SOLUTION: Suppose X and Y are independent random variables. X is exponentially distributed with mean 1/a) and Y is exponentially distributed with mean 1/b. For a random variable with mean µ

Question 1164362: Suppose X and Y are independent random variables. X is exponentially distributed with mean 1/a) and Y is exponentially distributed with mean 1/b. For a random variable with mean µ define the rate of the random variable as 1/µ. a) Find the pdf of the random variable Z = min(X,Y). b) What is the mean of this distribution? c) What is the relationship between the rate of min(X,Y) and the rates of X and Y.
Answer by CPhill(2261) (Show Source): You can put this solution on YOUR website!
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 $a + b$.
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