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

We are given two independent, exponentially distributed random variables, $X_1$ and $X_2$.
The mean of $X_1$ is ${\mu }_1=0.5=\frac{1}{2}$.
The mean of $X_2$ is ${\mu }_2=0.25=\frac{1}{4}$.

For an exponentially distributed random variable, the rate parameter $\lambda$ is the reciprocal of the mean ($\lambda =\frac{1}{\mu }$). Therefore:
The rate parameter of $X_1$ is ${\lambda }_1=\frac{1}{0.5}=2$.
The rate parameter of $X_2$ is ${\lambda }_2=\frac{1}{0.25}=4$.

We want to find the distribution of $Y=min(X_1,X_2)$.
Let us calculate the cumulative distribution function (CDF) of $Y$, denoted as $F_Y\left(y\right)=P(Y\le y)$ for $y\ge 0$.
It is easier to work with the survival function:
$P(Y>y)=P(min(X_1,X_2)>y)$

Since the minimum of $X_1$ and $X_2$ is greater than $y$ if and only if both individual variables are greater than $y$:
$P(Y>y)=P(X_1>y\text{and}{X}_2>y)$

Since $X_1$ and $X_2$ are independent, we can multiply their individual survival probabilities:
$P(Y>y)=P(X_1>y)\cdot P(X_2>y)$

For an exponentially distributed random variable with rate $\lambda$, the survival function is $P(X>x)=e^{-\lambda x}$ for $x\ge 0$.
Substituting these functions into our equation yields:
$P(Y>y)=e^{-{\lambda }_{1}y}\cdot e^{-{\lambda }_{2}y}=e^{-({\lambda }_1+{\lambda }_2)y}$

Therefore, the cumulative distribution function of $Y$ is:
$F_Y\left(y\right)=1-P(Y>y)=1-e^{-({\lambda }_1+{\lambda }_2)y}$

This is the CDF of an exponentially distributed random variable with rate parameter ${\lambda }_Y={\lambda }_1+{\lambda }_2$.
Substituting the values of ${\lambda }_1$ and ${\lambda }_2$:
${\lambda }_Y=2+4=6$

Thus, $Y$ is exponentially distributed with rate parameter $6$.
Its mean is the reciprocal of the rate parameter:
${\mu }_Y=\frac{1}{{\lambda }_Y}=\frac{1}{6}$

Therefore, the variable $Y$ is exponentially distributed with mean 1⁄6.