This problem involves understanding the properties of exponentially distributed random variables, specifically what happens when you take the minimum of two such independent variables. We are given two independent exponentially distributed random variables, X1 and X2, with their respective means.

Exponential Distribution Fundamentals

An exponential distribution is a continuous probability distribution that describes the time between events in a Poisson point process, i.e., a process in which events occur continuously and independently at a constant average rate. A key parameter of an exponential distribution is its rate parameter, often denoted by $\lambda$ (lambda). The relationship between the mean ($\mu$) and the rate parameter ($\lambda$) for an exponentially distributed random variable is fundamental and given by:

$$\mu = \frac{1}{\lambda}$$

Conversely, to find the rate parameter when the mean is known, we use the formula:

$$\lambda = \frac{1}{\mu}$$

This relationship is crucial for solving this problem, as we are given the means and need to find the rates.

Calculating Individual Rate Parameters

Let's determine the rate parameters for X1 and X2 based on their given means:

• For X1: The mean ($\mu_1$) is given as 0.5. To find its rate parameter ($\lambda_1$): $$\lambda_1 = \frac{1}{\mu_1} = \frac{1}{0.5} = 2$$
• For X2: The mean ($\mu_2$) is given as 0.25. To find its rate parameter ($\lambda_2$): $$\lambda_2 = \frac{1}{\mu_2} = \frac{1}{0.25} = 4$$

Minimum of Independent Exponential Variables

A very important property of exponentially distributed random variables is that if X1 and X2 are two independent exponentially distributed random variables with rate parameters $\lambda_1$ and $\lambda_2$ respectively, then their minimum, Y = min(X1, X2), is also an exponentially distributed random variable. The rate parameter of this new variable Y is the sum of the individual rate parameters.

So, the combined rate parameter for Y ($\lambda_Y$) is given by the sum of $\lambda_1$ and $\lambda_2$:

$$\lambda_Y = \lambda_1 + \lambda_2$$

Determining the Combined Rate and Mean of Y

Now, we can calculate the combined rate parameter for Y using the individual rates we found for X1 and X2:

• The combined rate for Y ($\lambda_Y$) is: $$\lambda_Y = 2 + 4 = 6$$

Since Y is also exponentially distributed with a rate parameter of $\lambda_Y = 6$, we can find its mean ($\mu_Y$) using the inverse relationship between mean and rate:

• The mean of Y ($\mu_Y$) is: $$\mu_Y = \frac{1}{\lambda_Y} = \frac{1}{6}$$

Conclusion for Y = min(X1, X2)

Therefore, Y = min(X1, X2) is an exponentially distributed random variable with a mean of $\frac{1}{6}$. This result is a direct application of the properties governing independent exponential distributions and their minimums.

The final answer is exponentially distributed with mean 1⁄6.