Let X and Y be two exponentially distributed and independent random variables…
2004
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
- A.
1/α+β
- B.
min(α ,β)
- C.
alpha beta/(alpha + beta)
- D.
α + β
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Correct answer: C
For two independent exponential random variables X and Y with means α and β, the minimum Z = min(X,Y) is also exponentially distributed. The rate parameter of Z is the sum of the individual rates: 1/α + 1/β. Therefore, the mean of Z is the reciprocal of this sum: 1/(1/α + 1/β) = αβ/(α + β). This matches Option C. Option A is incorrect because it adds the means instead of combining rates. Option B incorrectly assumes the minimum mean equals the smaller individual mean, which is not true for exponential distributions. Option D adds the means, which overestimates the result. Thus, C is correct.