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This problem involves a piecewise probability density function (PDF) \( f(x) \) defined as:

\[
f(x) = 
\begin{cases} 
\frac{\alpha - 1}{x^\alpha}, & x \geq 1 \\
0, & x < 1 
\end{cases}
\]
where \( \alpha > 2 \).

### Task 1: Show that \( f(x) \) is a valid probability density function.
A function is a PDF if:
1. \( f(x) \geq 0 \) for all \( x \), and
2. The integral of \( f(x) \) over all \( x \) equals 1:
   \[
   \int_{-\infty}^\infty f(x) \, dx = 1.
   \]

Since \( f(x) = 0 \) for \( x < 1 \), we only need to check \( x \geq 1 \):
\[
\int_1^\infty \frac{\alpha - 1}{x^\alpha} dx.
\]
Let’s compute this integral:

\[
\int_1^\infty \frac{\alpha - 1}{x^\alpha} dx = (\alpha - 1) \int_1^\infty x^{-\alpha} dx.
\]

For \( \alpha > 1 \), the integral \( \int x^{-\alpha} dx \) converges:
\[
\int x^{-\alpha} dx = \frac{x^{1-\alpha}}{1-\alpha}.
\]

Evaluate at the bounds \( 1 \) to \( \infty \):
\[
\int_1^\infty x^{-\alpha} dx = \left[ \frac{x^{1-\alpha}}{1-\alpha} \right]_1^\infty = \frac{1}{\alpha - 1}.
\]

Thus:
\[
\int_1^\infty f(x) dx = (\alpha - 1) \cdot \frac{1}{\alpha - 1} = 1.
\]

Hence, \( f(x) \) is a valid PDF.

---

### Task 2: Determine the cumulative distribution function \( F_X(x) \).
The cumulative distribution function \( F_X(x) \) is defined as:
\[
F_X(x) = \int_{-\infty}^x f(t) dt.
\]

For \( x < 1 \):
\[
F_X(x) = 0 \quad \text{(since \( f(t) = 0 \) for \( t < 1 \))}.
\]

For \( x \geq 1 \):
\[
F_X(x) = \int_1^x \frac{\alpha - 1}{t^\alpha} dt.
\]

Compute the integral:
\[
\int_1^x t^{-\alpha} dt = \left[ \frac{t^{1-\alpha}}{1-\alpha} \right]_1^x = \frac{x^{1-\alpha} - 1}{1-\alpha}.
\]

Thus:
\[
F_X(x) = (\alpha - 1) \cdot \frac{x^{1-\alpha} - 1}{1-\alpha} = 1 - x^{1-\alpha}, \quad x \geq 1.
\]

So:
\[
F_X(x) =
\begin{cases}
0, & x < 1, \\
1 - x^{1-\alpha}, & x \geq 1.
\end{cases}
\]

---

### Task 3: Calculate the expectation \( \mathbb{E}[X] \).
The expectation is given by:
\[
\mathbb{E}[X] = \int_{-\infty}^\infty x f(x) dx.
\]

Since \( f(x) = 0 \) for \( x < 1 \), we only integrate over \( x \geq 1 \):
\[
\mathbb{E}[X] = \int_1^\infty x \cdot \frac{\alpha - 1}{x^\alpha} dx = (\alpha - 1) \int_1^\infty x^{1-\alpha} dx.
\]

Using the earlier result for \( \int x^{-\alpha} dx \):
\[
\int_1^\infty x^{1-\alpha} dx = \frac{x^{2-\alpha}}{2-\alpha} \Big|_1^\infty = \frac{1}{\alpha - 2}.
\]

Thus:
\[
\mathbb{E}[X] = (\alpha - 1) \cdot \frac{1}{\alpha - 2} = \frac{\alpha - 1}{\alpha - 2}.
\]

---

### Final Results:
1. \( f(x) \) is a valid probability density function.
2. The cumulative distribution function \( F_X(x) \) is:
   \[
   F_X(x) =
   \begin{cases}
   0, & x < 1, \\
   1 - x^{1-\alpha}, & x \geq 1.
   \end{cases}
   \]
3. The expectation \( \mathbb{E}[X] \) is:
   \[
   \mathbb{E}[X] = \frac{\alpha - 1}{\alpha - 2}.
   \]

---

Do you have any questions or want detailed steps for a specific part?

### Related Questions:
1. What is the variance \( \text{Var}(X) \) of this distribution?
2. How does \( F_X(x) \) change as \( \alpha \to \infty \)?
3. What happens to \( \mathbb{E}[X] \) if \( \alpha \) approaches its lower bound (e.g., \( \alpha \to 2^+ \))?
4. Can we derive higher moments of \( X \), like \( \mathbb{E}[X^2] \)?
5. How does the behavior of \( f(x) \) compare to a uniform distribution for large \( \alpha \)?

### Tip:
When verifying if a function is a PDF, always confirm the normalization condition using proper integration techniques.

Soit \( \alpha > 2 \). On considère la fonction \( f \) définie par:
\[
f(x) = 
\begin{cases} 
\frac{\alpha - 1}{x^\alpha}, & x \geq 1 \\
0, & x < 1 
\end{cases}
\]
1. Montrer que \( f \) est une densité de probabilité d'une variable \( X \).
2. Déterminer la fonction de répartition \( F_X \).
3. Calculer l'espérance de \( X \).

Explore how to prove that a given function is a valid probability density function (PDF), determine its cumulative distribution function (CDF), and calculate the expected value. This problem is essential for university-level probability theory studies.

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