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: