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The problem asks to find and sketch the distribution functions for two random variables \( Y = \exp(X) \) and \( Z = \min(X, 3) \), where \( X \) is exponentially distributed with mean \( \lambda^{-1} = 1 \), so \( X \sim \text{Exp}(\lambda = 1) \).

### Step 1: Understanding the distribution of \( X \)

The probability density function (PDF) of \( X \), given that \( X \sim \text{Exp}(1) \), is:
\[
f_X(x) = \begin{cases}
e^{-x}, & x \geq 0 \\
0, & x < 0
\end{cases}
\]

### Step 2: Distribution of \( Y = \exp(X) \)

We aim to find the distribution of \( Y = \exp(X) \). First, note that \( Y \geq 1 \) since \( \exp(X) \geq \exp(0) = 1 \).

#### CDF of \( Y \):
The cumulative distribution function (CDF) of \( Y \) is given by:
\[
F_Y(y) = P(Y \leq y) = P(\exp(X) \leq y) = P(X \leq \ln(y))
\]
For \( y \geq 1 \), the CDF of \( X \) is:
\[
F_Y(y) = 1 - e^{-\ln(y)} = 1 - \frac{1}{y}
\]
Thus, the CDF of \( Y \) is:
\[
F_Y(y) = \begin{cases}
0, & y < 1 \\
1 - \frac{1}{y}, & y \geq 1
\end{cases}
\]

The PDF of \( Y \) can be obtained by differentiating the CDF:
\[
f_Y(y) = \frac{d}{dy} \left( 1 - \frac{1}{y} \right) = \frac{1}{y^2}, \quad y \geq 1
\]

### Step 3: Distribution of \( Z = \min(X, 3) \)

Next, consider the random variable \( Z = \min(X, 3) \). Here, \( Z \) takes values in \( [0, 3] \).

#### CDF of \( Z \):
For \( z \in [0, 3] \), the CDF of \( Z \) is:
\[
F_Z(z) = P(Z \leq z) = P(\min(X, 3) \leq z)
\]
If \( z < 3 \), then:
\[
F_Z(z) = P(X \leq z) = 1 - e^{-z}
\]
For \( z \geq 3 \), \( Z = 3 \), so:
\[
F_Z(z) = 1, \quad z \geq 3
\]

Thus, the CDF of \( Z \) is:
\[
F_Z(z) = \begin{cases}
1 - e^{-z}, & 0 \leq z < 3 \\
1, & z \geq 3
\end{cases}
\]

The PDF of \( Z \) for \( z < 3 \) is:
\[
f_Z(z) = \frac{d}{dz} \left( 1 - e^{-z} \right) = e^{-z}, \quad 0 \leq z < 3
\]
For \( z = 3 \), there is a point mass, which represents a discontinuity at \( z = 3 \).

### Step 4: Sketch the Distribution Functions

- For \( Y = \exp(X) \), the CDF is a curve that starts at 0 for \( y = 1 \) and asymptotically approaches 1 as \( y \to \infty \).
- For \( Z = \min(X, 3) \), the CDF behaves like the exponential distribution up to \( z = 3 \), after which it is a flat line at 1.

I can also sketch these distributions for you if needed.

Would you like to see the sketches? Also, let me know if you would like further explanations on any step. 

Here are 5 related questions that build on this material:

1. How does the exponential distribution's memoryless property influence \( Z = \min(X, 3) \)?
2. What is the relationship between the moments of \( X \) and \( Y = \exp(X) \)?
3. How does the PDF of \( Z \) behave as \( z \to 3 \)?
4. How would the distribution of \( Z \) change if \( Z = \min(X, 5) \)?
5. Can you generalize the approach for \( Z = \min(X, c) \) for any constant \( c \)?

**Tip:** Understanding how transformations of random variables affect their distributions is a key skill in probability theory, especially when dealing with nonlinear functions like \( \exp(X) \).

Let X be exponentially distributed with mean λ = 1. Find and carefully sketch the distribution functions for the random variables Y = exp(X) and Z = min(X, 3).

Explore the transformation of the exponential random variable X with λ = 1 into Y = exp(X) and Z = min(X, 3). Learn about the cumulative distribution functions (CDF) and probability density functions (PDF) of these variables, including detailed formulas and explanations.

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