Question

$\text{Let}{X}_1\text{and}{X}_2$ be two independent exponentially distributed random variables with means 0.5 and 0.25, respectively. Then $Y=min(X_1,X_2)$ is

• A. exponentially distributed with mean 1/6
• B. exponentially distributed with mean 2
• C. normally distributed with mean 3/4
• D. normally distributed with mean 1/6

Answer 1

The correct option is A exponentially distributed with mean 1/6

Probability distribution function for exponential random variable is defied as

$f\left(x\right)=\mu e^{-\mu x},x\ge 0\text{where}\mu$ is parameter and mean of exponential random variable is

$E\left(x\right)=\frac{1}{\mu }$

$E\left(x_1\right)=\frac{1}{{\mu }_1}=0.5\Rightarrow {\mu }_1=2$

$\text{and}E\left(x_2\right)=\frac{1}{{\mu }_2}=0.25\Rightarrow {\mu }_2=4$

So, $\because {\mu }_1+{\mu }_2=6$

So, $E\left(y\right)=\frac{1}{\mu }=\frac{1}{{\mu }_1+{\mu }_2}=\frac{1}{6}$