Let X₁ and X₂ be two independent exponentially distributed random variables with means 0.5 and 0.25, respectively. Then Y = min (X₁, X₂) is

The correct option is: exponentially distributed with mean 1⁄6

To understand why this is correct, we can derive the distribution of the minimum of two independent exponentially distributed random variables.

Step 1: Identify the rate parameters of the given distributions
An exponentially distributed random variable X with mean μ has a probability density function characterized by a rate parameter λ, where:

$\lambda =\frac{1}{\mu }$
For the first random variable X₁ with mean 0.5 (which is 1/2):

${\lambda }_1=\frac{1}{0.5}=2$
For the second random variable X₂ with mean 0.25 (which is 1/4):

${\lambda }_2=\frac{1}{0.25}=4$

Step 2: Determine the distribution of the minimum, Y = min(X₁, X₂)
Let us find the survival function (the probability that Y is greater than some value y) for y > 0:

$P(Y>y)=P(min(X_1,X_2)>y)$
For the minimum of two variables to be strictly greater than y, both individual variables must be strictly greater than y:

$P(Y>y)=P(X_1>yand{X}_2>y)$
Since X₁ and X₂ are independent, we can multiply their individual survival probabilities:

$P(Y>y)=P(X_1>y)·P(X_2>y)$
Using the survival function of exponential distributions, P(X_i > y) = e^-λ_iy:

$P(Y>y)=e^{-{\lambda }_{1}y}·e^{-{\lambda }_{2}y}=e^{-({\lambda }_1+{\lambda }_2)y}$

Step 3: Calculate the mean of the new distribution
The derived survival function P(Y > y) = e^{-(λ₁ + λ₂)y} shows that Y is also exponentially distributed with a combined rate parameter λ_Y:

${\lambda }_Y={\lambda }_1+{\lambda }_2=2+4=6$
The mean of this new exponential distribution Y is:

$E\left[Y\right]=\frac{1}{{\lambda }_Y}=\frac{1}{6}$

Therefore, Y is exponentially distributed with a mean of 1/6.