This might help:

Dark Mode

13,856 views

44 votes

Let $X$ and $Y$ be two exponentially distributed and independent random variables with mean $α$ and $β$, respectively. If $Z$ = min $(X, Y)$, then the mean of $Z$ is given by

- $\left(\dfrac{1}{\alpha + \beta}\right)$
- $\min (\alpha, \beta)$
- $\left(\dfrac{\alpha\beta}{\alpha + \beta}\right)$
- $\alpha + \beta$

41 votes

Best answer

**Answer is (C)**

$X$ is an exponential random variable of parameter λ when its probability distribution function is

$$f(x) = \begin{cases}\lambda e^{-\lambda x} & x \geq 0 \\ 0 & x < 0 \end{cases}$$

For a > 0, we have the cumulative distribution function

$$F_x(a) = \int_0^a f(x) dx = \int_0^a \lambda e^{-\lambda x} dx = -e^{-\lambda x} \mid_0^a = 1 - e^ {-\lambda a}$$

So,

$$P\left\{X < a \right \} = 1 - e^ {-\lambda a} $$ and

$$P\left\{X > a \right \} = e^ {-\lambda a} $$

Now, we use $P \left \{X > a \right \}$ for our problem because our concerned variable $Z$ is **min** of $X$ and $Y$.

For exponential distribution with parameter $\lambda$, mean is given by $\frac{1}{\lambda}$.

We have,

$P \left \{X > a \right \} = e^ {-\frac{1}{\alpha} a} $

$P \left \{Y > a \right \} = e^ {-\frac{1}{\beta} a} $

So, $\begin{align*}P\left \{Z > a \right \} &= P \left \{X > a \right \} P \left \{Y > a \right \} \left(\because \text{X and Y are independent events and } \\Z > \min \left(X, Y \right) \right)\\&=e^ {-\frac{1}{\alpha} a} e^ {-\frac{1}{\beta} a} \\&=e^{-\left(\frac{1}{\alpha} + \frac{1}{\beta} \right)a} \\&=e^{-\left(\frac{\alpha + \beta} {\alpha \beta} \right)a}\end{align*}$

This shows that $Z$ is also exponentially distributed with parameter $\frac{\alpha + \beta} {\alpha \beta}$ and mean $\frac{\alpha \beta} {\alpha + \beta}$.