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We are given that \( X \) is exponentially distributed with mean \( a = 1 \). This means that the probability density function (PDF) of \( X \) is:

\[
f_X(x) = \lambda e^{-\lambda x}, \quad x \geq 0
\]
Since the mean is \( 1 \), we have \( \lambda = 1 \), and the PDF simplifies to:

\[
f_X(x) = e^{-x}, \quad x \geq 0
\]

Now, let's proceed to find and sketch the distribution functions of the random variables \( Y = \exp(X) \) and \( Z = \min(X, 3) \).

### 1. Distribution of \( Y = \exp(X) \)

Let’s first find the cumulative distribution function (CDF) of \( Y \). The CDF of \( Y \) is given by:

\[
F_Y(y) = P(Y \leq y) = P(\exp(X) \leq y)
\]

Taking the natural logarithm on both sides, we get:

\[
P(\exp(X) \leq y) = P(X \leq \ln(y))
\]

Thus:

\[
F_Y(y) = P(X \leq \ln(y)) = F_X(\ln(y))
\]

For the exponential distribution, the CDF of \( X \) is:

\[
F_X(x) = 1 - e^{-x}, \quad x \geq 0
\]

Hence:

\[
F_Y(y) = 1 - e^{-\ln(y)} = 1 - \frac{1}{y}, \quad y \geq 1
\]

So, the PDF of \( Y \) is:

\[
f_Y(y) = \frac{d}{dy}F_Y(y) = \frac{1}{y^2}, \quad y \geq 1
\]

### 2. Distribution of \( Z = \min(X, 3) \)

The random variable \( Z \) is the minimum of \( X \) and 3, meaning it equals \( X \) if \( X \leq 3 \), and equals 3 otherwise. The CDF of \( Z \) is:

\[
F_Z(z) = P(Z \leq z)
\]

This can be broken down into two cases:

- For \( z < 3 \), we have \( Z = X \), so:

\[
F_Z(z) = P(X \leq z) = 1 - e^{-z}, \quad z < 3
\]

- For \( z \geq 3 \), \( Z = 3 \) with probability \( P(X > 3) \), and thus:

\[
F_Z(z) = P(X \leq 3) = 1 - e^{-3}, \quad z \geq 3
\]

The PDF of \( Z \) is therefore:

\[
f_Z(z) = 
\begin{cases} 
e^{-z}, & 0 \leq z < 3 \\
0, & z \geq 3
\end{cases}
\]

### Sketching the Distribution Functions

#### 1. CDF of \( Y = \exp(X) \):
- For \( y \geq 1 \), the CDF increases from 0 to 1 as \( y \) increases. The shape is hyperbolic.

#### 2. CDF of \( Z = \min(X, 3) \):
- The CDF increases exponentially from 0 at \( z = 0 \) to \( 1 - e^{-3} \) at \( z = 3 \). It remains constant at \( 1 - e^{-3} \) for \( z \geq 3 \).

Would you like further details on these steps or more explanation?

Here are some related questions to expand your understanding:
1. What is the connection between an exponential distribution and its mean?
2. How would the CDF of \( Y = \exp(X) \) change if the mean of \( X \) were not 1?
3. How do transformations of random variables like \( Y = \exp(X) \) generally affect their distributions?
4. Can the random variable \( Z = \min(X, 3) \) be interpreted as a truncated distribution?
5. What are the applications of minimum and exponential random variables in real-world problems?

**Tip:** When working with transformed random variables, always express the transformation in terms of the original variable before applying any probabilistic operations like finding CDFs or PDFs.

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

Learn how to derive and sketch the distribution functions for Y = exp(X) and Z = min(X, 3), where X is exponentially distributed with a mean of 1. Explore exponential distribution properties, CDFs, PDFs, and transformation of random variables.

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