Question 1164362
To solve for the distribution of the minimum of two independent exponential random variables, we first identify their parameters. 

For an exponential distribution, the **rate** $\lambda$ is the reciprocal of the **mean** $\mu$. 
* $X$ has mean $1/a$, so its rate $\lambda_X = a$.
* $Y$ has mean $1/b$, so its rate $\lambda_Y = b$.

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### **a) Find the PDF of $Z = \min(X, Y)$**

To find the probability density function (PDF), it is easiest to start with the **Survival Function** $P(Z > z)$. 
The minimum of two values is greater than $z$ if and only if **both** values are greater than $z$:
$$P(Z > z) = P(\min(X, Y) > z) = P(X > z \text{ and } Y > z)$$

Since $X$ and $Y$ are independent:
$$P(Z > z) = P(X > z) \cdot P(Y > z)$$
$$P(Z > z) = e^{-az} \cdot e^{-bz} = e^{-(a+b)z}$$

The Cumulative Distribution Function (CDF) is $F_Z(z) = 1 - P(Z > z) = 1 - e^{-(a+b)z}$. To find the PDF, we take the derivative with respect to $z$:
$$f_Z(z) = \frac{d}{dz} \left( 1 - e^{-(a+b)z} \right) = (a+b)e^{-(a+b)z} \quad \text{for } z \ge 0$$

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### **b) What is the mean of this distribution?**

The resulting PDF is clearly that of another exponential distribution with a new rate parameter $\lambda_Z = a + b$. The mean of an exponential distribution is the reciprocal of its rate:
$$E[Z] = \frac{1}{a + b}$$

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### **c) Relationship between the rates**

The **rate** of the minimum of two independent exponential random variables is the **sum** of the individual rates:
$$\text{Rate}(Z) = \text{Rate}(X) + \text{Rate}(Y)$$

**Intuition:** Imagine two independent lightbulbs. If the time it takes for bulb $A$ to burn out has rate $a$ and bulb $B$ has rate $b$, the "system" (which fails as soon as the first bulb goes out) fails at a combined, faster rate of $a + b$.