( X \) is defined by:

\[
M_X(t) = \mathbb{E}[e^{tX}] = \int_{-\infty}^{\infty} e^{tx} f(x) \, dx
\]

Given the p.d.f. \( f(x) \) of \( X \):

\[
f(x) =
\begin{cases} 
x, & 0 \leq x \leq 1 \\
2 - x, & 1 \leq x \leq 2 \\
0, & \text{otherwise}
\end{cases}
\]

We can break the m.g.f. into two integrals, one for each part of the p.d.f.

### Step 1: Set up the integral

The m.g.f. becomes:

\[
M_X(t) = \int_0^1 e^{tx} x \, dx + \int_1^2 e^{tx} (2 - x) \, dx
\]

### Step 2: Solve each integral

#### First integral: \( \int_0^1 e^{tx} x \, dx \)

We will solve this integral by parts.

Let:
- \( u = x \), so \( du = dx \)
- \( dv = e^{tx} dx \), so \( v = \frac{e^{tx}}{t} \)

Then applying integration by parts:

\[
\int_0^1 x e^{tx} \, dx = \left[\frac{x e^{tx}}{t}\right]_0^1 - \int_0^1 \frac{e^{tx}}{t} \, dx
\]

This simplifies to:

\[
\frac{e^t}{t} - \frac{1}{t^2}(e^t - 1)
\]

#### Second integral: \( \int_1^2 e^{tx} (2 - x) \, dx \)

This can be broken into two simpler integrals:

\[
\int_1^2 2 e^{tx} \, dx - \int_1^2 x e^{tx} \, dx
\]

For the first part \( \int_1^2 2 e^{tx} \, dx \):

\[
\int_1^2 2 e^{tx} \, dx = \left[\frac{2 e^{tx}}{t}\right]_1^2 = \frac{2e^{2t}}{t} - \frac{2e^t}{t}
\]

For the second part \( \int_1^2 x e^{tx} \, dx \), we again apply integration by parts.

Let:
- \( u = x \), so \( du = dx \)
- \( dv = e^{tx} dx \), so \( v = \frac{e^{tx}}{t} \)

Then applying integration by parts:

\[
\int_1^2 x e^{tx} \, dx = \left[\frac{x e^{tx}}{t}\right]_1^2 - \int_1^2 \frac{e^{tx}}{t} \, dx
\]

This simplifies to:

\[
\frac{2 e^{2t}}{t} - \frac{e^{t}}{t} - \frac{1}{t^2}(e^{2t} - e^t)
\]

### Step 3: Combine the results

Now we combine the results of both integrals to obtain the full m.g.f.

\[
M_X(t) = \left( \frac{e^t}{t} - \frac{1}{t^2}(e^t - 1) \right) + \left( \frac{2e^{2t}}{t} - \frac{2e^t}{t} - \frac{2e^{2t}}{t} + \frac{e^{t}}{t} + \frac{1}{t^2}(e^{2t} - e^t) \right)
\]

This expression simplifies further to give the moment generating function \( M_X(t) \).

Would you like me to simplify this fully or clarify any steps further?

Here are five related questions to deepen your understanding:
1. How do you derive the m.g.f. for a general piecewise-defined function?
2. What role does the moment generating function play in calculating moments?
3. How does integration by parts help in computing the m.g.f. of a function like this?
4. What are the first and second moments of this random variable based on the m.g.f.?
5. How can the m.g.f. be used to determine the distribution of a random variable?

**Tip:** The m.g.f. can help identify the distribution of a random variable by matching the moments with known distributions.

Learn how to calculate the moment generating function (m.g.f.) of a random variable X with a piecewise probability density function (p.d.f.). This detailed guide covers the integration steps for both parts of the p.d.f., involving key concepts like moment generating function and integration by parts. Suitable for advanced undergraduate students and those studying probability theory.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation