Discussion
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
Answer: A
Y is exponentially distributed with mean 1/6.
An exponential variable with mean m has rate 1/m, so the rates are \(\lambda_{1} = 2\) and \(\lambda_{2} = 4\). For independent variables,
\[ \begin{aligned} P(Y \gt y) &= P(X_{1} \gt y)\,P(X_{2} \gt y) \\ &= e^{-2y}e^{-4y} = e^{-6y} \end{aligned} \]
That is the survival function of an exponential variable with rate 6, so Y is exponential with mean 1/6.
The minimum of exponentials is always exponential, never normal, which removes C and D. Mean 2 would be a rate of 1/2, which is lower than either component's rate, impossible for a minimum, which must occur sooner on average than either variable alone.
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